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ROMANIAN MATHEMATICAL SOCIETY
Mehedinți Branch
ROMANIAN MATHEMATICAL MAGAZINE
R.M.M.
Nr.20-2018
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ROMANIAN MATHEMATICAL SOCIETY
Mehedinți Branch
DANIEL SITARU-ROMANIA EDITOR IN CHIEF ROMANIAN MATHEMATICAL MAGAZINE-PAPER VARIANT
ISSN 1584-4897 GHEORGHE CĂINICEANU-ROMANIA
EDITORIAL BOARD
DAN NĂNUȚI-ROMANIA EMILIA RĂDUCAN-ROMANIA MARIA UNGUREANU-ROMANIA DANA PAPONIU-ROMANIA GIMOIU IULIANA-ROMANIA ELENA RÎMNICEANU-ROMANIA DRAGA TĂTUCU MARIANA-ROMANIA DANIEL STRETCU-ROMANIA CLAUDIA NĂNUȚI-ROMANIA DAN NEDEIANU-ROMANIA GABRIELA BONDOC-ROMANIA OVIDIU TICUȘI-ROMANIA
ROMANIAN MATHEMATICAL MAGAZINE-INTERACTIVE JOURNAL ISSN 2501-0099 WWW.SSMRMH.RO
DANIEL WISNIEWSKI-USA EDITORIAL BOARD VALMIR KRASNICI-KOSOVO
ALEXANDER BOGOMOLNY-USA
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CUPRINS
Puncte coliniare.Drepte concurente-Ionică Constantin.............................................................................4
A retrospective of the Ionescu-Nesbitt inequality- D.M.Bătinețu-Giurgiu,Daniel Sitaru,Neculai Stanciu............................................................................................................. 10
About few inequalities in triangle-Vasile Jiglău...............................................................................14
Construction of some inequalities using Weierstrass’ theorem for compact sets–Daniel Sitaru,Claudia Nănuți…………………………………………………………………………………………………………..18
On the problem 11984/American Mathematical Monthly/May 2017/Proposed by Daniel Sitaru –Minimal and maximal bounds for the sum 풂ퟔ + 풃ퟔ + 풄ퟔ -Marius Drăgan,Neculai Stanciu.……………………………………………………………………………………………………………..…….……...….22
Some inequalities in four variables with sum 1 –Ștefan Andrei Mihalcea.............................23
Some algebraic inequalities proved by using a Fermat-Toricelli’s configuration –Daniel Sitaru…………………………………………………………………………………………………………………………………..24
About some famous inequalities-D.M.Bătinețu-Giurgiu, Daniel Sitaru..................................25
Proposed problems………………………………………….………………………………………………………………...32
Index of proposers and solvers RMM-20 Paper Magazine.………………………………………….…………80
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PUNCTE COLINIARE. DREPTE CONCURENTE
By Ionică Constantin-Romania
Prin acest articol mi-am propus să vin în sprijinul pregătirii elevilor din clasele terminale de gimnaziu. Problemele de geometrie plană, care au ca cerință demonstrarea coliniarității unor puncte sau a concurentei unor drepte, cu un grad sporit de dificultate, ele fiind probleme de demonstrație care necesită raționamente precise și o gamă variată de tehnici specifice și solicită din partea elevului multă inventivitate și perspicacitate. Între problemele de coliniaritate și cele de concurentă există o strânsă legătură. Astfel, pentru a dovedi că dreptele a,b,c sunt concurente, se consideră că 푎 ∩ 푏 = {퐴} și punctele 퐵,퐶 ∈ 푐. Stabilirea concurentei celor trei drepte se reduce la a arăta că punctele 퐴,퐵,퐶 sunt coliniare.
Pentru demonstrarea coliniarității unor puncte se pot folosi următoarele metode:
P1. Demonstarea coliniarității folosind Axioma lui Euclid, care constă în următoarele: dacă dreptele 퐵퐴 și 퐵퐶 sunt paralele cu o dreaptă dată 푑, atunci punctele 퐴,퐵,퐶 sunt coliniare.
P2.Demonstrarea coliniarității cu ajutorul unghiului alungit(unghiuri adiacente suplimentare). Această metodă constă în a utiliza rezultatul: dacă punctele 퐴 și 퐶 sunt situate de o parte și de a dreptei 퐵퐷 și dacă
푚(∢퐴퐵퐷) +푚(∢퐶퐵퐷) = 180° , atunci punctele 퐴,퐵,퐶 sunt coliniare.
P3.Demonstrarea coliniarității folosind rezultatul care afirmă că: dacă punctul 퐵 ∈ 퐷퐸, iar 퐴 și 퐶 sunt de o parte și de alta a dreptei 퐷퐸 și
∢퐴퐵퐷 ≡ ∢퐶퐵퐸, atunci punctele 퐴,퐵,퐶 sunt coliniare. (Reciproca teoremei unghiurilor opuse la vârf).
P4. Demonstrarea coliniarității folosind reciproca teoremei lui Menelaus.
P5. Demonstrarea coliniarității utilizând rezultatul care afirmă că: dacă punctele distincte 퐵 și 퐶 sunt situate de aceeași parte a dreptei 퐴퐷 și dacă ∢퐷퐴퐵 ≡ ∢퐷퐴퐶 , atunci punctele 퐴,퐵,퐶 sunt coliniare.
P6. Demonstrarea coliniarității utilizând identitatea 퐴퐵 + 퐵퐶 = 퐴퐶
P7. Demonstrarea coliniarității folosind rezultatul care afirmă: dintr-un punct exterior mai drepte se poate duce o perpendiculară și numai una pe acea dreaptă.
Pentru demonstrarea concurentei a trei drepte se pot utiliza următoarele metode:
Q1.Demonstrarea concurenței folosind proprietățile liniilor importante în triunghi. Această metodă constă în a găsi un triunghi în care dreptele respective sunt sau înălțimi,sau mediane, sau bisectoare, sau mediatoare.
Q2. Demonstrarea concurenței utilizând reciproca teoremei lui Ceva.
Q3. Demonstrarea concurenței folosind unicitatea mijlocului unui segment. Această metodă constă în următoarele: pe dreapta a se identifică punctele 퐴 și 퐵, iar pe dreapta 푏 se identifică punctele 퐶 și 퐷 astfel încât segmentele [퐴퐵] și [퐶퐷] să aibă același mijloc.
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Q4. Demonstrarea concurenței prin coliniaritate (așa cum s-a arătat la începutul paragrafului). Probleme care ilustrează metodele menționate
1. Punctul de intersecție al diagonalelor unui paralelogram se află pe dreapta ce unește mijloacele a două laturi opuse ale paralelogramului.
Soluție: Fie paralelogramul 퐴퐵퐶퐷, 푂 punctul de intersecție al diagonalelor 퐴퐶 și 퐵퐷, iar 푀 și 푁 mijloacele laturilor [퐴퐵] și [퐶퐷]. În 훥퐴퐵퐶, [푂푀] este linie mijlocie ⇒ 푂푀 ∥ 퐵퐶 (1), iar în 훥퐵퐶퐷, [푂푁] este linie mijlocie ⇒ 푂푁 ∥ 퐵퐶 (2). Din relațiile (1) și (2) ⇒ 푀,푂,푁 - coliniare (conform 푃 ).
2. Fie 휟푨푩푪 dreptunghic în 푨,푨푫 ⊥ 푩푪,푫 ∈ (푩푪), iar 푬 și 푭 simetricele punctului 푫 față de 푨푩, respectiv 푨푪. Să se arate că punctele 푬,푨,푭 sunt coliniare.
Soluție: Din faptul că punctul 퐸 este simetricul lui 퐷 față de 퐴퐵 ⇒ ∢퐸퐴퐵 ≡ ∢퐵퐴퐷 și 퐹 fiind simetricul lui 퐷 față de 퐴퐶 ⇒ ∢퐹퐴퐶 ≡ ∢퐶퐴퐷. Cum 푚(∢퐵퐴퐷) + 푚(∢퐶퐴퐷) = 90° ⇒ 푚(∢퐸퐴퐵) +푚(∢퐵퐴퐷) +푚(∢퐶퐴퐷) +푚(∢퐹퐴퐶) = 180° ⇒ 푚(∢퐸퐴퐹) = 180° ⇒ punctele 퐸,퐴,퐹 sunt coliniare (conform 푃 )
3. În rombul 푨푩푪푫 se consideră punctele 푴 și 푵 mijloacele segmentelor [푨푩], respectiv, [푪푫], iar punctul 푶 este intersecția diagonalelor 푨푪 și 푩푫. Să se arate că punctele 푴,푶,푵 sunt coliniare.
Soluție: 훥퐴푂푀 ≡ 훥퐶푂푁 (LUL) ⇒ ∢퐴푂푀 ≡ ∢퐶푂푁 și cum 퐴,푂,퐶 sunt coliniare ⇒ 푀,푂,푁 sunt coliniare (푃 ).
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4. O dreaptă taie laturile 푩푪,푨푪 și 푨푩 ale unui 휟푨푩푪 în punctele 푨 ,푩 respectiv 푪 . Dacă punctele 푴,푵,푷 sunt simetricele punctelor 푨 ,푩 ,푪 față de mijlocul laturii pe care este situat fiecare, arătați că 푴,푵,푷 sunt coliniare.
Soluție: Cum punctele 퐴 ,퐵 ,퐶 sunt coliniare ⇒ ⋅ ⋅ = 1 (Teorema Menelaus)
Dacă 퐷 este mijlocul [퐵퐶] și 푀 este simetricul lui 퐴 față de punctul
퐷 ⇒ = . Analog pentru punctele 퐸 și 퐹 mijloacele segmentelor [퐴퐶], respectiv [퐴퐵], avem
că: = și = ⇒ ⋅ ⋅ = 1 ⇒ 푀,푁,푃 sunt coliniare conform reciprocei teoremei lui Menelaus (푃 ).
5. Fie punctul 푬 interior pătratului 푨푩푪푫 și 푭 exterior, astfel încât triunghiurile 푨푩푬 și 푩푪푭 să fie echilaterale. Să se arate că punctele 푫,푬,푭 sunt coliniare.
Soluție: [퐴퐷] ≡ [퐴퐸] ⇒ 훥퐴퐷퐸 este isoscel și cum 푚(∢퐷퐴퐸) = 30° ⇒ 푚(∢퐴퐷퐸) = 75° ⇒
⇒ 푚(∢퐶퐷퐸) = 15°. Din [퐶퐷] = [퐶퐹] ⇒ 훥퐷퐶퐹 este isoscel și cum 푚(∢퐷퐶퐹) = 150° ⇒푚(∢퐶퐷퐹) = 15° ⇒ ∢퐶퐷퐸 ≡ ∢퐶퐷퐹 ⇒
punctele 퐷,퐸,퐹 sunt coliniare (푃 ).
6. Fie 푨푩푪푫 un patrulater convex și punctul 푬 așa încât 푪 și 푬 sunt de o parte și de alta a lui 푨푩, iar triunghiurile 푨푩푬 și 푨푫푪 sunt asemenea. Să se demonstreze că punctele 푪,푩,푬 sunt coliniare, dacă și numai dacă, are loc egalitatea: 푨푩 ⋅ 푪푫 + 푨푫 ⋅ 푩푪 ≡ 푨푪 ⋅ 푩푫
Soluție: Din asemănarea triunghiurilor 퐴퐵퐸 și 퐴퐷퐶 ⇒ = și prin schimbarea mezilor
⇒ = . Cum ∢퐶퐴퐸 ≡ ∢퐷퐴퐵 (deoarece ∢퐷퐴퐶 ≡ ∢퐵퐴퐸 și ∢퐵퐴퐶 este unghi comun)
⇒ 훥퐸퐴퐶 ∼ 훥퐵퐴퐷. Din 훥퐴퐵퐸 ∼ 훥퐴퐷퐶 ⇒ = = ⇒
⇒ 퐵퐸 = ⋅ , iar din 훥퐸퐴퐶 ∼ 훥퐵퐴퐷 ⇒ = = ⇒ 퐸퐶 = ⋅ . Punctele 퐶,퐵,퐸 sunt coliniare, dacă și numai dacă, 퐶퐵 + 퐵퐸 = 퐶퐸, adică
퐶퐵 + ⋅ = ⋅ ⇔ 퐴퐵 ⋅ 퐷퐶 + 퐵퐶 ⋅ 퐴퐷 = 퐴퐶 ⋅ 퐵퐷 (푃 )
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7. În trapezul isoscel 푨푩푪푫 fie 푶 punctul de intersecție al diagonalelor, iar 푴 și 푵 mijloacele bazelor [푨푩], respectiv [푪푫]. Să se arate că 푴,푶,푵 sunt puncte coliniare.
Soluție: Din 훥퐴푂퐵 isoscel și [푂푀] – mediană ⇒ [푂푀] – înălțime, adică 푂푀 ⊥ 퐴퐵 (1). Analog 푂푁 ⊥ 퐶퐷 (2). Din (1) și (2) și 퐴퐵 ∥ 퐶퐷 ⇒ 푀,푂 ,푁 sunt coliniare. (푃 )
8. Pe catetele 푨푩 și 푨푪 ale triunghiului dreptunghic 푨푩푪(풎(∢푨) = ퟗퟎ°) se construiesc în exterior pătratele 푨푩푭푮 și 푨푪푫푬. Să se arate că dreptele 푩푫 și 푪푭 se intersectează pe înălțimea 푨푯 a triunghiului 푨푩푪.
Soluție: Fie punctul 퐼 intersecția dreptelor 퐷퐸 și 퐹퐺. Din 훥퐺퐴퐼 ≡ 훥퐴퐵퐶 (c.c)
⇒ ∢퐺퐴퐼 ≡ ∢퐴퐵퐶 ≡ ∢퐶퐴퐻 și cum 퐺,퐴,퐶 sunt coliniare ⇒ 퐼,퐴,퐻 – coliniare, deci 퐼 este situat pe înălțimea 퐴퐻, adică 퐼퐻 ⊥ 퐵퐶. Cum 훥퐴퐵퐼 ≡ 훥퐵퐹퐶 (L.U.L) și 퐴퐵 ⊥ 퐵퐹, respectiv 퐼퐴 ⊥ 퐵퐶 ⇒ 퐶퐹 ⊥퐵퐼. Din 훥퐵퐸퐷 ≡ 훥퐼퐷퐶 (c.c) și
퐵퐸 ⊥ 퐼퐷, respectiv 퐸퐷 ⊥ 퐷퐶 ⇒ 퐵퐷 ⊥ 퐼퐶. Prin urmare 퐼퐻,퐵퐷 și 퐶퐹 sunt înălțimi în 훥퐵퐼퐶 ⇒ sunt concurente într-un punct 푀(푄 ).
9. Fie 푶 punctul de intersecție al diagonalelor trapezului 푨푩푪푫,푬 și 푭 mijloacele bazelor 푨푩 și 푪푫, iar 푮 și 푯 mijloacele diagonalelor 푨푪 și 푩푫. Dacă 푰 și 푱 sunt simetricele punctului 푶 în raport cu 푮 și 푯, să se arate că dreptele 푬푭, 푰푯 și 푮푱 sunt concurente.
Soluție: Cum 퐼 este simetricul lui 푂 față de 퐺 ⇒ [푂퐺] ≡ [퐺퐼] și analog [푂퐻] ≡ [퐻퐽] ⇒
⇒ [퐼퐻], [퐽퐺] sunt mediane în 훥푂퐼퐽 ⇒ [푂퐾] este a treia mediană în 훥푂퐼퐽, deci 퐸퐹, 퐼퐻 și 퐺퐽 sunt concurente (푄 )
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10. În vârfurile triunghiului 푨푩푪 se duc tangentele la cercul circumscris lui și care formează triunghiul 푨 푩 푪 . Dacă 푨ퟏ,푩ퟏ,푪ퟏ sunt centrele cercurilor înscrise în triunghiurile 푨 푩푪,푩 푪푨 și 푪 푨푩, să se demonstreze că dreptele 푨푨ퟏ,푩푩ퟏ,푪푪ퟏ sunt concurente.
Soluție: Din 퐴 - centrul cercului înscris 훥퐴 퐵퐶 ⇒ [퐵퐴 este bisectoarea ⇒ ∢퐴 퐵퐶 ⇒
⇒ ∢퐴 퐵퐶 ≡ ∢퐴 퐵퐴 ⇒ 퐴 este mijlocul arcului 퐵퐶 ⇒ [퐴퐴 este bisectoarea ∢퐵퐴퐶. Analog se arată că [퐵퐵 și [퐶퐶 sunt bisectoarele ∢퐴퐵퐶, respectiv ∢퐴퐶퐵, de unde rezultă concurența dreptelor 퐴퐴 ,퐵퐵 și 퐶퐶 (푄 )
11. Să se arate că perpendiculare duse din mijloacele laturilor unui triunghi pe laturile triunghiului ortic (format unind picioarele înălțimilor triunghiului dat), sunt concurente.
Soluție: Fie 퐴퐷 ⊥ 퐵퐶,퐵퐸 ⊥ 퐴퐶,퐶퐹 ⊥ 퐴퐵 ,퐷 ∈ [퐵퐶],퐸 ∈ [퐴퐶],퐹 ∈ [퐴퐵] și 퐴 ,퐵 ,퐶 mijloacele laturilor [퐵퐶], [퐶퐴], respectiv [퐴퐵]. Construim 퐴 푀 ⊥ 퐸퐹,퐶 푁 ⊥ 퐸퐷 și 퐵 푃 ⊥ 퐹퐷. În triunghiul dreptunghic 퐵퐸퐶 avem 퐸퐴 mediană relativă la ipotenuză ⇒ 퐸퐴 = , iar triunghiul
dreptunghic 퐵퐹퐶 avem [퐹퐴 ] mediană relativă la ipotenuză ⇒ 퐹퐴 = ⇒ 훥퐴 퐸퐹 este isoscel și cum 퐴 푀 ⊥ 퐸퐹 ⇒ 퐴 푀 este mediatoarea laturii 퐸퐹. Analog se arată că 퐶 푁 și 퐵 푃 sunt mediatoarele laturilor 퐸퐷 și 퐹퐷 ⇒ dreptele 퐴 푀,퐶 푁 și 퐵 푃 fiind mediatoarele laturilor triunghiului 퐷퐸퐹 sunt concurente într-un punct 푄. (푄 )
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12. Pe cateta 푨푪 a triunghiului dreptunghic 푨푩푪 se ridică în 푪 perpendiculara [푪푪 ] ≡[푨푪], iar pe cateta 푨푩 se ridică perpendiculara [푩푩 ] ≡ [푨푩]. Să se arate că dreptele 푩푪 și 푩 푪 se întâlnesc pe înălțimea [푨푨 ].
Soluție: Fie 퐴퐵 ∩ 퐵 퐶 = {퐷} și 퐴퐶 ∩ 퐶 퐵 = {퐸}. În 훥퐴퐵퐶 dreptunghic avem
퐴퐵 = 퐵퐶 ⋅ 퐶퐴 și 퐴퐶 = 퐵퐶 ⋅ 퐶퐴 ⇒ = (1) cum 훥퐶 퐶퐸 ∼ 훥퐵퐴퐸 (U.U) ⇒ = , dar [퐶퐶 ] ≡ [퐴퐶] ⇒ = (2). Din 훥퐷퐴퐶 ∼ 훥퐷퐵퐵 (U.U)
⇒ = și cum [퐵퐵 ] ≡ [퐴퐵] ⇒ = (3). Înmulțind membru cu membru relațiile (1), (2)
și (3) avem ⋅ ⋅ = ⋅ ⋅ = 1 ⇒ dreptele 퐵퐶 ,퐶퐵 și 퐴퐴 sunt concurente, conform reciprocei teoremei lui Ceva. (푄 )
13. Fie rombul 푨푩푪푫 cu 푨푪 diagonala mare și 푨ퟏ,푨ퟐ,푪ퟏ,푪ퟐ proiecțiile punctelor 푨 și 푪 pe laturile opuse. Să se demonstreze că dreptele 푨ퟏ푪ퟏ,푨ퟐ푪ퟐ,푨푪 și 푩푫 sunt concurente.
Soluție:Cum 퐴퐶 ∥ 퐴 퐶,퐴퐴 ∥ 퐶퐶 și 푚(∢퐴 ) = 90° ⇒ 퐴퐴 퐶퐶 este dreptunghi ⇒ 퐴 퐶 trece prin mijlocul 푂 a lui 퐴퐶. Analog se arată că și 퐴 퐶 trece prin mijlocul 푂 a lui 퐴퐶 ⇒ 퐴 퐶 ,퐴 퐶 ,퐴퐶 și 퐵퐷 sunt concurente (푄 )
14. Fie 푨푩푪푫 și 푨푩 푪 푫 două pătrate având laturile de aceeași lungime. Să se demonstreze că dreptele 푩푩 ,푪푪 și 푫푫 sunt concurente.
Soluție : Fie 퐵퐵 ∩ 퐶퐶 = {푃};훥퐵퐴퐵 ≡ 훥퐷퐴퐷 (L.U.L) ⇒ ∢퐴퐵퐵 ≡ ∢퐴퐷퐷 ⇒
⇒ 푚(∢퐶퐵푃) +푚(∢퐶퐷푃) = 180° ⇒ 푃퐵퐶퐷 este patrulater inscriptibil
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⇒ 푚(∢퐶퐵푃) = 푚(∢퐶퐷푃) = 45° = 푚(∢퐶퐵퐷) = 푚(∢퐶푃퐷). Cum 훥퐵퐴퐵 este isoscel ⇒ ∢퐴퐵퐵 ≡ ∢퐴퐵 퐵 ⇒ 푚(∢퐶 퐷 푃) + 푚(∢퐶 퐵 푃) = 180° ⇒
patrulaterul 푃퐵 퐶 퐷 este inscriptibil ⇒ 푚(∢퐶 푃퐷 ) = 푚(∢퐶 퐵 퐷 ) = 45° =
= 푚(∢퐶 퐷 퐵 ) = 푚(∢퐶 푃퐵 ) ⇒ ∢퐶푃퐷 ≡ ∢퐶 푃퐷 și cum 퐶,푃,퐶 sunt coliniare ⇒ 퐷,푃,퐷 sunt coliniare deci dreapta 퐷퐷 trece prin 푃, care este punctul de intersecție al dreptelor 퐵퐵 și 퐶퐶 ⇒ 퐵퐵 ,퐶퐶 și 퐷퐷 sunt concurente. (푄 ).
A RETROSPECTIVE OF THE IONESCU-NESBITT INEQUALITY
By D. M. Bătineţu-Giurgiu, Daniel Sitaru and Neculai Stanciu-Romania
1.Introduction
In Romanian Mathematical Gazette, Volume XXXII (September 15, 1926 - August 15, 1927), at page 120 Ion Ionescu - one of the founders and pillars of Mathematical Gazette published the
problem.
3478. If 푥,푦, 푧 are positive, show that:
+ + ≥ (*)
In the same volume, at pp. 194-196, are presented two solutions to this problem, as well as a generalization. From [17] yields that Nesbitt published the inequality (1) in 1903.It is
appropriate to say here that the problem.
In any triangle 푨푩푪, with usual notations holds the inequality
풂ퟐ + 풃ퟐ + 풄ퟐ ≥ ퟒ√ퟑ푺, (I-W),
was published first by Ion Ionescu in 1897 (see [1]), and also published by Roland Weitzenböck in 1919. However, this inequality has long been called ’’Weitzenböck’s inequality’’, and after the
appearance of paper [1] is called Ionescu-Weitzenböck inequality (IW)
Compared to the above we suggest that inequality (*) to be called Nesbitt-Ionescu inequality. In the next, we will do a retrospective of our results on Nesbitt-Ionescu inequality and we shall give some generalizations of problem 752 from The Pentagon journal (see [19]).
2. Ionescu-Nesbitt type inequality for two variables
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If 푥 , 푥 ∈ ℝ∗ , then: 푨(풙ퟏ,풙ퟐ) = 풙ퟏ
풙ퟐ+ 풙ퟐ
풙ퟏ≥ ퟐ, (see for e.g. [7,12])
If 푎, 푏, 푥 , 푥 ∈ ℝ∗ ,푋 = 푥 + 푥 and 푎푋 > 푏max{푥 , 푥 }, then: 푩(풙ퟏ,풙ퟐ) = 풙ퟏ
풂푿ퟏ 풃풙ퟏ+ 풙ퟐ
풂푿ퟐ 풃풙ퟐ≥ ퟐ
ퟐ풂 풃 (see [7,12])
3. Ionescu – Nesbitt type inequality for three variables
If 푥 , 푥 , 푥 ∈ ℝ∗ , then: 푪(풙ퟏ,풙ퟐ,풙ퟑ) = 풙ퟏ
풙ퟐ 풙ퟑ+ 풙ퟐ
풙ퟑ 풙ퟏ+ 풙ퟑ
풙ퟏ 풙ퟐ≥ ퟑ
ퟐ, (Nesbitt, 1903, see [17,12]).
If 푎, 푏, 푥 , 푥 , 푥 ∈ ℝ∗ , then: 푫(풙ퟏ,풙ퟐ,풙ퟑ) = 풙ퟏ
풂풙ퟐ 풃풙ퟑ+ 풙ퟐ
풂풙ퟑ 풃풙ퟏ+ 풙ퟑ
풂풙ퟏ 풃풙ퟐ≥ ퟑ
풂 풃, (see [4,12]).
If 푎 ,푏 , 푥 ∈ ℝ∗ , 푘 = 1,3 and 푎 + 푏 = 푎 + 푏 = 푎 + 푏 = 푡, then: 푬(풙ퟏ,풙ퟐ,풙ퟑ) = 풙ퟏ
풂ퟏ풙ퟐ 풃ퟏ풙ퟑ+ 풙ퟐ
풂ퟐ풙ퟑ 풃ퟐ풙ퟏ+ 풙ퟑ
풂ퟑ풙ퟏ 풃ퟑ풙ퟐ≥ ퟑ
풕, (see [4,12]).
If 푎, 푏, 푥 , 푥 , 푥 ∈ ℝ∗ , 푥 + 푥 + 푥 = 푋 and 푎푋 > 푏max{푥 , 푥 , 푥 }, then: 푭(풙ퟏ,풙ퟐ,풙ퟑ) = 풙ퟏ
풂푿ퟑ 풃풙ퟏ+ 풙ퟐ
풂푿ퟑ 풃풙ퟐ+ 풙ퟑ
풂푿ퟑ 풃풙ퟑ≥ ퟑ
ퟑ풂 풃, (see [4,12]).
If 푎, 푏, 푥 , 푥 , 푥 ∈ ℝ∗ , 푥 + 푥 + 푥 = 푋 , such that 푎푋 > 푏max{푥 ,푥 , 푥 } and 푚,푝 ∈ [1,∞), then:
푮(풙ퟏ,풙ퟐ,풙ퟑ) = 풙ퟏ풎
(풂푿ퟑ 풃풙ퟏ)풑 + 풙ퟐ풎
(풂푿ퟑ 풃풙ퟐ)풑 + 풙ퟑ풎
(풂푿ퟑ 풃풙ퟑ)풑 ≥ퟑ 풎 풑 ퟏ
(ퟑ풂 풃)풑 ⋅ 푿ퟑ풎 풑, (see [4,12]).
If 푎, 푏, 푥 , 푥 , 푥 ∈ ℝ∗ , 푥 + 푥 + 푥 = 푋 and 푚 ∈ 푅 , then: 푯(풙ퟏ,풙ퟐ,풙ퟑ) = 풙ퟏ
(풂푿ퟑ 풃풙ퟏ)풎 + 풙ퟐ(풂푿ퟑ 풃풙ퟐ)풎 + 풙ퟑ
(풂푿ퟑ 풃풙ퟑ)풎 ≥ퟑ풎
(ퟑ풂 풃)풎⋅푿ퟑ풎 ퟏ, (see [4]).
If 푎, 푏, 푐, 푥,푦, 푧 ∈ ℝ∗ , then:
푰(풙,풚,풛) =풂풙풚 + 풛
+풃풚풛 + 풙
+풄풛
풙 + 풚≥
(풂풙 + 풃풚 + 풄풛)ퟐ
(풂 + 풃)풙풚+ (풃 + 풄)풚풛 + (풄 + 풂)풛풙≥
≥ ퟑ(풂풃풙풚 풃풄풚풛 풄풂풛풙)(풂 풃)풙풚 (풃 풄)풚풛 (풄 풂)풛풙
, (see [2,5]).
If 푎, 푏, 푥,푦, 푧 ∈ 푅∗ and 푚 ∈ 푅 , then: 푱(풙,풚,풛) = 풙풎 ퟏ
(풂풚 풃풛)ퟐ풎 ퟏ + 풚풎 ퟏ
(풂풛 풃풙)ퟐ풎 ퟏ + 풛풎 ퟏ
(풂풙 풃풚)ퟐ풎 ퟏ ≥ퟑ풎 ퟏ
(풂 풃)ퟐ풎 ퟏ(풙 풚 풛)풎, (see [3]).
4. Ionescu – Nesbitt type inequality for 풏 variables
If 푥 ∈ ℝ∗ ,∀푘 = 1,푛,푋 = ∑ 푥 , then: 푵풏 = ∑ 풙풌
푿풏 풙풌풏풌 ퟏ ≥ 풏
풏 ퟏ, (see for e.g. [16], the case 풏 = ퟒ and also [12]).
If 푛 ∈ ℕ∗ − {1,2},푚,푝 ∈ [1,∞),푎, 푏, 푥 ∈ ℝ∗ ,∀푘 = 1,푛,푋 = ∑ 푥 , and 푎푋 >푏max 푥 , then:
푲풏 = ∑ 풙풌풏
(풂푿풏 풃풙풌)풑풏풌 ퟏ ≥ 풏 풎 풑 ퟏ
(풂풏 풃)풑 ⋅ 푿풏풎 풑, (see [3]).
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If 푎, 푏, 푥 ∈ ℝ∗ , 푐,푦 ∈ ℝ ,∀푘 = 1,푛,푛 ∈ ℕ∗ − {1}, 푋 = ∑ 푥 and 푎푋 > 푏max 푥 ,푦 ∈ 0, 푋 ,∀푘 = 1,푛 and 푚 ∈ ℝ , then:
푳풏 = ∑ 풙풌풎 ퟏ
(풂푿풏 풃풙풌 풄풚풌)ퟐ풎 ퟏ풏풌 ퟏ ≥ 풏풎 ퟏ
(풂풏 풃 풄)ퟐ풎 ퟏ푿풏풎, (see [3]).
If 푛 ∈ ℕ∗ − {1,2},푥 ∈ 푅 , 푏, 푐,푑, 푥 ∈ ℝ∗ ,∀푘 = 1,푛,푋 = ∑ 푥 , and 푐푋 > 푑max 푥 , then:
푴풏 = ∑ 풂푿풏 풃풙풌풄푿풏 풅풙풌
풏풌 ퟏ ≥ (풂풏 풃)⋅풏
풄풏 풅, (see [6] and [18]).
If 푎 ∈ ℝ , 푏, 푐,푑, 푥 ∈ ℝ∗ ,∀푘 = 1,푛,푋 = ∑ 푥 ,푚 ∈ [1,∞) and 푐푋 > 푑max 푥 , then:
푼풏 = ∑ 풂푿풏 풃풙풌풄푿풏풎 풅풙풌
풎풏풌 ퟏ ≥ (풂풏 풃)풏풎
풄풏풎 풅푿풏ퟏ 풎, (see [8] and also [11]).
If 푎,푚 ∈ ℝ , 푏, 푐,푑, 푥 ∈ ℝ∗ ,∀푘 = 1,푛,푋 = ∑ 푥 ,푝 ∈ [1,∞) and 푐푋 > 푑max 푥 , then:
푽풏 = ∑ 풂푿풏 풃풙풌풄푿풏풎 풅풙풌
풎 풑풏풌 ퟏ ≥ (풂풏 풃)풏풎풑
(풄풏풎 풅)풑푿풏ퟏ 풎풑, (see [9]).
If 푛 ∈ ℕ∗ − {1},푎 ∈ ℝ , 푏, 푐,푑, 푥 ∈ ℝ∗ ,∀푘 = 1,푛,푋 = ∑ 푥 , 푚,푝, 푟, 푠 ∈ [1,∞), such that 푐푋 > 푑max 푥 , then:
푾풏 = ∑ 풂푿풏풓 풃풙풌풓 풔
풄푿풏풎 풅풙풌풎 풑
풏풌 ퟏ ≥ (풂풏풓 풃)풔
(풄풏풎 풅)풑풏풎풑 풓풔 ퟏ푿풏
풓풔 풎풑, (see [10,13])
If 푛 ∈ ℕ∗ − {1},푚,푝 ∈ ℝ∗ , 푥 ∈ ℝ∗ ,∀푘 = 1,푛, and we denote 푋 , = ∑ 푥 , 푋 , = ∑ 푥 , then:
풀풏 = ∑ 풙풌풎
푿풏,풑 풙풌풑
풏풌 ퟏ ≥ 풏
풏 ퟏ⋅ 푿풏,풎푿풏,풑
, (see [15]).
If 푛 ∈ ℕ∗ − {1},푎 ∈ ℝ , 푏, 푐,푑,푚,푝 ∈ 푅∗ , 푥 ∈ 푅∗ , 푘 = 1,푛, 푋 , = ∑ 푥 ,푋 , = ∑ 푥 , such that
푐 ⋅ 푋 , > 푑 ⋅ max 푥 , then
풁풏 = ∑ 풂⋅푿풏,풎 풃⋅풙풌풎
풄⋅푿풏,풑 풅⋅풙풌풑
풏풌 ퟏ ≥ 풏⋅(풂풏 풃)
풄풏 풅⋅ 푿풏,풎푿풏,풑
, (see [15]).
If 푛 ∈ ℕ∗ − {1},푎, 푣 ∈ ℝ , 푏, 푐,푑,푚,푝 ∈ 푅∗ , 푥 ∈ ℝ∗ ,푘 = 1,푛, 푡 ∈ [1,∞),푋 , = ∑ 푥 ,푋 , = ∑ 푥 , such that
푐 ⋅ 푋 , > 푑 ⋅ max 푥 , then
∑ 풂⋅푿풏,풎 풃⋅풙풌풎 풕
풄⋅푿풏,풑 풅⋅풙풌풑 풗
풏풌 ퟏ ≥ 풏풗 풕⋅(풂풏 풃)풕
풄풏 풅⋅ 푿풏,풎
풕
푿풏,풑풗 , (see [14]).
See also [16].
New results:
If 푎 ∈ ℝ and 푏, 푐,푑, 푥,푦, 푧 ∈ 푅∗ ,푋 = 푥 + 푦 + 푧, 푐푋 > 푑max{푥,푦, 푧}, then:
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풂푿 + 풃풙풄푿− 풅풙
풄풚풄
≥ퟑ(ퟑ풂+ 풃)ퟑ풄 − 풅 .
If 푎 ∈ ℝ ,푚 ∈ [1,∞) and 푏, 푐,푑, 푥,푦, 푧 ∈ ℝ∗ ,푋 = 푥 + 푦 + 푧, 푐푋 > 푑max{푥,푦, 푧}, then 풂푿 + 풃풙
(풄푿 − 풅풙)풎 +풂푿 + 풃풚
(풄푿 − 풅풚)풎 +풂푿 + 풃풛
(풄푿 − 풅풛)풎 ≥ퟑ풎(ퟑ풂+ 풃)
(ퟑ풄 − 풅)풎푿풎 ퟏ.
If 푎,푚 ∈ ℝ and 푏, 푐,푑, 푥,푦, 푧 ∈ ℝ∗ ,푋 = 푥 + 푦 + 푧,퐶푋 > 푑max{푥,푦, 푧}, then: 풂푿 + 풃풙풄푿 − 풅풙
풎 ퟏ
+풂푿+ 풃풚풄푿 − 풅풚
풎 ퟏ
+풂푿+ 풃풛풄푿 − 풅풛
풎 ퟏ
≥ퟑ(ퟑ풂+ 풃)풎 ퟏ
(ퟑ풄 − 풅)풎 ퟏ .
If 푛 ∈ ℕ∗ − {1},푎 ∈ ℝ = [0,∞), 푏, 푐,푑, 푥 ∈ ℝ∗ = (0,∞), 푘 = 1,푛, 푋 = ∑ 푥 and 푋 > 푑max 푥 , then
풂푿풏 + 풃풙풌풄푿풏 − 풅풙풌
풏
풌 ퟏ
≥(풂풏 + 풃)풏풄풏 − 풅
.
If 푛 ∈ ℕ∗ − {1},푎,푚 ∈ ℝ = [0,∞), 푏, 푐,푑, 푥 ∈ ℝ∗ = (0,∞), 푘 = 1,푛, 푋 = ∑ 푥 and 푐푋 > 푑max 푥 , then
(풂푿풏 + 풃풙풌)풎 ퟏ
(풄푿풏 − 풅풙풌)ퟐ풎 ퟏ
풏
풌 ퟏ
≥(풂풏 + 풃)풎 ퟏ풏풎 ퟏ
(풄풏 − 풅)ퟐ풎 ퟏ푿풏풎
For proofs of the above results we used the following inequalities: The inequality of H. Bergström is
If 풙풌 ∈ ℝ,풚풌 ∈ 푹∗ ,∀풌 = ퟏ,풏,풏 ∈ ℕ∗ − {ퟏ}, then ∑ 풙풌ퟐ
풚풌풏풌 ퟏ ≥
∑ 풙풌풏풌 ퟏ
ퟐ
∑ 풚풌풏풌 ퟏ
.
The inequality of J. Radon is If 풙풌 ∈ ℝ∗ ,풚풌 ∈ ℝ∗ ,∀풌 = ퟏ,풏,풏 ∈ ℕ∗ − {ퟏ},풎 ∈ 푹 , then
풙풌풎 ퟏ
풚풌풎
풏
풌 ퟏ
≥(∑ 풙풌풏
풌 ퟏ )풎 ퟏ
∑ 풚풌풏풌 ퟏ
풎
References:
[1] Bătineţu-Giurgiu, D.M., Stanciu, N., Inegalităţi de tip Ionescu-Weitzenböck, Gazeta Matematică, 118(1), 2013.
[2] Bătineţu-Giurgiu, D.M., Stanciu, N., O extindere şi o rafinare a inegalităţii lui Nesbitt, Articole și Note Matematice, 4(1), 2011.
[3] Bătineţu-Giurgiu, D.M., Stanciu, N., Noi generalizări ale inegalității lui Nesbitt Articole și Note Matematice, 4(1), 2011.
[4] Bătineţu-Giurgiu, D.M., Bencze, M., Stanciu, N., New generalizations and new approaches for Nesbitt’s inequality, Mathematical Education in the Current European Context, 2(1), 2011.
[5] Bătineţu-Giurgiu, D.M., Stanciu, N.,O extindere şi o rafinare a inegalităţii lui Nesbitt, R.M.T., 91(1), 2012.
[6] Bătineţu-Giurgiu, D.M., Stanciu, N., Încă patru demonstraţii ale problemei L:155 din Sclipirea minţii nr. VII 2011, Sclipirea Minţii, 5(9), 6, 2012.
[7] Bătineţu-Giurgiu, D.M., Stanciu, N., Inegalitatea lui Nesbitt, Didactica Matematică, 2(1), 2012.
[8] Bătineţu-Giurgiu, D.M., Stanciu, N., A generalization of some remarkable inequalities, Experienţe Utile în Predarea-Învăţarea Matematicii, 1(1), 2012.
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[9] Bătineţu-Giurgiu, D.M., Stanciu, N., Problem 11634, The American Mathematical Monthly, 119(3), 2012.
[10] Bătineţu-Giurgiu, D.M., Stanciu, N., Nuevas generalizaciones y aplicaciones de la desigualdad de Nesbitt, Revista Escolar de la Olimpiada Iberoamericana de Matematicá, 11(47), 2012.
[11] Bătineţu-Giurgiu, D.M., Stanciu, N., Generalizations of some remarkable inequalities, The Teaching of Mathematics, 16(1), 2013.
[12] Bătineţu-Giurgiu, D.M., Stanciu, N., About Nesbitt’s inequality, Octogon Mathematical Magazine, 20(2), 2012.
[13] Bătineţu-Giurgiu, D.M., Stanciu, N., New generalizations and new applications for Nesbitt’s inequality, Journal of Science and Arts, 12(4), 2012.
[14] Bătineţu-Giurgiu, D.M., Stanciu, N., A New generalization of Nesbitt’s inequality, Journal of Science and Arts, 13(3), 2013.
[15] Bătineţu-Giurgiu, D.M., Stanciu, N., Zvonaru, T., Generalizarea problemei VIII. 169 din RecMat nr. 2/2013, Recreaţii Matematice, 15(1), 2014.
[16] Bătineţu-Giurgiu, D.M., Stanciu, N., Una nota sobre una desigualdad de Nesbitt-Ionescu, Revista Escolar de la Olimpiada Iberoamericana de Matematica, Numero 53, Julio-Diciembre, 2015.
[17] Nesbitt, A.M., Problem 15114, Educational Times, 3(2), 1903. [18] Zvonaru, T., Stanciu, N., Şase soluţii pentru problema L:155 din Sclipirea Minţii, nr.
VII, 2011, Sclipirea Minţii, 4(8), 2011. [19] Problem 725, The Pentagon, Fall 2014, p.
ABOUT FEW INEQUALITIES IN TRIANGLE
By Vasile Jiglău – Romania
In an arbitrary triangle 퐴퐵퐶 denote by 푙 ,푚 ,ℎ respectively the lengths of the internal angle-bisector, the median and the altitude corresponding to the side 푎 = 퐵퐶 of triangle. Prove that:
a. 풍풂ퟐ
풉풂ퟐ+ 풍풃
ퟐ
풉풃ퟐ + 풍풄ퟐ
풉풄ퟐ≥ ퟐ ⋅ 풍풂
풉풂⋅ 풍풃풉풃⋅ 풍풄풉풄
+ ퟏ.
b. 풎풂ퟐ
풉풂ퟐ+ 풎풃
ퟐ
풉풃ퟐ + 풎풄ퟐ
풉풄ퟐ≤ ퟐ ⋅ 풎풂
풉풂⋅ 풎풃풉풃⋅ 풎풄풉풄
+ ퟏ
c. explain why each of a. and b. are equivalent to the fundamental inequality of the triangle.
Proof: We’ll prove that each of the two inequalities from enunciation is equivalent to the fundamental inequality of the triangle. Denote by 푎, 푏, 푐 the sides of the given triangle and by
풙 = 풑 − 풂,풚 = 풑− 풃, 풛 = 풑 − 풄 (1)
The sums and the products which will be used below will be cyclic.
a. Using the well-known formulas which give the lengths of the angle-bisector and the altitude corresponding to the side 푎 = 퐵퐶, one obtain:
풍풂ퟐ
풉풂ퟐ= ퟒ풃풄풑(풑 풂)
(풃 풄)ퟐ⋅ 풂ퟐ
ퟒ푺ퟐ= ퟒ푹
풓⋅ 풂(풑 풂)
(풃 풄)ퟐ (2)
and by summation
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∑ 풍풂ퟐ
풉풂ퟐ= ퟒ푹
풓∑ 풂(풑 풂)
(풃 풄)ퟐ= ퟒ푹
풓⋅ ∑풙(풚 풛)(풙 풚 ퟐ풛)(풙 ퟐ풚 풛)
∏(풃 풄)ퟐ (3)
Using (2), one obtain immediately that
∏ 풍풂풉풂
= ퟏퟔ푹ퟐ풑∏(풃 풄)
(4)
Denote the numerator of (3) by 푈 = ∑푥(푦 + 푧)(푥 + 푦 + 2푧)(푥 + 2푦+ 푧) and the denominator of (3) by 푉 = ∏(푏 + 푐) = ∏(2푥 + 푦 + 푧). The inequality from enunciation becomes equivalent to:
4푅푟⋅
푈∏(푏 + 푐) ≥
32푅 푝∏(푏 + 푐) + 1 ⇔ 4푅푈 ≥ 32푅 푟푝 (푏 + 푐) + 푟 (푏 + 푐) ⇔
ퟒ푹푼 ≥ ퟑퟐ푹ퟐ풓풑푽+ 풓푽ퟐ (5)
Let’s write 푈 and 푉 as functions of 푅, 푟,푝
a.Let’s first write 푈 as function of the fundamental symmetric polynomials of 푥,푦, 푧. The following identity can be easily verified:
푼 = ퟓ(∑풙)ퟒ(∑풙풚) + ퟏퟗ풙풚풛(풙)ퟑ − ퟐ(풙)ퟐ(∑풙풚)ퟐ − ퟑ풙풚풛(∑풙)(∑풙풚)− ퟑ풙ퟐ풚ퟐ풛ퟐ (6)
The proof is given in the annex.
Using (6), let’s write 푈 as function of 푅, 푟,푝, applying the following:
∑푥 = ∑(푝 − 푎) = 푝, ∑푥푦 = ∑(푝 − 푎)(푝 − 푏) = 푟(4푅 + 푟), 푥푦푧 = ∏(푝 − 푎) = 푝푟
We obtain:
푼 = ퟓ풑ퟐ ퟒ푹풓+ 풓ퟐ + ퟏퟗ풑ퟒ풓ퟐ − ퟐ풑ퟐ ퟒ푹풓+ 풓ퟐퟐ− ퟑ풑ퟐ풓ퟐ ퟒ푹풓+ 풓ퟐ −
−ퟑ풑ퟐ풓ퟒ = 풑ퟐ풓 풑ퟐ(ퟐퟎ푹 + ퟐퟒ풓)− ퟑퟐ푹ퟐ풓+ ퟐퟖ푹풓ퟐ + ퟖ풓ퟑ (7)
2. Let’s now write 푉 = ∏(푏 + 푐) as function of 푅, 푟,푝. We have:
∏(풃+ 풄) = (∑풂)(∑풂풃) −풂풃풄 = ퟐ풑 풑ퟐ + 풓ퟐ + ퟒ푹풓 − ퟒ푹풓풑 = ퟐ풑 풑ퟐ + 풓ퟐ + ퟐ푹풓 (8)
(we have used the known identity ∑푎푏 = 푝 + 푟 + 4푅푟)
Taking into account (7) and (8), it results that (5), and as consequence the inequality from enunciation, become equivalent to:
4푅푟푝 [푝 (20푅+ 24푟) − (32푅 푟 + 28푅푟 + 8푟 )] ≥
≥ 64푅 푟푝 (푝 + 푟 + 2푅푟) + 4푟푝 (푝 + 푟 + 2푅푟)
which becomes, after computation, equivalent to:
푝 − 푝 (4푅 + 20푅푟 − 2푟 ) + 푟(4푅 + 푟) ≤ 0
which is the fundamental inequality of the triangle, because
푝 − 푝 (4푅 + 20푅푟 − 2푟 ) + 푟(4푅 + 푟) = 퐸 퐸 , where:
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퐸 = 푝 − 2푅 + 10푅푟 − 푟 − 2 푅(푅 − 2푟) ≥ 0
퐸 = 푝 − 2푅 + 10푅푟 − 푟 + 2 푅(푅 − 2푟) ≤ 0
b. The inequality from enunciation is equivalent to:
풎풂ퟐ
풉풂ퟐ⋅ 풎풃
ퟐ
풉풃ퟐ ⋅
풎풄ퟐ
풉풄ퟐ≥ ퟏ
ퟒ풎풂ퟐ
풉풂ퟐ+ 풎풃
ퟐ
풉풃ퟐ + 풎풄
ퟐ
풉풄ퟐ− ퟏ
ퟐ (9)
Using the formula 푚 = , one can easily verify the (known) identity:
푚ℎ
− 1 =(푏 − 푐 )
16푆
Denote by:
푚ℎ
− 1 =(푏 − 푐 )
16푆= 푢,
푚ℎ
− 1 =(푐 − 푎 )
16푆= 푣,
푚ℎ
− 1 =(푎 − 푏 )
16푆= 푤
Then 푢, 푣,푤 are obviously positive, and (9) becomes equivalent to:
4(푢 + 1)(푣 + 1)(푤 + 1) ≥ (푢 + 푣 + 푤 + 2) ⇔
⇔ 4푢푣푤 + 4(푢푣 + 푣푤 + 푤푢) + 4(푢 + 푣 + 푤) + 4 ≥
(푢 + 푣 +푤 + 2푢푣 + 2푣푤 + 2푤푢) + 4(푢 + 푣 +푤) + 4 ⇔
⇔ 4푢푣푤 + 2(푢푣 + 푣푤 + 푤푢) ≥ 푢 + 푣 + 푤 ⇔
⇔ ퟒ풖풗풘 + [ퟐ(풖풗+ 풗풘 + 풘풖) − 풖ퟐ − 풗ퟐ − 풘ퟐ] ≥ ퟎ (11)
We see that the above right parenthesis equals zero, because:
2(푢푣 + 푣푤 + 푤푢)− 푢 − 푣 − 푤 = 0 ⇔
⇔ 2 (푎 − 푏 ) (푎 − 푐 ) − (푎 − 푏 ) = 0
identity which can be easily verified by computation. In fact, from Heron’s identity:
2(푥 푦 + 푦 푧 + 푧 푥 )− 푥 − 푦 − 푧 =
= (푥 + 푦 + 푧)(−푥 + 푦+ 푧)(푥 − 푦 + 푧)(푥 + 푦 − 푧)
it results that the polynomial 2(푥 푦 + 푦 푧 + 푧 푥 )− 푥 − 푦 − 푧
is divisible by 푥 + 푦 + 푧, so 2∑(푎 − 푏 ) (푎 − 푐 ) −∑(푎 − 푏 ) is divisible by (푎 − 푏 ) +(푏 − 푐 ) + (푐 − 푎 ) = 0; the conclusion is the same, so (11) is equivalent to 4푢푣푤 ≥ 0, which
is true, then (9) is true, too.
Because 푢푣푤 ≥ 0 is equivalent to the fundamental inequality (see [1] – pag. 3), it results that this second inequality from the enunctiation of the problem is equivalent to the fundamental inequality of the triangle. From here, the equalities in a. and b. hold for the isosceles triangles
(wide – or tall – isosceles, see [2], theorem 1)
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c. it results immediately from the above solutions.
Remark: I added this last point because it can exist several proofs both for a. and for b. – I found one for each of the two – and I wanted to emphasise the equivalence with the foundamental
inequality.
Reference:
[1] D. Mitrinovic, s.a. – Recent advances in geometric inequalities
[2] T. Barsan – Bounds for elements of a triangle expressed by 푹, 풓,풑, Forum Geometricum, 2015, p. 99-103.
Annex – the proof of (6)
푈 = 푥(푦 + 푧) (푥 + 푦 + 2푧)(푥 + 2푦 + 푧) = 푥(푦 + 푧) [(푥 + 푦 + 푧) + 푧][(푥 + 푦 + 푧) + 푦]=
푥(푦 + 푧) 푥 + (푦 + 푧) 푥 + 푦푧 =
푥(푦 + 푧) 푥 + (푦 + 푧) 푥 + 푦 푧 + 2(푦 + 푧) 푥 + 2푦푧 푥
+ 2푦푧(푦 + 푧) 푥
and after the multiplication of 푥(푦 + 푧) with the terms of the right parenthesis, we obtain: 푈 = 퐴 + 퐵 + 퐶 +퐷 + 퐸 + 퐹, where:
퐴 = 푥(푦 + 푧) 푥 = 2 푥푦 푥
퐵 = 푥(푦+ 푧) 푥 = 푥 푥(푦 + 푧) = 푥 [푥(푥 + 푦 + 푧 − 푥) ]
= −2 푥 푥푦 + 푥 푥푦 + 5푥푦푧 푥
퐶 = 푥푦푧 푦푧 (푦 + 푧) = 푥푦푧 푥 푥푦 − 3푥푦푧 = 푥푦푧 푥 푥푦 − 3푥 푦 푧
퐷 = 2 푥 푥(푦 + 푧) = 2 푥 푥 푥푦 + 3푥푦푧 =
= 2 푥 푥푦 + 6푥푦푧 푥
퐸 = 푥(푦 + 푧) ⋅ 2푦푧 푥 = 4푥푦푧 푥
퐹 = 푥(푦 + 푧) ⋅ 2푦푧(푦+ 푧) 푥 = 2푥푦푧 푥 (푦 + 푧) =
= 4푥푦푧 푥 푥 − 푥푦 = 4푥푦푧 푥 − 4푥푦푧 푥 푥푦
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It results:
푈 = 2 푥 푥푦 − 2 푥 푥푦 + 푥 푥푦 +
+5푥푦푧 푥 + 푥푦푧 푥 푥푦 − 3푥 푦 푧 + 2 푥 푥푦 +
+6푥푦푧 푥 + 4푥푦푧 푥 + 4푥푦푧 푥 − 4푥푦푧 푥 푥푦
= 5 푥 푥푦 + 19푥푦푧 푥 − 2 푥 푥푦 −
−3푥푦푧(∑푥)(∑푥푦)− 3푥 푦 푧 (6)
CONSTRUCTION OF SOME INEQUALITIES USING WEIERSTRASS’S THEOREM FOR COMPACT SETS
By Daniel Sitaru, Claudia Nănuți-Romania
Abstract: In this article we show a construction method of some inequalities using Weierstrass’s theorem.
Weierstrass Theorem.The compactness is preserved by continuous functions, i.e. the image of the compact space under a continuous mapping is also compact. A subset of the real line is compact if and only if it is both closed and bounded. This implies the following generalization of the extreme value theorem: a continuous real-valued function on a nonempty compact space is bounded above and attains its supremum.
Corollary 1. Let be 푎 > 0,퐾 = [0,푎] × [0,푎];푓:퐾 → ℝ a continuous function. If 푓 is strictly convex in each variable (when the others one is fixed) then 푓 touches its maximum in one of the vertices of 퐾 square (because the square’s sides are parallel with the axes).
max 푓 = max{푓(0,0);푓(푎, 0);푓(0,푎);푓(푎,푎)} (see [4])
Corollary 2. Let be 푎 > 0;퐾 = [0,푎] × [0,푎] × [0,푎];푓:퐾 → ℝ a continuous function. If 푓 is strictly convex in each variable (when the others are fixed) then 푓 touches its maximum in one of the vertices of the cube (because the cube’s sides are parallel with the axes).
max 푓 = max 푓(0,0,0);푓(푎, 0,0);푓(0,푎, 0);푓(0,0,푎);푓(푎,푎, 0);푓(푎, 0,푎); 푓(0,푎,푎); 푓(푎,푎,푎) (see [4])
APPLICATIONS
Application 1. Prove that if 푎, 푏 ∈ [0,2] then:
푎푏 + 2
+푏
푎 + 2+ (2 − 푎)푏 ≤ 12
Daniel Sitaru
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Proof 1(Claudia Nănuți).
Let be 퐾 = [0,2] × [0,2];푓:퐾 → ℝ; 푓(푎, 푏) = + + (2− 푎)푏
The function 푓 is continuous on [0,1] × [0,1]. From Weierstrass theorem 푓 touches its maximum on
[0,1] × [0,1], which is a convex domain. We have 푓 ʹ = − ( ) − 푏 , 푓 ʹʹ = + ( ) >
0,푓 ʹ = − ( ) + + 2푏(2 − 푎) and 푓 ʹʹ = ( ) + + 2(2− 푎) > 0.
The function 푓 is strictly convex in 푎, 푏 and is defined on a square with the sides parallel with the axes. It follows that 푓 touches its maximum in one of the square’s vertices. Since 푓(0,0) = 0,
푓(0,2) = + + (2 − 0) ⋅ 2 = 12, 푓(2,0) = + + (2 − 2) ⋅ 0 = 2 and 푓(2,2) = +
+ (2 − 2) ⋅ 2 = 4 we find max( , )∈ 푓(푎, 푏) = 푓(0,2) = 12.
Proof 2 (Dang Thuanh Tuan,Vietnam).
Since 푏 ∈ [0,2] ⇒ 푏 ≤ 2푏 we obtain + − 2 =( )
( )( ) =( )( ) ( )( ) ( )( )
( )( ) ≤ 0.
So + + (2 − 푎)푏 ≤ + + + 2푏 − 푎푏 ≤ 2 + + 2 ⋅ 2 = 12.
The equality holds for 푎 = 0,푏 = 2.
Application 2. Let be 푥,푦, 푧 ∈ [0,1]. Prove that:
+ + + (1− 푥)(1 − 푦)(1− 푧) ≤ 1
Claudia Nănuți
Proof (Daniel Sitaru):
Let be 푓: [0,1] × [0,1] × [0,1] → ℝ;
푓(푥,푦, 푧) = + + + (1 − 푥)(1− 푦)(1− 푧)
The function 푓 is continuous on 푓: [0,1] × [0,1] × [0,1].From Weierstrass theorem 푓 touches its maximum on [0,1] × [0,1] × [0,1], which is a convexe domain.
푓 ʹ = − ( ) − ( ) − (1 − 푦)(1− 푧) , 푓 ʹʹ = ( ) + ( ) > 0
Analogous: 푓 ʹʹ > 0,푓 ʹʹ > 0.The function 푓 is strictly convexe on 푥,푦, 푧 and it is defined on a cube with the sides parallel with the axes. It follows that 푓 touches its maximum in one of the cube’s vertices.
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푓(1,1,1) =3
2018; 푓(1,1,0) =
22018
;푓(0,0,1) =1
2018
푓(1,0,1) =2
2018; 푓(0,1,0) =
12018
;푓(1,0,0) =1
2018
푓(0,1,1) =2
2018; 푓(0,0,0) = 1
It follows that max 푓(푥,푦, 푧) = 1.
Application 3. Let be 푥,푦, 푧 ∈ [0,1]. Prove that:
(푦 + 1)푒 + (푥 + 1)푒 +(1− 푥)푒푦 + 푧
≤ 4푒
Daniel Sitaru
Proof (Claudia Nănuți)
Let be 푓: [0,1] × [0,1] × [0,1] → ℝ; 푓(푥,푦, 푧) = (푦 + 1)푒 + (푥 + 1)푒 + ( )
The function 푓 is continuous on [0,1] × [0,1] × [0,1]. From Weierstrass theorem 푓 touches its maximum on [0,1] × [0,1] × [0,1] which is a convex domain.
푓 ʹ = 2푥(푦 + 1)푒 + 푒 − 푒 , 푓 ʹʹ = 2(푦 + 1) 푒 + 2푥 푒 > 0, 푓 ʹ = 푒 − (1 − 푥)푒
푓 ʹʹ = (1− 푥)푒 > 0, 푓 ʹ = 2푧(푥 + 1)푒 − (1 − 푥)푒
푓 ʹʹ = 2(푥 + 1) 푒 + 2푧 푒 + (1− 푥)푒 > 0
The function 푓 is strictly convex on 푥,푦, 푧 and it is defined on a cube with the sides parallel with the axes. It follows that 푓 touches its maximum in one of the cube’s vertices.
푓(0,0,0) = 2 + 푒 ; 푓(1,0,0) = 푒 + 2;푓(0,1,0) = 3 + 푒
푓(0,0,1) = 1 + 푒; 푓(1,1,0) = 2푒 + 2;푓(0,1,1) = 2 + 푒; 푓(1,0,1) = 3푒;푓(1,1,1) = 4푒
It follows that max 푓(푥,푦, 푧) = 4푒.
Application 4. Prove that if 푎, 푏 ∈ [0,1] then:
푎 + 12
+푏 + 3
3+ (1 − 푎)푒 ≤ 푒 + 4
Claudia Nănuți
Proof (Daniel Sitaru).
Let be 푓: [0,1] × [0,1] → ℝ; 푓(푥) = + + (1− 푎)푒
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The function 푓 is continuous on [0,1] × [0,1]. From Weierstrass theorem 푓 touches its maximum on [0,1] × [0,1], which is a convex domain.
푓 ʹ = 2 − (푏 + 2)3 ln 3− 푒 , 푓 ʹʹ = (푏 + 3)푒 ln 3 > 0
푓 ʹ = −(푎 + 1)2 ln 2 + 3 − (1− 푎)푒 , 푓 ʹʹ = (푎 + 1)2 ln 2 + (1− 푎)푒 > 0
The function 푓 is strictly convex on 푎, 푏 and it is defined on a square with the sides parallel with the axes. It follows that 푓 touches its maximum in one of the square’s vertices. Since 푓(0,0) = 푒 +4;푓(1,0) = 3;푓(0,1) = and 푓(1,1) = , it follows that max 푓 = 푒 + 4 . So we obtain
+ + (1− 푎)푒 ≤ 푒 + 4.
PROPOSED PROBLEMS
Problem 1. Prove that if 푎, 푏 ∈ [0,1] then:
1 ≤1
푎 + 푏 + 1+
1푏 + 푐 + 1
+1
푐 + 푎 + 1≤ 3
Daniel Sitaru
Problem 2. Prove that if 푎, 푏, 푐 ∈ [0,1] then:
34≤
1푎 + 푏 + 2
+1
푏 + 푐 + 2+
1푐 + 푎 + 2
+ (1− 푎)(1− 푏)푒 ≤52
Daniel Sitaru
References:
[1] Daniel Sitaru, Math Phenomenon, Ed. Paralela 45, 2016.
[2] Radu Gologan, Daniel Sitaru, Leonard Giugiuc, 300 Romanian Mathematical Challenges, Ed. Paralela 45, 2016.
[3] Daniel Sitaru, Leonard Giugiuc, Claudia Nănuți, Diana Trăilescu, Inegalități – Inequalities, Ecko – Print Publishing House, 2015.
[4] G. H. Hardy, J.E. Littlewood, G. Polya, Inequalities, Cambridge University Press, 1934.
[5] Gazeta Matematică, Seriile A și B.
[6] Octogon Mathematical Magazine.
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ON THE PROBLEM 11984/AMERICAN MATHEMATICAL MONTHLY-May 2017 PROPOSED BY DANIEL SITARU
Minimal and maximal bounds for the sum 풂ퟔ + 풃ퟔ + 풄ퟔ
By Marius Drăgan, Neculai Stanciu – Romania
Theorem. In any triangle 퐴퐵퐶 with the sides 푎, 푏, 푐, circumradius 푅, inradius 푟 and 푑 =√푅 − 2푅푟 holds the following inequality:
2 (푅 + 푟 − 푑) [4(푅 − 푟 + 푑) + 푅 ] ≤ 푎 + 푏 + 푐 ≤ 2 (푅 + 푟 + 푑) [4(푅 − 푟 − 푑) + 푅 ]
Proof. We use the following
Lemma (fundamental inequality or Blundon’s inequality). For any triangle 퐴퐵퐶 the inequality 푠 ≤ 푠 ≤ 푠 hold where 푠 , 푠 represents the semiperimeter of two isosceles triangles 퐴 퐵 퐶 and 퐴 퐵 퐶 which have the same circumradius 푅 and inradius 푟 as the triangle 퐴퐵퐶 with the sides
푎 = 2 (푅 + 푟 − 푑)(푅 − 푟 + 푑), 푏 = 푐 = 2푅(푅 + 푟 − 푑)
푎 = 2 2(푅 + 푟 + 푑)(푅 − 푟 − 푑), 푏 = 푐 = 2 2푅(푅 + 푟 + 푑) where
푑 = √푅 − 2푅푟. A proof of this lemma is given in [1] .
From the identity ∑ 푥 = 3푥푦푧 + ∑ 푥 ∑ 푥 − ∑ 푦푧 if we replace
푥 = 푎 ,푦 = 푏 , 푧 = 푐 then we obtain
푎 = 3푎 푏 푐 + 푎 푎 − 푏 푐
We consider the functions:
푓,푔,ℎ,퐹: [푠 푠 ] → 푅,푓(푠) = 3푎 푏 푐 = 3(4푅푟푠) , 푔(푠) = 2(푠 − 푟 − 4푅푟),
ℎ(푠) = 푎 − (푏푐) =
= 푎 − 4 푎푏 푎 + 6푎푏푐 푎 + 푎푏 =
= 푠 − 2푟(4푅 + 7푟)푠 + (4푅 + 푟) 푟 , with ℎ (푠) = 4푠(푠 − 4푅푟 − 7푟 ), 퐹(푠) = 푓(푠) + 푔(푠)ℎ(푠).
Since 푠 ≥ 푠 ≥ 16푅푟 − 5푟 ≥ 4푅푟 + 7푟 it results that ℎ is increasing on [푠 , 푠 ]. Also, 푓 and 푔 are increasing on [푠 , 푠 ]. Hence, 퐹 is increasing on [푠 , 푠 ], then 퐹(푠 ) ≤ 퐹(푠) ≤ 퐹(푠 ) or
푎 + 푏 + 푐 ≤ 푎 + 푏 + 푐 ≤ 푎 + 푏 + 푐 (1).
If we replace the values of 푎 , 푏 , 푐 ,푎 , 푏 , 푐 from the lemma in (1), then we obtain the desired inequality and we are done.
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Reference:
[1] M. Drăgan, N. Stanciu, A new proof of the Blundon inequality, Rec. Mat, 2(2017), 100-104.
[2] D. Sitaru, Problem 11984, The American Monthly, Vol. 124, No. 5 (May 2017), p. 466.
SOME INEQUALITIES IN FOUR VARIABLES WITH SUM 1
By Ștefan Andrei Mihalcea-Romania
Let 푎, 푏, 푐,푑 ≥ 0, with 푎 + 푏 + 푐 + 푑 = 1. First, we start with ∑푎푏푐. We can associate the
members in a nice way, like this: ∑푎푏푐 = 푎푑(푏 + 푐) + 푏푐(푎 + 푑) ≤ ( )( ) ( )( ) =
( )( ) (1), using that their sum is 1. Also, ∑푎푏푐 ≤ ( )( ) (2) and
∑푎푏푐 ≤ ( )( ) (3). If we add (1) and (2) it results: ∑푎푏푐 ≤ ( )( ) ( )( ).
But 푑 + 푐 = 1− 푎 − 푏 and 푏 + 푐 = 1− 푎 − 푑. So, ∑푎푏푐 ≤( ) ( )
.
→ 16∑푎푏푐 ≤ 2 + 2푎 − 2푐 − (1 + 푎 − 푐) = 1 − (푎 − 푐) . Analogous,
16∑ 푎푏푐 ≤ 1 − (푏 − 푑) . From this, it results that max{(푎 − 푐) , (푏 − 푑) } + 16∑푎푏푐 ≤ 1.
Summing (1), (2), (3) → ∑푎푏푐 ≤ ( )( ) ( )( ) ( )( ), but this equal with
∑→ ∑푎푏푐 ≤
∑ (4). Coming back, we have
(푎 + 푐)(푏 + 푑) + (푎 + 푏)(푐 + 푑) + (푎 + 푑)(푏 + 푐) = 2(푎 + 푏 + 푐) − [(푎 + 푏) + (푎 + 푐) + (푏 + 푐) ]. So,
36∑푎푏푐 ≤ 3(2− 2푑)− (2 − 2푑) = 2(1− 푑)(1 + 2푑). From this, it results that ∀푥 ∈
{푎, 푏, 푐,푑},∑푎푏푐 ≤ ( )( ). Making 푥 = 푎, … , 푥 = 푑 , then summing → 72∑푎푏푐 ≤ 3 +
4∑ 푎푏. We can prove the last inequality using (4): we will prove that ∑
≤ 3 +
4∑ 푎푏 ↔ ∑ 푎푏 ≤ ↔ ∑푎 ≥ , which is true by 퐶 − 퐵 − 푆 and 푎 + 푏 + 푐 + 푑 = 1. From
Viette Relations → 푎, 푏, 푐,푑 are solutions for 푥 − 푥 + ∑ 푎푏 푥 − 푥 + 푎푏푐푑 = 0. Making
푥 = 푎, … , 푥 = 푑, then summing → ∑푎 + ∑ 푎푏 (∑푎 ) + 4푎푏푐푑 = ∑푎 +∑푎푏푐. From (4) →
푎 + 푎푏푐 ≥ 푎 + 6 푎푏푐 푎 + 4푎푏푐푑 ↔
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푎 (1− 푎) ≥ 4푎푏푐푑 + 푎푏푐 5 − 12 푎푏
SOME ALGEBRAIC INEQUALITIES PROVED BY USING A FERMAT-
TORICELLI’S GEOMETRIC CONFIGURATION
By Daniel Sitaru-Romania
Abstract: In this paper it’s developed a method to construct algebraic inequalities by constructing a triangle configuration. Famous inequalities are redesigned using the properties of Fermat – Torricelli point of this triangle.
Keywords: Fermat – Torricelli; inequalities
Main result: If 푥,푦, 푧 > 0 then:
1. ∏(풙ퟐ + 풙풚 + 풚ퟐ)ퟑ ≥ 풙풚 + 풚풛 + 풛풙
2. ∑ 풙ퟐ 풙풚 풚ퟐ
∏ 풙ퟐ 풙풚 풚ퟐ≤ ퟑ
풙풚 풚풛 풛풙
3. ∏(ퟐ풙ퟐ + 풙(풚 + 풛) − 풚풛) ≤ ∏(풙ퟐ + 풙풚 + 풚ퟐ)
4. ퟑ∑ 풚ퟐ 풚풛 풛ퟐ
풙ퟐ 풙풛 풛ퟐ≥ (ퟐ∑풙ퟐ + ∑풙풚)∑ ퟏ
풙ퟐ 풙풚 풚ퟐ
5. (풙풚 + 풚풛 + 풛풙)∑ 풙ퟐ + 풙풚+ 풚ퟐ ≤ ퟑ ∏(풙ퟐ + 풙풚 + 풚ퟐ)
6. (풙 + 풚 + 풛)ퟐ ≤ ∑ (풙ퟐ + 풙풚 + 풚ퟐ)(풚ퟐ + 풚풛+ 풛ퟐ)
7. ퟑ(∑풙풚)ퟐ
∑ 풙ퟐ 풙풚 풚ퟐퟐ ≤
∏ 풙ퟐ 풙풚 풚ퟐ
∑ 풙ퟐ 풙풚 풚ퟐ≤ ∏ 풙ퟐ 풙풚 풚ퟐ
ퟑ(∑풙풚)ퟐ
Let 푇 be a point in a plane and 퐴,퐵,퐶 such that 푇퐴 = 푥;푇퐵 = 푦;푇퐶 = 푧 and ∢퐴푇퐵 =∢퐵푇퐶 = ∢퐶푇퐴 = 120°
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푇 is the Fermat – Torricelli’s point of 훥퐴퐵퐶. By cosine’s rule in 훥퐵푇퐶; 훥퐶푇퐴; 훥퐴푇퐵:
푎 = 퐵퐶 = 푦 + 푦푧 + 푧
푏 = 퐶퐴 = 푧 + 푧푥 + 푥
푐 = 퐴퐵 = 푥 + 푥푦 + 푦
Area of triangle 퐴퐵퐶:
푆 = [퐴퐵퐶] =12푇퐵 ⋅ 푇퐶 sin 120° +
12푇퐶 ⋅ 푇퐴 sin 120° +
+12푇퐴 ⋅ 푇퐶 sin 120° =
√34
(푥푦 + 푦푧 + 푧푥)
Proof 1: By Carlitz’s inequality (AMM – 1961) in 훥퐴퐵퐶:
푎 푏 푐 ≥4√3
푆
(푥 + 푥푦 + 푦 ) ≥4√3
⋅√34
(푥푦 + 푦푧 + 푧푥)
(푥 + 푥푦 + 푦 ) ≥ 푥푦 + 푦푧 + 푧푥
Observation: Inequality (1) it is known as Wu’s inequality.
Proof 2: By Mitrinovic’s inequality in 훥퐴퐵퐶:
푠 ≤3√3
2푅
2푠 ≤ 3√3푅 ⇔ 푎 + 푏 + 푐 ≤ 3√3푎푏푐4푆
4푆(푎 + 푏 + 푐) ≤ 3√3푎푏푐
4 ⋅√35
푥푦 푥 + 푥푦 + 푦 ≤ 3√3 푥 + 푥푦 + 푦
√3 ⋅ 푥푦 ⋅ 푥 + 푥푦 + 푦 ≤ 3√3 푥 + 푥푦 + 푦
∑ 푥 + 푥푦 + 푦∏ 푥 + 푥푦 + 푦
≤3
푥푦 + 푦푧 + 푧푥
Proof 3: cos퐴 = =( )( )
=
cos퐴 =2푥 + 푥(푦 + 푧)− 푦푧
2 (푧 + 푧푥 + 푥 )(푥 + 푥푦 + 푦 )
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It’s known that: cos퐴 cos퐵 cos 퐶 ≤
2푥 + 푥(푦 + 푧)− 푦푧2 (푧 + 푧푥 + 푥 )(푥 + 푥푦 + 푦푧)
≤18→∏(2푥 + 푥(푦 + 푧)− 푦푧)
∏(푥 + 푥푦 + 푦 ) ≤ 1
(2푥 + 푥(푦 + 푧) − 푦푧) ≤ (푥 + 푥푦 + 푦 )
Proof 4: It’s known Walker’s inequality in any 훥퐴퐵퐶:
3푎푏
+푏푐
+푐푎
≥ (푎 + 푏 + 푐 )1푎
+1푏
+1푐
3푦 + 푦푧 + 푧푥 + 푥푧 + 푧
≥ (푥 + 푥푦 + 푦 ) ⋅1
푥 + 푥푦 + 푦
3푦 + 푦푧 + 푧푥 + 푥푧 + 푧
≥ 2 푥 + 푥푦 ⋅1
푥 + 푥푦 + 푦
Proof 5: It’s known Curry’s inequality (AMM-1967). In any triangle 퐴퐵퐶:
4푆√3 ≤9푎푏푐
푎 + 푏 + 푐
4 ⋅√34
(푥푦 + 푦푧 + 푧푥) ⋅ √3 ≤9∏ 푥 + 푥푦 + 푦∑ 푥 + 푥푦 + 푦
(푥푦 + 푦푧 + 푧푥) 푥 + 푥푦 + 푦 ≤ 3 (푥 + 푥푦 + 푦 )
Proof 6: It’s known Hadwiger – Finsler’s inequality.In any triangle 퐴퐵퐶:
푎 + 푏 + 푐 ≥ 4푆√3 + (푎 − 푏) + (푏 − 푐) + (푐 − 푎)
2(푎푏 + 푏푐 + 푐푎) ≥ 4푆√3 + 푎 + 푏 + 푐
2 (푥 + 푥푦 + 푦 )(푦 + 푦푧 + 푧 ) ≥ 4√3 ⋅√34
푥푦 + (푥 + 푥푦 + 푦 )
2 (푥 + 푥푦 + 푦 )(푦 + 푦푧 + 푧 ) ≥ 3 푥푦 + 2 푥 + 푥푦
푥 + 2 푥푦 ≤ (푥 + 푥푦 + 푦 )(푦 + 푦푧 + 푧 )
(푥 + 푦 + 푧) ≤ (푥 + 푥푦 + 푦 )(푦 + 푦푧 + 푧 )
Proof 7: It’s known Klamkin’s inequality.In any 훥퐴퐵퐶:
4푟 ≤푎푏푐
푎 + 푏 + 푐≤ 푅
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4 ⋅푆푠
≤푎푏푐
푎 + 푏 + 푐≤푎 푏 푐
16푆
4 ⋅(∑푥푦)
∑ 푥 + 푥푦 + 푦≤∏ 푥 + 푥푦 + 푦∑ 푥 + 푥푦 + 푦
≤∏(푥 + 푥푦 + 푦 )
16 ⋅ (∑푥푦)
3(∑푥푦)
∑ 푥 + 푥푦 + 푦≤
∏(푥 + 푥푦 + 푦 )∑ 푥 + 푥푦 + 푦
≤∏(푥 + 푥푦 + 푦 )
3(∑푥푦)
Observation:All inequalities (1-7) becomes equalities for 푥 = 푦 = 푧.
Reference:
[1] – Daniel Sitaru – “Math Phenomenon”- “Editura Paralela 45” Publishing House – Pitești – Romania – 2016
[2] – “Romanian Mathematical Magazine”, www.ssmrmh.ro
ABOUT SOME FAMOUS INEQUALITIES
By D.M. Bătinețu-Giurgiu, Daniel Sitaru – Romania
Abstract: In this paper we develop few famous inequalities with the same start point in a lemma.
Keywords: Bergstrom, Hadwigwer-Finsler, Cebyshev, Mitrinovic, Doucet, Tsintsifas, Hölder, Minkowski
If 푛 ∈ ℕ; 푛 ≥ 2;푥 ∈ (0, ∞); 푘 ∈ 1,푛 we denote 푥 = ∑ 푥 .
Lemma: If 푚 ∈ ℕ;푦 ∈ [0, ∞) then:
푚 + 푦 ≥ (푚 + 1)푦 (1)
Proof: If 푚 = 0 its obviously.
If 푚 ≠ 0 by AM-GM: 푚 + 푦 ≥ (푚 + 1) 1 ⋅ 1 ⋅ … ⋅ 1" "
⋅ 푦 = (푚 + 1)푦
Theorem 1: If 푚,푛 ∈ ℕ;푛 ≥ 2;푥 ∈ (0, ∞), 푘 ∈ 1,푛;푎 ∈ [0, ∞); 푐푋 > 푑max 푋 ;
푏, 푐,푑 ∈ (0, ∞) then: 푚푛 + ∑ ≥ ( )( ) (2)
Proof:
퐴 (푚) = 푚푛 +푎푋 + 푏푋푐푋 − 푑푋
= 푚 +푎푋 + 푏푋푐푋 − 푑푋
≥( )
≥ (푚 + 1) ⋅ ∑ = (푚 + 1)퐵 (3)
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퐵 =푎푋 + 푏푋푐푋 − 푑푋
⇔ 푑퐵 =푎푑푋 + 푏푑푋푐푋 − 푑푋
⇔ 푑퐵 + 푛푏 =
=푎푑푋 + 푏푑푋푐푋 − 푑푋
+ 푏 = (푎푑 + 푏푐)푋 ⋅1
푐푋 − 푑푋≥
≥ (푎푑 + 푏푐)푋 ⋅푛
∑ (푐푋 − 푑푋 ) = (푎푑 + 푏푐)푋 ⋅푛
푐푛푋 − 푑 ∑ 푥=
= (푎푑 + 푏푐)푋 ⋅푛
푐푛푋 − 푑 ∑ 푥= (푎푑 + 푏푐)푋 ⋅
푛푐푛푋 − 푑푋
=(푎푑 + 푏푐)푛푐푛 − 푑
⇔
⇔ 푑퐵 ≥(푎푑 + 푏푐)푛푐푛 − 푑
− 푏푛 =(푎푑 + 푏푐)푛 − 푏푐푛 + 푏푑푛
푐푛 − 푑=
(푎푛 + 푏)푑푛푐푛 − 푑
⇔
⇔ 퐵 ≥ ( ) (4)
By (3); (4): 퐴 (푚) ≥ ( )( )
Theorem 2: If 푚,푛,푝 ∈ ℕ; 푛 ≥ 2;푋 ∈ (0, ∞); 푘 ∈ 1,푛;푎 ∈ [0, ∞);푏, 푐,푑 ∈ (0, ∞) such that 푐푋 > 푑max 푥 then:
∑ (푚 + (푎푋 + 푏푋 ) ) ⋅ 푝 + ( ) ≥ (푚 + 1)(푝+ 1) ⋅ ( ) (5)
Proof:
퐶 (푚) = (푚 + (푎푋 + 푏푥 ) ) 푝 +1
(푐푋 − 푑푥 ) ≥
≥ (푚 + 1)(푎푋 + 푏푥 ) (푝 + 1) ⋅1
푐푋 − 푑푥=
(푚 + 1)(푝 + 1)∑ = (푚 + 1)(푝 + 1)퐵 (6)
퐵 =푎푋 + 푏푋푐푋 − 푑푋
=(푎푋 + 푏푥 )
(푐푋 − 푑푥 )(푎푋 + 푏푥 ) =(푎푋 + 푏푥 )
푎푐푋 + (푏푐 − 푎푑)푋 푥 − 푏푑푥≥
≥∑ ( )
∑ ( ) = ( )( ) ∑ (7)
∑ 푋 ≥ . By (7): 퐵 ≥ ( )
( )= ( )
( ) =
= ( )( ) ( ) = ( )
( )( ) = ( ) (8)
By (6); (7): 퐶 (푚) ≥ (푚 + 1)(푝 + 1) ( )
Theorem 3: If 푚,푛,푝 ∈ ℕ; 푛 ≥ 2; 푎 ,푏 ∈ [0, ∞); 푘 ∈ 1,푛 then:
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∑ 푚 + 푎 ( ) ⋅ ∑ 푝 + 푏 ( ) ≥ (푚 + 1)(푝+ 1)(∑ 푎 푏 ) (9)
Proof: By lemma:
퐷 (푚,푝) = 푚 + 푎 ( ) ⋅ 푝 + 푏 ( ) ≥ (푚 + 1)푎 (푝 + 1)푏 =
= (푚 + 1)(푝 + 1) 푎 푏 ≥ (푚 + 1)(푝+ 1) 푎 푏
Theorem 4: If 푚,푛,푝, 푟, 푠 ∈ ℕ,푛 ≥ 2;푎 ,푏 ∈ (0, ∞); 푘 ∈ 1,푛 then:
(푚 + |푎 | ) (푝 + |푎 | ) ⋅ (푟 + |푏 | ) (푠 + |푏 | ) ≥
≥ (푚 + 1)(푝+ 1)(푟 + 1)(푠 + 1)(∑ 푎 푏 ) (10)
Proof: By lemma:
퐸 (푚,푝, 푟, 푠) = (푚 + |푎 | )(푝 + |푎 | ) (푟 + |푏 | (푠 + |푏 | ) ≥
≥ (푚 + 1)|푎 |(푝+ 1)|푎 | (푟 + 1)|푏 |(푠+ 1)|푏 | =
= (푚 + 1)(푝 + 1)(푟 + 1)(푠 + 1) |푎 | |푏 | ≥ (푚 + 1)(푝 + 1)(푟 + 1)(푠 + 1) 푎 푏
Theorem 5: If 푚,푛 ∈ ℕ;푛 ≥ 2;푎 ∈ (0, ∞); 푘 ∈ 1,푛:
푚푛 + ∑ 푎 ≥ (푚 + 1)푛 ∏ 푎 (11)
Proof:
퐹 (푚) = 푚푛 + 푎 = 푚 + 푎 ≥ (푚 + 1) 푎 ≥ (푚 + 1)푛 푎
Theorem 6: If 푚,푛 ∈ ℕ;푛 ≥ 2;푎 , 푏 ∈ (0, ∞); 푘 ∈ 1,푛;푝, 푞 ∈ (0, ∞); + = 1 then:
푚푛 + ∑ 푎 ( ) 푚푛 +∑ 푏 ( ) ≥ (푚 + 1)∑ 푎 푏 (12)
Proof:
푚푛 + 푎 ( ) 푚푛 + 푏 ( ) ≥ 푚 + 푎 ( ) ⋅ 푚 + 푏 ( ) ≥
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≥ (푚 + 1)푎 ⋅ (푚 + 1)푏 = (푚 + 1) ⋅ 푎 ⋅ 푏 =
= (푚 + 1) 푎 푏 ≥ö
(푚 + 1) 푎 푏
Theorem 7: If 푚,푛 ∈ ℕ;푛 ≥ 2;푎 , 푏 ∈ (0, ∞); 푘 ∈ 1,푛;푝 ∈ (1, ∞) then:
푚푛 + ∑ 푎 ( ) + ∑ 푚 + 푏 ( ) ≥ (푚 + 1) (∑ (푎 + 푏 ) ) (13)
Proof: 퐺 (푚) = 푚푛 + ∑ 푎 ( ) + ∑ 푚 + 푏 ( ) =
= 푚 + 푎 ( ) + 푚 + 푏 ( ) ≥( )
(푚 + 1)푎 + (푚 + 1)푏 =
= (푚 + 1) 푎 + 푏 ≥ (푚 + 1) (푎 + 푏 )
Theorem 8: If 푚,푛,푝 ∈ ℕ; 푛 ≥ 2; 푎 ,푏 ∈ (0, ∞); 푘 ∈ 1,푛; (푎 ) ; (푏 ) same monotony then:
푛∑ 푚 + 푎 푝 + 푏 ≥ (푚 + 1)(푝 + 1)(∑ 푎 )(∑ 푏 ) (14)
Proof: By lemma: 퐻 (푚,푝) = 푛∑ 푚 + 푎 푝 + 푏 ≥ 푛∑ (푚 + 1)푎 (푝+ 1)푏 =
= 푛(푚 + 1)(푝+ 1) 푎 푏 ≥ (푚 + 1)(푝+ 1) 푎 푏
Theorem 9: If 푚 ∈ ℕ;푢, 푣,푤, 푥,푦, 푧 ∈ (0, ∞) then:
3푚 + + + ≥ (푚 + 1) ⋅ ( ) − (푚 + 1)(푥 + 푦 + 푧 ) (15)
Proof: 퐾(푚) = 3푚 + + + =
= 푚 +푢푥푣 + 푤 + 푚 +
푣푦푤 + 푢 + 푚 +
푤푧푢 + 푣 ≥ (푚 + 1)
푢푥푣 + 푤 +
푣푦푤 + 푢 +
푤푧푢 + 푣 =
= (푚 + 1)푢푥푣 +푤
+ 푥 − (푚 + 1)(푥 + 푦 + 푧 ) =
= (푚 = 1)(푢 + 푣 + 푤)푥
푣 +푤− (푚 + 1)(푥 + 푦 + 푧 ) ≥
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≥ (푚 + 1)(푢 + 푣 +푤) ⋅(푥 + 푦 + 푧)∑ (푣 + 푤) −
(푚 + 1)(푥 + 푦 + 푧 ) =
= (푚 + 1)12
(푥 + 푦 + 푧) − (푥 + 푦 + 푧 )
In 훥퐴퐵퐶; 푠 - semiperimeter; 퐹 - area; 푟 – inradii, 푅 - circumradii. Its known that:
푎 + 푏 + 푐 = 2(푠 − 푟 − 4푅푟) (16)
푎푏 + 푏푐 + 푐푎 = 푠 + 푟 + 4푅푟 (17)
Doucet’s inequality: 4푅 + 푟 ≥ 푠√3 (D)
If in (15) we take 푥 = 푎;푦 = 푏; 푧 = 푐:
3푚 +푢푎푣 +푤
≥ (푚 + 1) ⋅(푎 + 푏 + 푐)
2− (푚 + 1)(푎 + 푏 + 푐 ) =
= (푚 + 1) 2푠 − 2(푠 − 푟 − 4푅푟) = 2(푚 + 1)푟(4푅 + 푟) ≥( )
2(푚 + 1)푟푠√3 = 2(푚 + 1)√3퐹
If 푚 = 0 in (18):
+ + ≥ 2√3퐹 (T)
which is Tsintisifas’ inequality.
Theorem 10: If 푚,푛 ∈ ℕ; 푢, 푣,푤, 푥,푦, 푧 ∈ (0, ∞); 푢 + 푣 +푤 = 푈; 푑, 푡 ∈ (0, ∞); 푑 ≥ 푡;
푝 ∈ (0, ∞) then:
푚 +푑푈 − 푡푢푝푈 + 푡푢
⋅ 푛 + 푋 ( ) + 푚+푑푈 − 푡푣푝푈 + 푡푣
⋅
⋅ 푛 + 푦 ( ) + 푚 +푑푈 − 푡푤푝푈 + 푡푤
⋅ 푛 + 푧 ( ) ≥
≥ ( )( ) (푑 + 푝)(푥 + 푦 + 푧) − (3푝+ 푡)(푥 + 푦 + 푧 ) (19)
Proof: Let 퐿(푚,푛) be the LHS in (19). By lemma:
퐿(푚,푛) ≥ (푚 + 1) ⋅푑푈 − 푡푢푝푈 + 푡푢
(푛 + 1)푋 = (푚 + 1)(푛 + 1)푑푈 − 푡푢푝푈 + 푡푢
⋅ 푥 ⇔
⇔ 퐿(푚,푛) + (푚 + 1)(푛 + 1)(푥 + 푦 + 푧 ) ≥ (푚 + 1)(푛+ 1)푑푈 − 푡푢푝푈 + 푡푢
+ 1 푥 =
= (푚 + 1)(푛 + 1)(푑+ 푝)푈 ⋅푥
푝푈 + 푡푢≥ (푚 + 1)(푛+ 1)(푑 + 푝)푈 ⋅
(푥 + 푦 + 푧)∑ (푝푈 + 푡푢) =
Romanian Mathematical Society-Mehedinți Branch 2018
32 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
=(푚 + 1)(푛 + 1)(푑 + 푝)푈 ⋅ (푥 + 푦 + 푧)
3푝푈 + 푡 ∑ 푈=
(푚 + 1)(푛 + 1)(푑 + 푝)(푥 + 푦 + 푧)3푝+ 푡
⇔
⇔ 퐿(푚,푛) ≥(푚 + 1)(푛 + 1)
3푝 + 푡(푑 + 푝)(푥 + 푦 + 푧) − (3푝 + 푡)(푥 + 푦 + 푧 )
If 푚 = 푛 = 0;푝 = 0; 푑 = 푡 = 1 by (19):
푥 + 푦 + 푧 ≥ (푥 + 푦 + 푧) − (푥 + 푦 + 푧 ) (20)
푥 + 푦 + 푧 ≥ (푥 + 푦 + 푧) − (푥 + 푦 + 푧 ) (21)
If in (21); 푥 = 푎; 푦 = 푏; 푧 = 푐 then in 훥퐴퐵퐶:
푣 +푤푢
푎 +푤 + 푢푣
푏 +푢 + 푣푤
푐 ≥ (푎 + 푏 + 푐) − (푎 + 푏 + 푐 ) =
= 4푠 − 2(푠 − 푟 − 4푅푟) = 2푠 + 2푟(4푅 + 푟) ≥
≥ 2푠 ⋅ 3√3푟 + 2푟(푟 + 4푅) ≥
≥ 6√3푟푠 + 2푟푠√3 = 8√3푟푠 = 8√3퐹
This is called D.M. Bătinețu-Giurgiu’s inequality: 푎 + 푏 + 푐 ≥ 8√3퐹 (B-G)
Finally we add a proof by Hadwiger-Finsler’s inequality:
푎 + 푏 + 푐 ≥ 4√3퐹 + (푎 − 푏) + (푏 − 푐) + (푐 − 푎) (H-F)
Proof: (퐻 − 퐹) ⇔ 퐵 = 푎 + 푏 + 푐 − 2(푎 + 푏 + 푐 − 푎푏 − 푏푐 − 푐푎) ≥ 4√3퐹
퐵 = 푎 + 푏 + 푐 − 2(푎 + 푏 + 푐 ) + 2(푎푏 + 푏푐 + 푐푎) =
= 2(푎푏 + 푏푐 + 푐푎)− (푎 + 푏 + 푐 ) =( ) ( )
2(푠 + 푟 + 4푅푟)− 2(푠 − 푟 − 4푅푟) =
= 4푟(푅 + 푟) ≥ 4푟푠√3 = 4√3푟푠 = 4√3퐹
PROPOSED PROBLEMS
5-CLASS-STANDARD
V.1. Se consideră numerele naturale 푎 și 푏. Împărțindu-l pe 푎 la 푏 se obține câtul 5 și restul 푟. Împărțindu-l pe 푎 la (푏 + 1) se obține restul 푟. Aflați 푎 și 푏.
Proposed by Constantin Ionică-Romania
V.2. Să se determine numerele naturale 푛 care împărțite la 5 dau câtul 푎 și restul 푏, iar împărțite la 9 dau câtul 푏 și restul 푎. Proposed by Constantin Ionică-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
33 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
V.3. Să se determine 푥,푦, 푧 ∈ ℕ știind că ele verifică relația 3 ⋅ 4 ⋅ 5 + 4푦 = 4 + 101
Proposed by Constantin Ionică-Romania
V.4. Pe o scară a unui bloc sunt 17 apartamente cu două, trei și respectiv patru camere. Câte apartamente de fiecare fel dacă sunt în total 52 de camere?
Proposed by Constantin Ionică-Romania
V.5. Pentru ce număr natural 푛, numărul 푁 = 1 ⋅ 2 ⋅ 3 ⋅ … ⋅ 푛 + 417 este pătrat perfect?
Proposed by Constantin Ionică-Romania
V.6. Să se rezolve ecuația 2 + 2 + 2 + 2 = 1856, știind că 푥,푦, 푧, 푡 sunt numere naturale distincte. Proposed by Constantin Ionică-Romania
V.7. Aflați două numere naturale 푎 și 푏 cu 푎 < 푏, iar produsul lor și cel mai mic multiplu comun al acestora sunt egale cu 13585. Proposed by Constantin Ionică-Romania
V.8. Aflați două numere naturale 푎 și 푏 cu 푎 > 푏, știind că suma lor este 144, iar cel mai mic multiplu comun al lor este 420. Proposed by Constantin Ionică-Romania
V.9. Aflați numerele naturale 푎, 푏, 푐,푑 știind că verifică relațiile: 푎푏 + 푐 + 푑 = 1200, 푎푏 = 3(푐 + 푑), cel mai mic multiplu comun al numerelor 푎 și 푏 este 60, iar cel mai mare divizor comun al numerelor 푐 și 푑 este 25. Proposed by Constantin Ionică-Romania
V.10. Aflați numerele prime de forma abc, știind că: c ⋅ ab = b ⋅ ac + 50.
Proposed by Constantin Ionică-Romania
V.11. Să se determine numărul natural 퐴 = 5 ⋅ 7 , știind că 5퐴 are cu 8 divizori mai mult decât 퐴, iar 7퐴 are cu 6 divizori mai mult decât 퐴. Proposed by Constantin Ionică-Romania
V.12. Să se arate că numerele: 푎 = 1 ⋅ 2 ⋅ 3 ⋅… ⋅ 95 ⋅ 96 și 푏 = 1 ⋅ 2 ⋅ 3 ⋅ … ⋅ 97 ⋅ 98 dau același rest la împărțirea cu 1901. Proposed by Constantin Ionică-Romania
V.13. Să se determine cifrele 푥,푦,푧 pentru care șirul 4526,7523,8720, continuă cu 5푥푦푧.
Proposed by Gheorghe Căiniceanu-Romania
V.14. O minipiscină în formă de paralelipiped dreptunghic are dimensiunile 퐿 = 220 cm. 푙 = 150 cm, ℎ = 60 cm. Se umple jumătate de piscină. a) Câți litri de apă avem în piscină?
b) Scufundăm în piscină o piatră paralelipipedică având dimensiunile 44, 50, 20 cm. Cu cât crește nivelul apei în piscină? Proposed by Gheorghe Căiniceanu-Romania
V.15. a) Prove that )1(
11
11
nnnn
. b) Write the fraction 2010
1 as a sum of 2011 subunit
fractions with the numerator equal to 1.
Proposed by D. M. Bătinețu-Giurgiu, Neculai Stanciu-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
34 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
V.16. Let yx, , z be nonzero natural numbers. Find )min( zxyzxyzyxA and a triplet ),,( zyx such that 2015A . Is A a prime number?
Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania V.17. Prove that exists an infinity of triplets ),,( zyx of natural numbers which verify the relationship cba zyx , where 1),( cab .
Proposed by D.M. Bătineţu-Giurgiu,Neculai Stanciu-Romania V.18. Find the prime numbers a b c such that: 2101 121 15 2018a b c .
Proposed by Marin Chirciu-Romania
V.19. Prove that the number 2 45 n can be written of a sum of three nonzero perfect squares, for any natural number n. Proposed by Marin Chirciu-Romania
V.20. If ba, are nonzero digits in ten base, such that 10 ba , and Nc such that 25 cab , then prove that the number cba 20152015 is a perfect square.
Proposed by D.M. Bătineţu-Giurgiu, Neculai Stanciu-Romania V.21. Prove that,if ,a b and c are real numbers such that acb )(2 , bac )(2 and
cba )(2 , then: (i) 0 cba ; (ii) 0 cabcab .
Proposed by D.M.Bătinețu-Giurgiu,Neculai Stanciu-Romania
V.22. Let be the number 2015 1 2 3 ... 2014A . Prove that the number A can be written
as a sum of five nonzero distinct perfect squares. Proposed by Marin Chirciu-Romania
V.23. Let be the number 111 1 2 3 ... 110A . Prove that the number A can be written as a
sum of five perfect squares, nonzero distinct. Proposed by Marin Chirciu-Romania
V.24. Prove that if 7 ab and7 abc then 7 a cba . Proposed by Marin Chirciu-Romania
V.25. Prove that: 340 221 3572 3 2 . Proposed by Marin Chirciu-Romania
V.26. Let n be a nonzero natural number. Solve the equation:
21 2 3 ... 2 1 2 1nx nx nx nx n n .
Proposed by Marin Chirciu-Romania
V.27. Let , , ,a b x y be nonzero natural numbers such that 2 1a b a and ,2 1 1b a .
Prove that 1ax a y se divide cu 2 1b a a if and only if
2 1bx a b y se divide cu 2 1b a a .Proposed by Marin Chirciu-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
35 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
V.28. Find the natural numbers pairs for which the sum between their product and their difference
is 2005. Proposed by Marin Chirciu-Romania
V.29. Let A N be a set having simultaneously the properties:
(1)1 A , (2)If x A then 3x A , (3)If 4x A then x A .Prove that 0,6,180,182 A .
Proposed by Marin Chirciu-Romania
V.30. Prove that 2006n it can be written as a sum of three perfect nonzero, distinct squares, for any *n N . Proposed by Marin Chirciu-Romania
V.31. Prove that 2001n can be written as a sum of three nonzero, distinct perfect squares, for any *n N . Proposed by Marin Chirciu-Romania
V.32. Prove that 30n can be written as a sum of three nonzero distinct perfect squares, for any *n N . Proposed by Marin Chirciu-Romania
V.33. Prove that 21n can be written as a sum of three distinct nonzero perfect squares for any *n N . Proposed by Marin Chirciu-Romania
V.34. Prove that 35n can be written as a sum of three nonzero distinct perfect squares, for any *n N . Proposed by Marin Chirciu-Romania
V.35. Dacă 푎, 푏, 푐 ∈ ℕ∗ astfel ca , ,a b a c b c să fie simultan pătrate perfecte, arătaţi că cel
puţin două din numerele a, b, c sunt pare. Proposed by Dan Nedeianu-Romania
6-CLASS-STANDARD
VI.1. Suma cifrelor unui număr natural este 푦, iar produsul cifrelor este un număr 푧 (nenul). Să se afle 푥 știind că 푥 + 푦 + 푧 = 139. Proposed by Constantin Ionică-Romania
VI.2. Aflați numerele naturale 푎, 푏, 푐 știind că: = = și 푎 + 푏 + 3 = 2푐
Proposed by Constantin Ionică-Romania
VI.3. Aflați cea mai mică și cea mai mare valoare a sumei (푎 + 푏 + 푐), unde 푎, 푏, 푐 ∈ ℕ și verifică relația: = = . Proposed by Constantin Ionică-Romania
VI.4. Numerele naturale nenule 푎, 푏, 푐 sunt direct proporționale cu numerele 2,3 respectiv 4, iar numerele 푐 și 푑 sunt invers proporționale cu numerele și . a. Știind că (푏 + 푐)(푐 + 푑) = 63푎, calculați numerele 푎, 푏, 푐,푑. b. Cât la sută din (푏 + 푑) reprezintă (푎 + 푐)?
Proposed by Constantin Ionică-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
36 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
VI.5. Fie 푎 ,푎 ,푎 , … ,푎 o rearanjare a numerelor 1,2,3, … ,2017. Să se arate că produsul (푎 − 1) ⋅ (푎 − 2) ⋅ (푎 − 3) ⋅ … ⋅ (푎 − 2017) este număr par.
Proposed by Constantin Ionică-Romania
VI.6. Fie 푎, 푏, 푐 ∈ ℤ astfel încât 9푎 + 6푏 − 5푐 = 0. Să se arate că: 푐(푏 − 푎) se divide cu 15.
Proposed by Constantin Ionică-Romania VI.7. Să se determine numerele întregi 푥 și 푦 care verifică relația: 22푥 + 39푦 = 1001.
Proposed by Constantin Ionică-Romania
VI.8. Numerele 푎, 푏, 푐 ∈ ℕ∗ sunt invers proporționale cu numerele 푏 + 푐, 푐 + 푎,푎 + 푏. Alfați numerele 푎, 푏, 푐 dacă ele verifică relația: 푎 푏 + 푏 푐 + 푐 푎 = 16(푎푏 + 푏푐 + 푐푎).
Proposed by Constantin Ionică-Romania
VI.9. Determinați 푥,푦 ∈ 푁∗ din ecuația ( ) = , 푥 > 2
Proposed by Constantin Ionică-Romania
VI.10. Aflați 푛 ∈ ℕ încât prin împărțirea numerelor 3399, 2197 și 1287 la numerele 4푛, 3푛 respectiv 2푛 se obțin resturile 39, 37 și 23. Proposed by Constantin Ionică-Romania
VI.11. Fie unghiul 퐴푂퐵 cu măsura de 60° și semidreapta [푂퐶 în interiorul ∢퐴푂퐵 încât 푚(∢퐴푂퐶) =20°. Se consideră [푂퐷 ⊥ [푂퐴, [푂퐷 și [푂퐴 în același semiplan față de [푂퐵, iar [푂퐸 este opusă [푂퐶 și [푂퐹 este bisectoarea ∢퐵푂퐸. a. Calculați 푚(∢퐷푂퐸) și 푚(∢퐵푂퐹) b. Arătați că [푂퐸 este bisectoarea ∢퐷푂퐹 c. Dacă [푂퐷ʹ este opusă [푂퐷 cercetați dacă ∢퐵푂퐶 ≡ ∢퐹푂퐷ʹ
Proposed by Constantin Ionică-Romania
VI.12. Să se arate că oricare ar fi patru numere de forma 7 ( ) + 8 ( ) + 9 ( ) cu 푥,푦, 푧 ∈ ℕ∗ există cel puțin două numere a căror diferență se divide cu 10.
Proposed by Constantin Ionică-Romania
VI.13. Fie 훥퐴퐵퐶 cu 퐴퐵 = 9 cm, 퐴퐶 = 15 cm, 퐵퐶 = 18 cm și punctele 푀 ∈ [퐵퐶],푁 ∈ [퐵퐶],
푀 ∈ (퐵푁) astfel încât = și = , iar punctele 퐸 și 퐹 sunt mijloacele segmentelor [퐴푀], respectiv [퐴푁]. a. Calculați lungimile segmentelor [퐵푀], [푀푁], [푁퐶] b. Arătați că ∢퐴퐹퐵 ≡ ∢퐴퐸퐶.
Proposed by Constantin Ionică-Romania
VI.14. In 1st of January, 2015, Nicu borrows 50€ from his friends Mitică, Titu and Marius. He agrees to pay the whole borrowed sum at 31 December 2015. The annual interest rate for the three credits is 0,8%, 1% respectively 1%. After calculus, he is going to pay an interest
rate of 0,96% to the whole borrowed sum. If the borrowed sum from Titu is 211 bigger than
the one borrowed from Mitică, how much does Nicu borrowed from each of his friends? Proposed by D.M.Bătinețu-Giurgiu,Neculai Stanciu-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
37 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
VI.15. Solve for integers: 281 9 8 47x xy y . Proposed by Marin Chirciu-Romania
VI.16. Prove that: if ,a b and c are real numbers such that acb )(2 , bac )(2 and
cba )(2 , then, 0)( abcbaab .
Proposed by D.M.Bătinețu-Giurgiu,Neculai Stanciu-Romania
VI.17. Prove that if abc 0 and cbba )(3 , then b
aca
cbc
ba 222222
.
Propsed by D.M.Bătinețu-Giurgiu,Neculai Stanciu-Romania
VI.18. Prove that if ,,,,,,, zyxdcba and t are nonzero integers such that
tzczybyxa ,, and xtd , then the numbers adbccdab , and
bdac cannot be simultaneous prime numbers. Proposed by D.M.Bătinețu-Giurgiu,Neculai Stanciu-Romania
VI.19. If a and b are real positive numbers and 1x x a x a , compute x x .
Proposed by Marin Chirciu-Romania
VI.20. We consider the nonzero natural numbers n and a . Prove that the following is a fraction in
its lowest terms 1
22 2 1
n
n
aa
. Proposed by Marin Chirciu-Romania
VI.21. Let be *nN , n given. Solve in integers the system of equations: 3 13 2
xy z nx yz n
Proposed by Marin Chirciu-Romania
VI.22. Solve in real numbers the system of equations:
111
xy yyz zzx x
Proposed by Marin Chirciu-Romania
VI.23. Solve in integers the equation: 12 2 3 1 0x xx x
Proposed by Marin Chirciu-Romania
VI.24. Prove that 13 divides 8 121 8n , where n N . Proposed by Marin Chirciu-Romania
VI.25. Prove that 17 divides 8 181 64n , where n N . Proposed by Marin Chirciu-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
38 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
VI.26. If ,a bZ such that 25 12a b is a multiple of 11, then 2 3a b is multiple of 11.
Proposed by Marin Chirciu-Romania
VI.27. Let be 0, 0x y such that 2 2 2x y . Prove that :13
xyxy x y
.
Proposed by Marin Chirciu-Romania
VI.28. Let be ,x y prime given numbers and the sets
, , 1A xa a B ya b C xy c c N N N .Find the set A B C .
Proposed by Marin Chirciu-Romania
VI.29. Prove that if ,a b Z such that 29 16a b is a multiple of 13, then 6 7a b is also a multiple
of 13. Mutually is true? Proposed by Marin Chirciu-Romania
VI.30. Prove that the equation: 2 2 2007x y doesn’t have solutions in integers.
Proposed by Marin Chirciu-Romania
VI.31. Let x and y be natural numbers such that: 3 349 49 2 7x y x y . Prove that 7 7x y it
divides with 181. Proposed by Marin Chirciu-Romania
VI.32. Let be , , ,a b x yZ . Prove that if 2 2a b divides ax by , then 2 2a b divides 2 2x y .
Proposed by Marin Chirciu-Romania
VI.33. Se consideră, în jurul punctului O , unghiurile adiacente ∢퐴 푂퐴 ,∢퐴 푂퐴 , … ,∢퐴 푂퐴
, toate cu măsurile egale cu 1 și o mulțime B cu 7 elemente, {1, 2,3,..., 21}B .
Să se arate că există două unghiuri ∢퐴 푂퐴 ,∢퐴 푂퐴 , cu , , ,m n p q B , care au aceeași
bisectoare. Proposed by Dan Nedeianu-Romania
7-CLASS-STANDARD
VII.1. Să se arate că:
푥 − 2푥 + 122푥 − 1
+푦 − 4푦 + 148
푦 − 2+푧 − 6푧 + 178
푧 − 3≥ 72,∀푥 > 1,푦 > 2,푧 > 3
Proposed by Constantin Ionică-Romania
VII.2. Determinați 푥,푦 ∈ 푍 care verifică relația: 푥푦 + 1 = 3푥 + 2푦.
Proposed by Constantin Ionică-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
39 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
VII.3. Să se arate că dacă 푥푦푧 = 1, atunci − − = 1, 푥,푦, 푧 ∈ ℝ∗
Proposed by Constantin Ionică-Romania
VII.4. Determinați mulțimea 퐴 = 푥 ∈ 푁|푥 = 푎 + 푏 + 푐, = = ∈ 푁
Proposed by Constantin Ionică-Romania
VII.5. Determinați numerele naturale prime diferite 푎, 푏, 푐 care verifică relația:
푎 + 푏 + 푐 = 384
Proposed by Constantin Ionică-Romania
VII.6. If a, b, c ≥ 0, a + b + c = ab + bc + ca then:
(1 + a)(1 + b)3(1 + a )
+(1 + b)(1 + c)
3(1 + b )+
(1 + c)(1 + a)3(1 + c ) ≤ 1 + a + b + c
Proposed by Daniel Sitaru-Romania
VII.7. If a, b, c, d > 0, a + b + c + d = 4 then :
(2a + b + c)(2b + c + d)(2c + d + a)(2d + a + b)(a + b + c + d) ≤ 16
Proposed by Daniel Sitaru-Romania
VII.8. If 푎, 푏, 푐 > 0 then:
(∑푎 푏 )(∑푎 푏 ) ∑ ∑ ≥ (∑푎)(∑푎 ) ∑ ∑
Proposed by Daniel Sitaru-Romania
VII.9. If 푥,푦, 푧 > 0 then:
(푥 + 푦 + 푧)∑(푥 + 푦) + 2(푥 + 푦)(푦 + 푧)(푧 + 푥)4(푥 + 푦 + 푧) ≥
1327
Proposed by Daniel Sitaru-Romania
VII.10. Prove that if 푥,푦, 푧 > 0 then:
푥푦
+ 2푦푧
+ 3푧푥≤ 6
푥푦
+2푦푧
+3푧푥
Proposed by Daniel Sitaru-Romania
VII.11 If 푎, 푏, 푐 > 0,푎 + 푏 + 푐 = 3 then: ∑푎(푎 + 1)(푎 + 2)(푎 + 3) ≥ 72
Proposed by Daniel Sitaru-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
40 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
VII.12. If a, b, c, d, e > 0,푎 + 푏 + 푐 + 푑 + 푒 = 5 then:
b + c + d + e2a + b + c + d + e
≥103
Proposed by Daniel Sitaru-Romania
VII.13. If a, b, c > 0 then:
4ab−
ab
+ 4bc−
bc
+ 4ca−
ca
≤ 9
Proposed by Daniel Sitaru-Romania
VII.14. If 푥,푦, 푧 > 0;푥 + 푦 + 푧 = 1 then:
푥푧
+푦푥
+푧푦
+ 2푥푦푧
+푦푧푥
+푧푥푦
≥ 3
Proposed by Daniel Sitaru-Romania
VII.15. If 푥,푦, 푧 > 0, 푥 + 푦 + 푧 = 1 then:
푥푧
+푦푥
+푧푦
+푥푦
+푦푥
+푦푧
+푧푦
+푧푥
+푥푧≥ 7
Proposed by Daniel Sitaru-Romania
VII.16. Prove that if 푎, 푏, 푥,푦, 푧 ∈ (0,∞) then:
푦푧(푎 푦 + 푏 푧)푥
+푧푥(푎 푧 + 푏 푥)
푦+푥푦(푎 푥 + 푏 푦)
푧≥
23푎푏(푥 + 푦+ 푧)
Proposed by D. M. Bătinețu – Giurgiu; Neculai Stanciu – Romania-RMM Summer Edition-2016
VII.17. Prove that if: 풂,풃, 풄 > 0;푎 + 푏 + 푐 = 3 then:
풂ퟏ풃ퟑ
+ퟏ풄ퟑ
≥ퟏퟖ
풂ퟑ + 풃ퟑ + 풄ퟑ
Proposed by Daniel Sitaru – Romania-RMM Summer Edition 2016
VII.18. If 푎, 푏, 푐 are the length’s sides in any triangle the following relationship doesn’t holds:
푎푏
+푏푐
+푐푎
=23푏푎
+푐푏
+푎푐
Proposed by Redwane El Mellas – Morroco-RMM Summer Edition 2016
VII.19. Să se demonstreze că triunghiul cu laturile 푎, 푏, 푐 pentru care:
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41 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
푎 푏 + 푏 푐 + 푐 푎 = 푎푏 + 푏푐 + 푐푎 și 푎푏푐 = 1 este isoscel.
Proposed by Carmen Victorița – Chirfot – Romania
VII.20. Să se demonstreze că triunghiul cu laturile 푎, 푏, 푐 pentru care
푎 푏 + 푏 푐 + 푐 푎 = 푎푏 + 푏푐 + 푐푎 și 푎푏푐 = 1 este isoscel.
Proposed by Carmen Victorița – Chirfot – Romania
VII.21. Fie ≥ 3 ,푛 ∈ ℕ și o mulțime A formată din n numere raționale cu suma egală cu 1. Dacă
suma oricăror 1n elemente din A este strict pozitivă, să se determine cea mai mică valoare întreagă a celui mai mic element al lui A . Proposed by Dan Nedeianu-Romania
8-CLASS-STANDARD
VIII.1. Aflați 푥,푦 ∈ 푍 care verifică relația: 푥 − 3푥 = 푦 − 5푦
Proposed by Constantin Ionică-Romania
VIII.2. Rezolvați în mulțimea numerelor reale ecuația: |2푥 + 1| = |3푥 + 2|
Proposed by Constantin Ionică-Romania
VIII.3. Să se arate că dacă 푎 + 푏 − 6푎 + 12푏 + 29 = 0, atunci 푎 > 푏.
Proposed by Constantin Ionică-Romania
VIII.4. Fie 푎, 푏, 푐 numerele reale pozitive și 푎 ⋅ 푏 ⋅ 푐 = 1. Să se arate că
1 +1푎
1 +1푏
1 +1푐
≥ 8
Proposed by Constantin Ionică-Romania
VIII.5. Să se determine mulțimea: 퐴 = (푥,푦) ∈ 푍 × 푁|√4푥 + 2017 + √푥 + 505 = 푦
Proposed by Constantin Ionică-Romania
VIII.6. Să se determine valoarea maximă a expresiei: 퐸(푥,푦) = , 푥,푦 ∈ 푁∗
Proposed by Constantin Ionică-Romania
VIII.7. If a, b, c > 0, a + b + c = abc then:
4(a + b)(a + c)(b + c) +
4(b + c)(b + a)(c + a) +
4(c + a)(c + b)(a + b) ≤ 3 + a + b + c
Proposed by Claudia Nănuți,Daniel Sitaru-Romania
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VIII.8. If x, y, z, t, a, b, c ∈ (0,∞), xyzt = a then: 3a∑ ≥ 4 a + a
Proposed by Claudia Nănuți,Daniel Sitaru-Romania
VIII.9. If 푀 ∈ 퐼푛푡(훥퐴퐵퐶) then: 푀퐴+ 푀퐵 + 푀퐶 ≥
Proposed by Claudia Nănuți,Daniel Sitaru-Romania
VIII.10. If ba, and c are the dimensions of and rectangular parallelepiped and d is its diagonal
prove that: ac
cb
bad
333
.
Proposed by D.M.Bătinețu-Giurgiu,Neculai Stanciu-Romania VIII.11. Let ba, and c be the lengths of the edges of a rectangular parallelepiped. Prove
that: Nmncbacbacba mnmnmnmmmnnn
,,
3))((
, if and only if the
rectangular parallelepiped is a cube. Proposed by D.M.Bătinețu-Giurgiu, Neculai Stanciu-Romania
VIII.12. Solve in the set RR the system of equations:
4
1322
33
xyyxyx
.
Proposed by D.M.Bătinețu-Giurgiu, Neculai Stanciu-Romania
VIII.13. Solve in *R the system of equations:
361111
11
xyzzyx
xyx
Proposed by Roxana Mihaela Stanciu,Neculai Stanciu-Romania VIII.14. Find all triplets (푚,푛,푝) where 푚,푛 are two natural numbers and 푝 is a prime number, satisfying the equation: 푚 = 4(푝 − 1).
Proposed by Nguyen Viet Hung – Vietnam-RMM Autumn Edition 2016
VIII.15. Prove that if 푥,푦, 푧 > 0, 푥푦푧 = 8 then: 푥 + 푦 + 푧 ≥ 2푥 푦 + 푧 + 2푦√푧 + 푥 + 2푧 푥 + 푦
Proposed by Iuliana Trașcă – Romania-RMM Autumn Edition 2016
VIII.16. If 푥,푦, 푧 > 0, then prove that: (푥 푦 + 푦 푧 + 푧 푥 )( ) +
( )+ ( ) ≥
Proposed by D. M. Bătinețu – Giurgiu, Neculai Stanciu – Romania-RMM Autumn Edition 2016
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VIII.17. Să se determine numerele reale pozitive 푎, 푏, 푐 pentru care: 푎 + 푏 + 푐 = 푎 + 푏 + 푐 , știind că 푎푏푐 = 1. Proposed by Carmen Victorița – Chirfot – Romania
VIII.18. Să se determine numerele reale pozitive 푎, 푏, 푐 pentru care 푎 + 푏 + 푐 = 푎 + 푏 + 푐 , știind că 푎푏푐 = 1. Proposed by Carmen Victorița – Chirfot – Romania
VIII.19. Să se determine numerele reale pozitive 푎, 푏, 푐 pentru care 푎 + 푏 + 푐 = 푎 + 푏 + 푐 , știind că 푎푏푐 = 1. Proposed by Carmen Victorița – Chirfot – Romania
VIII.20. Să se determine numerele reale pozitive 푎, 푏, 푐 pentru care 푎 푏 + 푏 푐 + 푐 푎 =(푎푏 + 푏푐 + 푐푎), știind că 푎푏푐 = 1. Proposed by Carmen Victorița – Chirfot – Romania
VIII.21. Să se determine numere reale pozitive 푎, 푏, 푐 pentru care:
푎 푏 + 푏 푐 + 푐 푎 = 푎푏 + 푏푐 + 푐푎, știind că 푎푏푐 = 1.
Proposed by Carmen Victorița – Chirfot – Romania
VIII.22. Să se demonstreze că triunghiul cu laturile 푎, 푏, 푐 pentru care:
푎 푏 + 푏 푐 + 푐 푎 = 푎푏 + 푏푐 + 푐푎 și 푎푏푐 = 1 este isoscel.
Proposed by Carmen Victorița – Chirfot – Romania
VIII.23. Fie 푛 ∈ ℕ∗,푛 ≥ 3, n impar. Dacă 1 2 3, , ,..., na a a a sunt numere întregi impare, să se arate
că există푏 ∈ ℕ∗ pentru care 2 2 2 2 21 2 3 ... na a a a b să fie pătrat perfect.
Proposed by Dan Nedeianu-Romania
9-CLASS-STANDARD
IX.1. Let 푎, 푏 and 푐 be the side lengths of a triangle 퐴퐵퐶 with incenter 퐼. Prove that:
1퐼퐴
+1퐼퐵
+1퐼퐶
≥ 31푎
+1푏
+1푐
Proposed by George Apostolopoulos – Messolonghi – Greece
IX.2. Let 퐴퐵퐶 be a triangle with inradius 푟 and circumradius 푅. Prove that
(a) √ ≤ + + ≤ √ , (b) 9 ≤ cos + cos + cos ≤
Proposed by George Apostolopoulos – Messolonghi – Greece
IX.3. Let ABC be a triangle with circumradius R and inradius r. Prove that
27rR
≤ sin A + sin B + sin C ≤278
1 −rR
Proposed by George Apostolopoulos – Messolonghi – Greece
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IX.4. Let 퐴퐵퐶 be a triangle with circumradius 푅 and inradius 푟. Prove that
4 ≤ sec퐴2
+ sec퐵2
+ sec퐶2≤
2푅푟
Proposed by George Apostolopoulos – Messolonghi – Greece
IX.5. Let 푎, 푏, 푐 be the lengths of the sides of a triangle with perimeter 3 and inradius 푟. Prove that:
288푟 ≤(푎 + 푏)푎 + 푏
+(푏 + 푐)푏 + 푐
+(푐 + 푎)푐 + 푎
≤2푟
Proposed by George Apostolopoulos – Messolonghi – Greece
IX.6. Let 푎, 푏, 푐 be the side lengths of a triangle 퐴퐵퐶 with incentre 퐼, circumradius 푅 and inradius 푟. Prove that:
√퐴퐼푎
+√퐵퐼푏
+√퐶퐼푐
≤√22⋅√푅 + 푟
푟
Proposed by George Apostolopoulos – Messolonghi – Greece
IX.7. Let 푎, 푏, 푐 be the side lengths of a triangle 퐴퐵퐶 with inradius 푟 and circumradius 푅. Prove that (푏 + 푐 ) sin 퐴 + (푐 + 푎 ) sin 퐵 + (푎 + 푏 ) sin 퐶 ≤ (3푅 − 16푟 ).
Proposed by George Apostolopoulos – Messolonghi – Greece
IX.8. Let 푥,푦, 푧 be positive real numbers with 푥푦푧 = 1. Prove that:
√푥 + 1 + 푦 + 1 + √푧 + 1푥 + 푦 + 푧
≤ √2
Proposed by George Apostolopoulos – Messolonghi – Greece
IX.9. Let 푥,푦, 푧 > 0 be positive real numbers. Then:
1푥 + 푦
+1
푦 + 푧+
1푧 + 푥
≥4 3푥푦푧(푥 + 푦 + 푧)
(푥 + 푦)(푦 + 푧)(푧 + 푥)
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Skopje
IX.10. Let 푥,푦, 푧 > 0 be positive real numbers and 퐹 be the area of the triangle 퐴퐵퐶. Then:
푎푏 + 푏푐 + 푐푎 ≥ 8√3퐹.
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Skopje
IX.11. Let 푡,푢, 푣, 푥,푦, 푧 > 0 be positive real numbers, 푡 ≥ max{푢, 푣} and 푆 = 푥 + 푦 + 푧. Then:
푡푆 − 푢푦 − 푣푧푢푦 + 푣푧
푎 +푡푆 − 푢푧 − 푣푥푢푧 + 푣푥
푏 +푡푆 − 푢푥 − 푣푦푢푥 + 푣푦
푐 ≥4(3푡 − 푢 − 푣)√3푆
푢 + 푣
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Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Skopje
IX.12. Let 푥,푦, 푧 > 0 be positive real numbers and 퐹 be the area of the triangle 퐴퐵퐶. Then:
푥 + 푦푧
푎 +푦 + 푧푥
푏 +푧 + 푥푦
푐 ≥ 8√3퐹
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Skopje
IX.13. Let 푥,푦, 푧 > 0 be positive real numbers and 퐹 be the area of the triangle 퐴퐵퐶 with
circumradius 푅. Then: sin + sin + sin ≥ √
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Skopje
IX.14. Let 푥,푦, 푧 > 0 be positive real numbers and 퐹 be the area of the triangle 퐴퐵퐶. Then:
(푦 + 푧) 푎푥
+(푧 + 푥) 푏
푦+
(푧 + 푥) 푐푧
≥ 64퐹
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Skopje
IX.15. Let 푚, 푥,푦, 푧 > 0 be positive real numbers and 퐹 be the area of the triangle 퐴퐵퐶. Then:
푎 푥(푦 + 푧) +
푏 푦(푧 + 푥) +
푐 푧(푥 + 푦) ≥
2
√3퐹
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Skopje
IX.16. If 푥,푦, 푧 > 0 then in 훥퐴퐵퐶 the following relationship holds:
(2푥 + 푦)푎푦 + 2푧
+(2푦 + 푧)푏푧 + 2푥
+(2푧 + 푥)푐푥 + 2푦
≥ 4√3푆
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
IX.17. If 푥,푦, 푧 > 0 then in any 훥퐴퐵퐶 the following relationship holds:
푦푧푎푥
+ 푧푥푏푦
+ 푥푦푐푧
≥ 4√3푆
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
IX.18. If 푥,푦, 푧 > 0 then in any 훥퐴퐵퐶 the following relationship holds:
푥푦 + 푧
tan퐴2
+푦
푧 + 푥tan
퐵2
+푧
푥 + 푦tan
퐶2≥ 2−
(4푅 + 푟)2푠
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
IX.19. In 훥퐴퐵퐶 the following relationship holds:
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푎푟 + ℎ
+푏
푟 + ℎ+
푐푟 + ℎ
≥3√3
2
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
IX.20. In 훥퐴퐵퐶 the following relationship holds:
푎푏푅푏 + 푟푐
+푏푐
푅푐 + 푟푎+
푐푎푅푎 + 푟푏
≥4√3푆푅 + 푟
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
IX.21. In 훥퐴퐵퐶 the following relationship holds:
푟 푟푟 + 푟
+푟 ⋅ 푟푟 + 푟
+푟 ⋅ 푟푟 + 푟
≥푠2
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
IX.22. If 푎, 푏, 푐 > 0, 푎 + 푏 + 푐 + 푎푏푐 = 4, then: |(푎푏 − 1)(푏푐 − 1)(푎푐− 1)| ≤ 1− 푎 푏 푐
Proposed by Vadim Mitrofanov-Ukraine
IX.23. In any ∆퐴퐵퐶 the following relationship holds: ≥ 푎 sin + 푏 sin + 푐 sin ≥ .
Proposed by Vadim Mitrofanov-Ukraine
IX.24. In any ∆퐴퐵퐶 the following relationship holds:
cos퐴4
+ cos퐵4
+ cos퐶4≤√3 − 1
4√6푡 +
11√3 + 378√6
, 푡 =푟푅
Proposed by Vadim Mitrofanov-Ukraine
IX.25. In any ∆퐴퐵퐶 the following relationship holds: 푟 sin퐴 + 푟 sin퐵 + 푟 sin 퐶 ≤ 2√3(2푅 − 푟).
Proposed by Vadim Mitrofanov-Ukraine
IX.26. Dacă [푥] + [2푥] + [3푥] + ⋯+ [푛푥] = ( ) 푥 pentru orice număr natural nenul 푛, atunci demonstrați că numărul 푥 este întreg. ([푦] reprezintă partea întreagă a numărului real 푦)
Proposed by Nicolae Papacu – Romania
IX.27. Rezolvați în numere reale sistemul [푥] ⋅ [푦] = 푥 + 푦[푥] + [푦] = [푥푦].
Proposed by Nicolae Papacu – Romania
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IX.28. Rezolvați în numere reale sistemul [푥] ⋅ [푦] = 푥 + 푦[푥] + [푦] = 푥푦 . ([푎] reprezintă partea întreagă a
numărului real 푎) Proposed by Nicolae Papacu – Romania
IX.29. Se consideră parabola de ecuație 푦 = 3푥 − 2푥 − 1 și dreapta de ecuație 푦 = 푚푥 − 2, unde 푚 este parametru real. Să se determine valoarea lui 푚 pentru care dreapta este tangentă la parabolă. Determinați coordonatele punctului de tangentă și reprezentați grafic.
Proposed by Gheorghe Căiniceanu-Romania
IX.30. In 훥퐴퐵퐶;퐺 - centroid, 푑 = 푑(퐺 ,퐵퐶);푑 = 푑(퐺,퐴퐶), 푑 = 푑(퐺,퐴퐵); 푥,푦 > 0. If 푀 ∈ 퐵퐶;푁 ∈ 퐶퐴;푃 ∈ 퐴퐵 then: ∑ ⋅ ⋅ + (푥 + 푦) ⋅ ∑∑ ≥ 2(푥 + 푦) √3 + 1
Proposed by D.M. Bătinețu – Giurgiu, Dan Nănuți – Romania
IX.31. In 훥퐴퐵퐶 the following relationship holds: + + ≥ 81푟
Proposed by D.M. Bătinețu – Giurgiu, Dan Nănuți – Romania
IX.32. If 푥,푦, 푧, 푡 > 0 then in 훥퐴퐵퐶 the following relationship holds:
푡 + 푥푦 + 푧
푎 +푥 + 푦 + 푧
푡푏 +
푡 + 푦 + 푧푥
푐 ≥ 8√3푆
Proposed by D.M. Bătinețu – Giurgiu, Dan Nănuți – Romania
IX.33. If 푎, 푏, 푐,푚,푛 > 0 then:
푚푎 + 푛푏푐
+푚푏 + 푛푐
푎+푚푐 + 푛푎
푏+ (푚 + 푛) ⋅
푎푏 + 푏푐 + 푐푎푎 + 푏 + 푐
≥ 4(푚 + 푛)
Proposed by D.M. Bătinețu – Giurgiu, Dan Nănuți – Romania
IX.34. Să se determine numerele reale pozitive 푎, 푏, 푐 astfel încât:
푎 푏 + 푏 푐 + 푐 푎 + 푎 + 푏 + 푐 = 6 și 푎푏푐 = 1.
Proposed by Carmen Victorița – Chirfot – Romania
IX.35. Solve in positive real numbers: + = 128(푥 + 푦 )
4푥 − 3푦 =
Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
IX.36. Let 푎, 푏, 푐 > 0 such that: 푎 + 푏 + 푐 = 3. Find the minimum of the expression:
푃 =푎
+ 5푏푐+
푏
+ 5푐푎+
푐
+ 5푎푏+
(푎 + 푏)(푏 + 푐)(푐 + 푎)16
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Proposed by Hoang Le Nhat Tung – Hanoi – Vietnam
IX.37. Let 푎, 푏, 푐 be positive real numbers such that: 푎푏 > 6
+ 3푏 + = + . Find the minimum
value of the expression: 푃 = 3푎 + 2푏 + 푐
Proposed by Do Quoc Chinh – Vinh Phuc – Vietnam
IX.38. Let 푥,푦, 푧 be positive real numbers satisfying 푥 + 푦 + 푧 = 3. Find the maximum value of the expression:푃 = + +
√
Proposed by Do Quoc Chinh – Vinh Phuc – Vietnam
IX.39. Să se determine numerele reale pozitive 푎, 푏, 푐 pentru care:
푎 푏 + 푏 푐 + 푐 푎 = 9(푎푏 + 푏푐 + 푐푎), știind că 푎푏푐 = 1.
Proposed by Carmen Victorița – Chirfot – Romania
IX.40. Prove that the quadrilateral 퐴퐵퐶퐷 is a parallelogram if and only if there exist a point with the
property that any straight line passing through this point divides the quadrilateral in two equivalent
polygons. Proposed by Cristian Heuberger-Romania
IX.41. Se consideră triunghiul ABC , punctele M AB , N AC , O mijlocul lui BC , 'O
mijlocul lui MN , I centrul cercului înscris în ∆퐴퐵퐶. Dacă 퐴퐼 ∥ 푂푂′ ,să se determine valoarea
raportului BMCN
. Proposed by Dan Nedeianu-Romania
10-CLASS-STANDARD
X.1. Let 퐴퐵퐶 be a triangle with circumradius 푅 and inradius 푟. Prove that:
316
≤ cos 퐴+ cos 퐵 + cos 퐶 ≤ 6푟푅
−123
8⋅푟푅
+518
Proposed by George Apostolopoulos – Messolonghi – Greece
X.2. Let 퐴퐵퐶 be a triangle with inradius 푟 and circumradius 푅. Prove that:
2√3푟푅
≤∑ sin 퐴∑ sin 퐴
≤√34
푅푟
1 −푟푅
Proposed by George Apostolopoulos – Messolonghi – Greece
X.3. Let 푎, 푏, and 푐 be the side lengths of a triangle with inradius 푟. Prove that
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1푎 + 2푏 푐
+1
푏 + 2푐 푎+
1푐 + 2푎 푏
≤√36푟
Proposed by George Apostolopoulos – Messolonghi – Greece
X.4. Let 퐴퐵퐶 be an acute triangle with orthocentre 퐻. Prove that
퐴퐻 ⋅ 퐵퐻 + 퐵퐻 ⋅ 퐶퐻 + 퐶퐻 ⋅ 퐴퐻 ≤ 6푅푟, where 푅 and 푟 are the circumradius and inradius respectively of triangle 퐴퐵퐶.
Proposed by George Apostolopoulos – Messolonghi – Greece
X.5. Let 푥,푦, 푧 > 0 be positive real numbers. Then in triangle 퐴퐵퐶 with semiperimeter 푠 and inradius 푟 the following relationship holds:
푥푦 + 푧
cot퐴2
+푦
푧 + 푥cot
퐵2
+푧
푥 + 푦cot
퐶2≥ 18−
푠2푟
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Skopje
X.6. Let 푚 > 0 and 퐹 be the area of the triangle 퐴퐵퐶. Then:
푎푏 + 푐 +
푏푐 + 푎 +
푐푎 + 푏 ≥ 2√3퐹
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Skopje
X.7. Let 푟 , 푟 , 푟 be the exradii, ℎ ,ℎ ,ℎ the altitudes and 푚 ,푚 ,푚 the medians of a triangle 퐴퐵퐶 with semiperimeter 푠, circumradius 푅 and inradius 푟. Then:
푟ℎ 푚
+푟
ℎ 푚+
푟ℎ 푚
≥54푟
푠 − 푟 − 4푅푟
Proposed by D.M.Bătinețu-Giurgiu-Romania,Martin Lukarevski-Skopje
X.8. If 푥,푦, 푧 ∈ 0, then in 훥퐴퐵퐶 the following relationship holds:
tan 푥 + tan푦sin 푧
+tan 푧
sin푥 + sin푦푎 > 10√3푆
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
X.9. If 푚,푛 > 0 then in 훥퐴퐵퐶 the following relationship holds:
푚푠 + 푛푟 + 4푛푅푟 ≥ (3푚 + 푛)√3푆
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
X.10. If 푥,푦, 푧 ∈ (0,1) then:
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tan푦 ⋅ tan 푧(1− 푥 ) sin푥
1(푥 + 푦) >
27√38
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
X.11. If 푚 ≥ 0;푥,푦, 푧 > 0 then:
(푥 + 푦)(푥 + 2푦 + 푧) ≥
3푥푦푧(푥 + 푦 + 푧)2
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
X.12. If 푥,푦, 푧,푢, 푣 > 0 then:
(푥 + 푦) + (푦 + 푧)(푧 + 푥)(푥 + 푦) 푢(푦 + 푧) + 푣(푧 + 푥)
≥9
(푢 + 푣)(푥 + 푦 + 푧)
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
X.13. If 푥,푦, 푧,푢, 푣,푤 > 0 then:
(푥 + 푦)(푢 + 푣) + (푦 + 푧)(푣 +푤) + (푧 + 푥)(푤 + 푢) ≥ 4 3푥푦푧푢푣푤(푥 + 푦 + 푧)(푢 + 푣 +푤)
Proposed by D.M. Bătinețu – Giurgiu; Claudia Nănuți – Romania
X.14. In any ∆퐴퐵퐶 the following relationship holds:
퐿 퐿 + 퐿 퐿 + 퐿 퐿 ≤ (푅 + 푟) , where 퐿 the bases of the bisector of angle 퐴.
Proposed by Vadim Mitrofanov-Ukraine
X.15. In any ∆퐴퐵퐶 the following relationship holds: 2푎푚 ≥ 푏ℎ + 푐ℎ Proposed by Vadim Mitrofanov-Ukraine
X.16. In any ∆퐴퐵퐶 the following relationship holds: 6 1 + ≥ cos + cos + cos ≥( )
.
Proposed by Vadim Mitrofanov-Ukraine
X.17. In any ∆퐴퐵퐶 the following relationship holds: 4푡 − 2 ≥ + + , 푡 = .
Proposed by Vadim Mitrofanov-Ukraine
X.18. In any ∆퐴퐵퐶 the following relationship holds:
16푡 − 5 ≤ + + ≤ ( ) , 푡 = , 푛 ,푛 ,푛 the Nagel cevians.
Proposed by Vadim Mitrofanov-Ukraine
X.19. Rezolvați ecuația (8 + 45 + 75 ) + + = (2 + 3 + 5 )
Proposed by Nicolae Papacu –Romania
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X.20. Arătați că ecuația 퐴 + 퐴 = 2 7 + 퐴 + 퐴 are o infinitate de soluții.
Proposed by Nicolae Papacu –Romania
X.21. Fie 푝 un număr natural nenul și mulțimile 푀 = 푧 ∈ 푍[푖] 푅푒 ≥ 1 , unde:
푍[푖] = {푎 + 푖푏|푎, 푏 ∈ 푍}.a) Să se arate că 퐶푎푟푑푀 ≤ 푝 , egalitate pentru 푝 ∈ {1,2}.
b) Să se arate că pentru orice 푝 ≥ 2, mulțimea 푀 conține trei numere complexe care reprezintă afixele a trei puncte coliniare și trei numere care reprezintă afixele unui triunghi dreptunghic.
Proposed by Nicolae Papacu –Romania
X.22. If 푥, 푦, 푧 > 0 then in 훥퐴퐵퐶 the following relationship holds:
푥푟푦 + 푧 +
푦푟푧 + 푥 +
푧푟푥 + 푦 ≥
4푠 − (4푅 + 푟)2
Proposed by D.M. Bătinețu – Giurgiu, Claudia Nănuți – Romania
X.23. If 푥,푦 > 0;퐺 - centroid in 훥퐴퐵퐶;푑 = 푑(퐺 ,퐵퐶);푑 = 푑(퐺,퐴퐶),푑 = 푑(퐺,퐴퐵) then:
푤푥푑 + 푦푑
+∑푤 푤
(푥 + 푦)∑푑≥
6 1 + √3푥 + 푦
Proposed by D.M. Bătinețu – Giurgiu, Claudia Nănuți – Romania
X.24. If 푎, 푏, 푐 > 0;푎푏푐 = 1;푥,푦, 푧 ∈ (0,1) then:
푎푏푥(1 + 푥 )(푏 + 푐 ) +
푏푐푦(1 − 푦 )(푐 + 푎 ) +
푐푎푧(1− 푧 )(푎 + 푏 ) ≥
27√32(푎 + 푏 + 푐 + 푎 + 푏 + 푐)
Proposed by D.M. Bătinețu – Giurgiu, Claudia Nănuți – Romania
X.25. If 푥,푦, 푧 > 0 then in 훥퐴퐵퐶 the following relationship holds:
(푦 + 푧)푟푥
+(푧 + 푥)푟
푦+
(푥 + 푦)푟푧
≥ 2푠
Proposed by D.M. Bătinețu – Giurgiu, Claudia Nănuți – Romania
X.26. Let 훥퐴퐵퐶, with area = 훥. Show that: 2훥 ≤ 3푅(푚 + 푚 + 푚 )
Proposed by Mihalcea Andrei Ștefan - Romania
X.27. Să se arate că în orice tetraedru are loc inegalitatea 푚 + 푚 + 푚 + 푚 ≥ √ 푆, unde 푚 ,푚 ,푚 ,푚 sunt medianele tetraedrului, iar 푆 aria totală.
Proposed by Marian Ursarescu – Romania
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X.28. Fie 퐴퐵퐶퐷 un tetraedru. Notăm cu 푆 aria 훥 format de 퐴,퐵 și mijlocul muchiei opuse lui 퐴퐵. Să se arate că: 4(푆 + 푆 + 푆 + 푆 + 푆 + 푆 ) ≥ 푆 , unde 푆 este aria totală a tetraedrului.
Proposed by Marian Ursarescu – Romania
X.29. Să se arate că în orice 훥퐴퐵퐶 are loc inegalitatea:
푟푟 푟 (푟 + 푟 ) +
푟푟 푟 (푟 + 푟 ) +
푟푟 푟 (푟 + 푟 ) ≥
92푝
Proposed by Marian Ursarescu – Romania
X.30. Să se arate că în orice 훥퐴퐵퐶 are loc inegalitatea:
ℎℎ ℎ (ℎ + ℎ ) +
ℎℎ ℎ (ℎ + ℎ ) +
ℎℎ ℎ (ℎ + ℎ ) ≥
92푝
Proposed by Marian Ursarescu – Romania
X.31. Să se arate că în orice 훥 are loc inegalitatea: 퐴푙 + 퐵푙 + 퐶푙 ≤ 푝
Proposed by Marian Ursarescu – Romania
X.32. Fie 퐴퐵퐶 un 훥 ascuțitunghic și 퐴ʹ,퐵ʹ ,퐶 ʹ punctele în care înălțimile 훥 intersectează cercul
circumscris. Să se arate că: a. ʹ ʹ ʹ = 8 cos퐴 cos퐵 cos 퐶 b. 푆 ʹ ʹ ʹ ≤ 휎
Proposed by Marian Ursarescu – Romania
X.33. Să se arate că în orice tetraedru are loc inegalitatea: ∑⋅
≤ √ ⋅
Proposed by Marian Ursarescu – Romania
X.34. Fie 퐴퐵퐶 un triunghi și 퐼 , 퐼 , 퐼 centrele cercurilor exînscrise triunghiului 퐴퐵퐶. Să se arate că: 푆 + 푆 + 푆 ≥ 3푆 . Proposed by Marian Ursarescu – Romania
X.35. Să se arate că ∀푝,푛 ∈ ℕ∗, 푝 + 푝 − 1 + 푝+ 푝 + 1 + 1 ⋮ 4
Proposed by Marian Ursarescu – Romania
X.36. Să se determine numerele reale pozitive nenule 푎, 푏, 푐 pentru care 푎 + 푏 + 푐 = 푎 + 푏 + 푐 , știind că 푎푏푐 = 1. Proposed by Carmen Victorița – Chirfot – Romania
X.37. Să se determine numerele reale pozitive 푎, 푏, 푐 pentru care 푎 푏 + 푏 푐 + 푐 푎 = 푎푏 + 푏푐 + 푐푎, știind că 푎푏푐 = 1. Proposed by Carmen Victorița – Chirfot – Romania
X.38. Să se demontreze că există o infinitate de numere reale pozitive 푎, 푏, 푐 astfel încât 푎 푏 +푏 푐 + 푐 푎 = 푎 + 푏 + 푐 și 푎푏푐 = 1.
Proposed by Carmen Victorița – Chirfot – Romania
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X.39. Let 훥퐴퐵퐶 , with 퐴 ∈ (퐵퐶),퐵 ∈ (퐶퐴),퐶 ∈ (퐴퐵). Prove that:
푎(퐵 퐶 ⋅ 퐶 퐴+ 퐵 퐴 ⋅ 퐶 퐵) ≥ 푎푏푐 + 2 퐴 퐵 + 퐴 퐶
Proposed by Mihalcea Andrei Ștefan-Romania
X.40. Să se determine că nu există numerele reale pozitive 푎, 푏, 푐 astfel încât:
푎 푏 + 푏 푐 + 푐 푎 = 3(푎 + 푏 + 푐 ) și 푎푏푐 = 3√3.
Proposed by Carmen Victorița – Chirfot – Romania
X.41. Fie 퐴퐵퐶 un triunghi și cevianele 퐴퐴 ,퐵퐵 și 퐶퐶 concurente cu 푂. Notăm cu 푅 ,푅 ,푅 ,푅 ,푅 ,푅 razele cercurilor circumscrise triunghiurilor 푂퐵퐴 ,푂퐶퐴 ,푂퐶퐵 ,푂퐴퐵 ,푂퐴퐶 respective 푂퐵퐶 . Să se arate că : 푅 푅 푅 = 푅 푅 푅 .
Proposed by Marian Ursarescu – Romania
X.42. Să se rezolve sistemul: √푥 + 1 + 푥 푦 + 1 + 푦 = 1
2√ + 2 = 4
Proposed by Marian Ursarescu – Romania
X.43. Să se rezolve ecuaţia 푠푖푛 √ + 푐표푠 √ = √ , 푥 ∈ ℝ.
Proposed by Dan Nedeianu-Romania
11-CLASS-STANDARD XI.1. Prove that:
1 1 1 1푥 푦 푧 푡푥 푦 푧 푡푥 푦 푧 푡
+
1 1 1 1푥 푦 푧 푡푥 푦 푧 푡푦푧푡 푥푧푡 푥푦푡 푥푦푧
= 0
where 푥,푦, 푧, 푡 ∈ 푅∗.
Proposed by Chirfot Carmen – Victoriţa-Romania
XI.2. Let 푎, 푏, 푐 be positive real numbers. Find the minimmum of the expression:
푃 =( )
+( )
+( )
+
Proposed by Hoang Le Nhat Tung-Hanoi-Vietnam
XI.3. Solve the system of equation:
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⎩⎪⎨
⎪⎧
1√푎
+1√푏
+1√푐
= 3
푎푏
+푏푐
+푐푎
=푎 + 푏
2+
푏 + 푐2
+푐 + 푎
2
Proposed by Hoang Le Nhat Tung-Hanoi-Vietnam
XI.4 In 훥퐴퐵퐶 the following relationship holds:
2 (1− 8 ⋅ 2 ) + 4 ⋅ 2 1 − 2 + 2 ⋅ 2 (8− 2 ) > 0
Proposed by Iuliana Trașcă – Romania
XI.5. Let 푚,푛 be positive real numbers. Prove that:
1푚
+1푛
≤4034− 2015푚푚 + 2017
+4034 − 2015푛푛 + 2017
+푚 + 푛 + 2009
2
Proposed by Iuliana Trașcă – Romania
XI.6. Se consideră matricele 퐴,퐵 ∈ 푀 (ℝ) cu 퐴퐵 ≠ 퐵퐴. a) Dacă 퐴 = 퐵 demonstrați că 푡푟(퐴 ) = 푡푟(퐵 ), pentru orice număr natural nenul 푛. b) Dați exemplu de două matrice 퐴,퐵 ∈ 푀 (ℝ) cu 퐴퐵 ≠ 퐵퐴 astfel încât 푡푟(퐴 ) = 푡푟(퐵 ) și 퐴 ≠ 퐵 , pentru orice număr natural nenul 푛. Proposed by Nicolae Papacu –Romania
XI.7. Let − ≤ 푎 < 푏 < 푐 ≤ . Prove that:
sin
sin≤
(푏 − 푎) 1 + sin(푎 + 푐)
(푐 − 푎) 1 + sin(푎 + 푏)
Proposed by Mihalcea Andrei Ștefan - Romania XI.8. Let 푎, 푏, 푐,푑 ≥ 0, with 푎√푎 + 푏√푏 + 푐√푐 + 푑√푑 = 1. Prove that:
9 √1 + 푎푏 ≤ 38 + 푎 + 푏 + 푐 + 푑
Proposed by Mihalcea Andrei Ștefan - Romania XI.9. Let 푓:퐷 → ℝ, a convexe function and 푎, 푏, 푐 ∈ 퐷, with 푎 + 푏 + 푐 = 1. Prove that 푓(푎) + 푓(푏) + 푓(푐) ≥ 푓(푞) + 2푓 . Using this, you can prove that if 푎, 푏, 푐 ≥ 0,푎 + 푏 + 푐 = 1,
then: ∑ (푎 + 푏)(푎 + 푐) ≥ ( )
Proposed by Mihalcea Andrei Ștefan - Romania
XI.10. 훺 = lim → ( ) −!
( )! ⋯ ( )
!
;푛 ∈ ℕ∗. Find: 훺 = lim → ∑ 훺
Proposed by Daniel Sitaru -Romania
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XI.11. 퐴 = 0 11 1 ,퐴 = 푎 푏
푐 푑 ,푛 ∈ ℕ∗, 훺 = lim → . Prove that: 훺 < 1.
Proposed by Daniel Sitaru -Romania
XI.12. Find:
훺 = lim→
푒
Proposed by Daniel Sitaru -Romania
XI.13. Find: 훺 = lim → + + ⋯+ ⋅ ln푛
Proposed by Daniel Sitaru -Romania
XI.14. Find:
훺 = lim→
푛 tan휋푒푛!
5
Proposed by Daniel Sitaru -Romania
XI.15. Există matrici 퐴 ∈ 푀 (ℝ) astfel încât 233 2 tA A I A ?
Proposed by Dan Nedeianu-Romania XI.16. Find:
훺 = lim→
2‼√3‼ √5‼ ⋅ … ⋅ (2푛− 1)‼
Proposed by D.M. Bătinețu – Giurgiu, N. Stanciu – Romania
XI.17. Let 푎 , 푏 ∈ (0,∞) and 푛 ≥ 1, lim → (푎 − 푎 ) = 푎, and
푏 = lim→
√2! ⋅ √3! ⋅ … ∙ √푛!
find:
훺 = lim→
푎 ⋅ 푏푛 + 1
−푎 ⋅ 푏푛
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
XI.18. If 푎, 푏 ∈ ℕ,푎 < 푏,훺(푎, 푏) = lim → ∑( )( ), then: 훺(푎, 푏) ≥ !
!.
Proposed by Daniel Sitaru – Romania
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XI.19. If 푎, 푏, 푐 > 0,푎 + 푏 + 푐 = 1, 훺(푎, 푏) = lim →√ √√ √
then:
푏(푎 + 1) ⋅ 훺(푎, 푏) ≤ 2
Proposed by Daniel Sitaru – Romania
XI.20. Find:
훺 = lim→∞
(푛 + 1)!− 푛! 1 +11!
+12!
+ ⋯+1푛!
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
XI.21. Find:
훺 = lim→∞
1푛!
((1− 푥) + cos 푛푥)푒 푑푥
Proposed by Daniel Sitaru – Romania
XI.22. If 푎, 푏 > 0, |푥| < 1, |푦| < 1, 훺(푥) = lim → ∑ ( ) then:
(푎 + 푏)훺푎푥 + 푏푦푎 + 푏
≤ 푎훺(푥) + 푏훺(푦)
Proposed by Daniel Sitaru – Romania
XI.23. Find:
퐿 = lim→
1푘
+1푘
+1푘
Proposed by Daniel Sitaru – Romania
XI.24. Evaluate:
lim→
2√1 + 푥 + 2√2 + 푥 + ⋯+ 2√푛 + 푥 − 푛(푛 + 1)푥
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
XI.25. Find:
훺 = lim→∞
∑ 푘 ⋅ 2푘푘
푛(푛 + 1)(2푛 + 1)
Proposed by Daniel Sitaru – Romania
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XI.26. If 훺(푎) = lim →∞ 푛 푒 − 푒 ,푎 > 0, then:
훺(푎)푏 + 푐
+훺(푏)푐 + 푎
+훺(푐)푎 + 푏
> 푎 + 푏 + 푐
Proposed by Daniel Sitaru – Romania
XI.27. 푓:ℝ → [푎, 푏],푎 < 푏. Find:
훺 = lim→
(푛 − 푘 + 1) 푓(푘)푘(1 + 2 +⋯+ 푛 )
Proposed by Daniel Sitaru – Romania
XI.28. Find:
훺 = lim→∞
(푛 − 푘 + 1)푒1 + 2 +⋯+ 푛
Proposed by Daniel Sitaru – Romania
XI.29 If 푎 > 0,푛 ≥ 1, lim →∞ 푎 = 푎, 푏, 푐 > 0 then find:
훺 = lim→∞
1푛
푎푏 + 푐푎
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania XI.30. Find:
훺 = lim→∞
푘(푘 + 1)!
푘(푘 + 2)
(푘 + 1)!
푘(푘 + 3푘 + 3)
(푘 + 1)!
Proposed by Daniel Sitaru – Romania
XI.31. Find:
훺 = lim→∞
1푛
푚 푝
Proposed by Daniel Sitaru – Romania
XI.32. Find:
훺 = lim→∞
푛(푛 + 1)(푛 + 2) ⋅… ⋅ (2푛 − 2) arctan1 ⋅ 3 ⋅ 5 ⋅ … ⋅ (2푛 − 3)
Proposed by Daniel Sitaru -Romania
XI.33 Find:
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58 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
훺 = lim→∞
13푛 + 1
1√푘!
Proposed by Adil Abdullayev-Baku-Azerbaidian
XI.34. Find:
훺 = lim→
1푛 + 3
푘√푘!
Proposed by Adil Abdullayev-Baku-Azerbaidian
XI.35. If 푎, 푏, 푐, 푥,푦, 푧 > 0,푎 + 푏 + 푐 = 1, 훺(푎) = lim → 푛 푎 ! ! ⋯ ! − 1
then: 훺(푎푥 + 푏푦 + 푐푧) ≥ 훺 푥 푦 푧
Proposed by Daniel Sitaru – Romania
XI.36. Find:
훺 = lim→∞
1 + 6 + 11 + 16 + ⋯+ (10푘 − 9)2푘 − 1
Proposed by Daniel Sitaru – Romania
XI.37. Să se determine funcțiile monotone 푓:ℝ → ℝ care verifică relația:
2푓(푥) + 푓(푎 ) + 푓(log 푥) = 1,∀푥 > 0unde 푎 ≥ 2.
Proposed by Marian Ursarescu – Romania
XI.38. Find:
훺 = lim→
1 +1푚
푝! (1 + 푝 )
Proposed by Daniel Sitaru – Romania
XI.39. Find:
훺 = 1 +1푒
+1푒
⋅
Proposed by Daniel Sitaru – Romania
XI.40. 푥 = 2, = 1 + . Find the closed form and lim → 푥
Romanian Mathematical Society-Mehedinți Branch 2018
59 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
Proposed by Maria Elena Panaitopol-Romania
XI.41. Fie șirul (푥 ) ∈ℕ astfel încât 푥 > 0, 푥 = 푥 + ,푝 ∈ ℕ∗ fixat. Să se calculeze:
lim→∞
푥 + 푥 + ⋯+ 푥푛 √푛
Proposed by Marian Ursarescu – Romania
XI.42. Să se calculeze:
lim→
푛! 푥 − ln(1 + sinh푥) ln(1 + sinh 2푥) … ln(1 + sinh푛푥)푥
Proposed by Marian Ursarescu – Romania
XI.43. Find:
Ω = lim→∞
e + √2 ⋅ e + √3 ⋅… ⋅ e + √nπ + √2 ⋅ π + √3 ⋅ … ⋅ π + √n
Proposed by Daniel Sitaru – Romania
XI.44. Fie 퐴,퐵 ∈ 푀 (ℝ). Dacă 퐴 + 퐵 = 푂 , atunci: det(퐴퐵 + 퐵퐴) ≥ det(퐴퐵 − 퐵퐴)
Proposed by Marian Ursarescu – Romania
XI.45. (푥 ) , (푦 ) and (푧 ) are three real positive numbers sequences, such that 2푥 = 푦 + , 2푦 = 푧 + , 2푧 = 푥 + . Prove that the sequences are convergent and
find their limits. Proposed by Cristian Heuberger-Romania
XI.46. Fie șirul (푥 ) astfel încât 푥 = , 4푥 = 5푥 + 3 푥 − 4,∀푛 ≥ 1. Să se calculeze:
lim→
푥2
Proposed by Marian Ursarescu – Romania
XI.47. Fie 푥 = , 푥 = , 푥 = ⋅ . Să se arate că 푥 este convergent și să se
calculeze lim →∞ 푥 . Proposed by Marian Ursarescu – Romania
XI.48. Fie șirul (푥 ) astfel încât 푥 > 0, 2푥 = 푥 + 푥 + 1 + 푥 − 푥 + 1
Să se calculeze lim →∞ √푛 + 푛 + 1− √푛 − 푛 + 1 .
Proposed by Marian Ursarescu – Romania
XI.49. Let be 푛,푚 ∈ ℕ∗ and the invertible matrix 퐴 ∈ ℳ (ℂ),퐴 = 푎 , , , , with
∑ 푎, = 1. Prove that det(퐴 − 퐼 ) ≠ 0. Proposed by Dana Heuberger-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
60 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
XI.50. Find 푎 > 0, such that ≥ (푎 ⋅ 푥) ,∀푥 > 0.
Proposed by Cristian Heuberger-Romania
XI.51. Find the functions 푓:ℝ → ℝ, twice derivable, such that 푓 ⋅ 푓 ʹʹ = (푓 )ʹ − 푓 ʹ
Proposed by Cristian Heuberger-Romania
XI.52. Let 푎, 푏 ∈ ℝ, with 푎 < 푏, 푓: [푎, 푏] → [푎, 푏] a continuous and surjective function and the sets 퐴 = {푥 ∈ [푎, 푏]|푓(푥) = 푥}, 퐵 = {푥 ∈ [푎, 푏]|푓(푥) = 푎 + 푏 − 푥}. Prove that the set 퐴 ∪ 퐵 has at least 3 elements. Proposed by Cristian Heuberger-Romania
XI.53. Let be 푎, 푏 ∈ ℝ,푎 < 푏, and 푓: [푎, 푏] → [푎, 푏] a continuous and surjective function. Prove that
at least one of the following sets 퐴 = {푥 ∈ [푎, 푏]|푓(푥) = 푥},퐵 = {푥 ∈ [푎, 푏]|푓(푥) = 푎 + 푏 − 푥} has
at least two elements. Proposed by Cristian Heuberger-Romania
XI.54. We consider 푘 ∈ (1, ∞) and the function 푓:ℝ → ℝ such that:
lim→∞
푥 푓 푥 = 푎 ∈ ℝ.
Let be 푎 = 푓(1) + 푓(2) + ⋯+ 푓(푛),푛 ∈ ℕ∗. a. Prove that lim →∞ 푓(푥) = 0. b. Prove that the
sequence (푥 ) converges, where 푥 = ⋯ ,∀푛 ∈ ℕ∗.
Proposed by Dana Heuberger, Cristian Heuberger-Romania
XI.55. For every 푛 ∈ ℕ∗, let 휎 , 휀 be two permutations of 푆 . Prove that the sequence
(푎 ) ,푎 = ∑ ( )( ) is bounded. Proposed by Dana Heuberger-Romania
XI.56. Let a, b ∈ ℕ∗ and (x ) be a real numbers sequence such that for all
n ∈ ℕ∗, x = ax + bx − 2. a. Prove that if a = 2, b = 3 and x = 1, then x ∈ ℕ∗.
b. Prove that the set A = {(a, b) ∈ ℕ∗ × ℕ∗|∃n ∈ ℕ∗, suchthatx ∉ ℕ∗,∀x ∈ ℕ∗} is infinite.
Proposed by Cristian Heuberger-Romania
XI.57. If 푥 ≥ 0, prove that the sequence (푥 ) , 푥 = ⋅ (푒 ⋅ 푥 ) converges and find its limit. Proposed by Cristian Heuberger-Romania
XI.58. Let 푓: (−1,∞) → (−1,∞) be a function such that for all 푥 > −1, 푓 1 + = 1 + ( )
Romanian Mathematical Society-Mehedinți Branch 2018
61 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
a. If lim →√ 푓(푥) exists, find its value. b. If lim → 푓(푥) = 1, prove that lim → 푓(푥) = −1.
Proposed by Dana Heuberger, Gheorghe Boroica-Romania
XI.59. Let (푎 ) and (푏 ) be two sequences of strictly positive real numbers such that ∀푛 ∈ ℕ∗, 푏 = 푎 − . If lim →∞ 푏 = 푙 > 0, prove that the sequence (푎 ) converges and
find its limit. Proposed by Dana Heuberger-Romania
12-CLASS-STANDARD
XII.1. Să se determine toate mulțimile 퐻 ⊂ 푀 (ℤ) cu trei elemente pentru care 퐻 este grup în raport cu înmulțirea matricelor. Proposed by Nicolae Papacu –Romania
XII.2. Să se determine toate mulțimile 퐻 ⊂ 푀 (ℤ) cu patru elemente pentru care 퐻 este grup în raport cu înmulțirea matricelor. Proposed by Nicolae Papacu –Romania
XII.3. Fie inelul (퐴, +,⋅) cu cel puțin două elemente care verifică proprietatea
(푃): există 푛,푝 ∈ ℕ, astfel încât 푥 = 푥 , pentru orice 푥 ∈ 퐴.
1) Demonstrați că 푥 = 푥, pentru orice 푥 ∈ 퐴. 2) Dați exemplu de inel cu cel puțin trei elemnte care verifică proprietatea (푃). 3) Există inele finite care nu verifică proprietatea (푃)?
Proposed by Nicolae Papacu –Romania
XII.4. Să se determine numerele naturale prime 푝 și 푞 știind că ∑ 푐푎푟푑 푈 ℤ = 푞 , unde 푈 ℤ reprezintă mulțimea elementelor inversabile ale inelului ℤ , +,⋅ .
Proposed by Nicolae Papacu –Romania
XII.5. Să se determine numerele prime 푝 și 푞 știind că∑ 푐푎푟푑 퐷 ℤ = 푞 , unde 퐷 ℤ reprezintă mulțimea divizorilor lui zero din inelul ℤ , +,⋅
Proposed by Nicolae Papacu –Romania
XII.6. Să se demonstreze că:
푒 푑푡 + 푒 푑푡 ≥ 푒 푑푡 + 푒 푑푡
Proposed by Daniel Sitaru, Claudia Nănuți – Romania
XII.7. Fie 푓: [0,1] → ℝ o funcție continuă și ∆퐴퐵퐶 de arie 푆 și semiperimetru 푝. Să se arate că:
Romanian Mathematical Society-Mehedinți Branch 2018
62 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
3 푓(푥) 푑푥 + 2푆1
sin퐴1
푓(푥) 푑푥 ≤ 4푝
Proposed by Daniel Sitaru, Claudia Nănuți – Romania
XII.8. Pe mulțimea 푀 = [0,1] definim legea de compoziție: 푥 ∗ 푦 = 푥 1 − 푦 + 푦√1 − 푥 . Notam: 푥 = 푥, 푥 = 푥 ∗ 푥, … … , 푥 = 푥 ∗ 푥 ∗ … ∗ 푥 , … .
„ ”unde 푥 ∈ 푀. Să se arate că dacă 푥 ∈ 푀 atunci
푥 = 푥 ∗ 푥 ∗ … ∗ 푥„ ”
∈ 푀
Proposed by Daniel Sitaru, Claudia Nănuți – Romania XII.9. Să se calculeze:
퐿 = lim→
푛1
1 + 푛푥푑푥
Proposed by Daniel Sitaru, Claudia Nănuți – Romania XII.10. Să se arate că (∀)훼 ∈ [2,7] avem:
arctg 푥 푑푥 ≤훼 − 2
5arctg 푥 푑푥
Proposed by Daniel Sitaru, Claudia Nănuți – Romania XII.11. Să se calculeze:
I =2x + 2n + 1
( ) + ( ) + 1dx; n ∈ ℕ∗
Proposed by Daniel Sitaru, Claudia Nănuți – Romania XII.12. Să se calculeze:
퐼 = ln 1 + tg19휋84
tg푥 푑푥
Proposed by Daniel Sitaru, Claudia Nănuți – Romania XII.13. Fie 푛 ∈ ℕ∗. Să se calculeze:
퐼 =√푛 + 1− 푥
√푛 + 1− 푥 + √푥 − 푛푒푑푥
Proposed by Daniel Sitaru, Dan Nănuți – Romania
Romanian Mathematical Society-Mehedinți Branch 2018
63 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
XII.14. Să se calculeze:
L = lim→
√2푛 − 푘 + √2푛 + 푘√4푛 − 푘
Proposed by Daniel Sitaru, Dan Nănuți – Romania XII.15. Să se calculeze:
퐿 = lim→
1푛
푥 + 4푥 + 12푥 + 9푥(푥 + 3) − 푥 − 243
푑푥
Proposed by Daniel Sitaru, Dan Nănuți – Romania XII.16. Să se demonstreze că:
1sin 푥 + sin푥 cos 푥 + cos 푥
푑푥 ≥휋4
Proposed by Daniel Sitaru, Dan Nănuți – Romania XII.17. Să se arate că: ∫ (cos(sin 푥) + sin(cos 푥))푑푥 ≤
Proposed by Daniel Sitaru, Dan Nănuți – Romania XII.18. Pe mulțimea 퐺 = (0,2) definim legea de compoziție: 푥 ∘ 푦 =
Să se calculeze:
퐿 = lim→
(푥 ∘ 푥 ∘ … ∘ 푥)„ ”
Proposed by Daniel Sitaru, Dan Nănuți – Romania
XII.19. Să se calculeze:
퐿 = lim→
1푛
푥 sin 휋푥푥 + (1− 푥)푘
푑푥
Proposed by Daniel Sitaru, Dan Nănuți – Romania XII.20. Să se arate că:
퐼 =cos 2015푥 − cos 2016푥
sin푥푑푥 > 0,0001
Proposed by Daniel Sitaru– Romania
Romanian Mathematical Society-Mehedinți Branch 2018
64 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
XII.21. Fie 푓:ℝ → ℝ∗ o funcție continuă și astfel încât: (∀)푥,푦 ∈ ℝ; 푓(푓(푥 + 푦) + 푥) = 푦 + 푓(2푥).
Să se arate că:
tg푥푓(푥) 푑푥 >
푓(푥)sin 푥
푑푥
Proposed by Daniel Sitaru -Romania
XII.22. Să se arate că:
11 (푥 + 1)(푥 + 2) ∙ … ∙ (푥 + 10)푑푥 < 6
Proposed by Daniel Sitaru -Romania
XII.23. Să se calculeze:
lim→
tg1푛
sin(1 + 푛) + sin 1 +2푛
+ ⋯+ sin 1 +푛푛
Proposed by Daniel Sitaru -Romania
XII.24. Să se arate că:
4 푥 푑푥 ≥ 푒
Proposed by Daniel Sitaru -Romania
XII.25. Să se arate că:
2푥 + 13(푥 + 4)(푥 + 9)푑푥 ≤
512
Proposed by Daniel Sitaru -Romania
XII.26. Să se arate că:
푥sin 푥
∙sin 푥푥
푑푥 >tg푥푥
∙푥
tg푥푑푥
Proposed by Daniel Sitaru -Romania
Romanian Mathematical Society-Mehedinți Branch 2018
65 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
XII.27. Fie 퐺 = 푀(푥) 푀(푥) =√
0√
0 1 0
√0
√
;푥 ∈ (−1,1) .Să se arate că: (퐺,∙) ⋍ (ℝ∗ ,∙)
Proposed by Daniel Sitaru -Romania
XII.28. Fie 푀(푥) =√
0√
0 1 0
√0
√
∈ 푀 (ℝ). Să se găsească 푥 ∈ (−1,1) astfel încât pentru
푛 ∈ ℕ∗/{1} să avem: 푀(푥) = 푀
Proposed by Daniel Sitaru -Romania
XII.29. Evaluate:
lim→∞
푒 푑푥
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
XII.30. Find:
훺 = lim→∞
푛푥 + arctan 푥
푒푑푥
Proposed by Daniel Sitaru -Romania
XII.31. Find: 훺 = lim →∞∫ √
∫ √, [∗] - great integer function.
Proposed by Daniel Sitaru -Romania
XII.32. Find:
Ω = lim→∞
1n
k arctankn
Proposed by Daniel Sitaru -Romania
XII.33. Find:
훺 = lim→
1푛
7 ⋅ 5
Proposed by Daniel Sitaru -Romania
Romanian Mathematical Society-Mehedinți Branch 2018
66 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
XII.34. Fie ecuația 푥 + 푎푥 + 푏푥 + 푎푥 + 푐 = 0, cu 푎, 푏, 푐 ∈ ℝ. Dacă ecuația are toate rădăcinile reale, atunci are loc inegalitatea: |1− 푏 + 푐| ≥ 1 + |푐|
Proposed by Marian Ursarescu-Romania
XII.35. If 푎, 푏 ∈ 0, and 푎 + 푏 = , then evaluate the integral:
ln(1 + tan 푥)푑푥
Proposed by D.M. Bătinețu-Giurgiu, Neculai Stanciu-Romania
XII.36. Find the continuous function 푓:ℝ → ℝ, such that:
푓(푡)푑푡 = 푥 ⋅ 푓(푥) − ln 1 + 푥 ,∀푥 ∈ ℝ.
Proposed by Cristian Heuberger-Romania
XII.37. Să se calculeze: ∫ 푑푥. Proposed by Marian Ursarescu-Romania
XII.38. We denote by 퐼 the set of the units and by 푁 the set of the non-invertible elements of the commutative ring with unity (퐴, +,⋅).We consider that the ring has the property:
∀푥,푦 ∈ 푁 ⇒ 푥 + 푦 ∈ 푁. For 푦 ∈ 푁, we define the functions 푓 :퐴 → 퐴, 푓 (푥) = 푥 + 푦 + 푥푦 and 푔 : 퐼 → 퐼, 푔 (푥) = 푓 (푥). Prove that 푓 and 푔 are two bijections.
Proposed by Dana Heuberger-Romania
XII.39. We consider 푛 ∈ ℕ,푛 ≥ 2 and set 푀 = 푥 ∈ ℤ |푥 = 3푥 + 1 . a. Find 푀 and 푀 .
b. If 푀 has and odd number of elements, find 푛. Proposed by Dana Heuberger-Romania
XII.40. Let (퐴, +,⋅) be a ring with unity, with 푐푎푟푑(퐴) ≥ 4, such that ∀푥,푦 ∈ 퐴 ∖ {0,1}, with 푥 ≠ 푦, we have 푥 = 푦 or 푦 = 푥. Find 푐푎푟푑(퐴). Proposed by Dana Heuberger-Romania
XII.41. Let (퐴, +,⋅) be a commutative ring with unity, having at least two elements, such that for all 푥 ∈ 퐴, 푥 is invertible if and only if 1− 푥 is non-invertible. We denote by 퐼 the set of the invertible elements of 퐴 and with 푁 the set of the non-invertible elements of 퐴. We define the operation 푥 ∗ 푦 = 1 + 푥 + 푦,∀푥,푦 ∈ 퐴. Prove that (푁, +) and (퐼,∗) are isomorphic groups.
Proposed by Dana Heuberger-Romania
XII.42. (퐴, +,⋅) is a ring with 0 ≠ 1. We consider the set 푀 = {푥 ∈ 퐴|푥 = 푥 + 1}. If the cardinal of 푀 is odd, prove that ∀푥 ∈ 푀, 푥 + 1 = 0. Proposed by Dana Heuberger-Romania
XII.43. We say that the nontrivial group (퐺 ,⋅) with the identity element 푒 ∈ 퐺 has the property 풫, if there exist two of its proper subgroups, 퐻 ,퐻 , with 퐻 ∩ 퐻 = {푒}, such that
∀푥 ∈ 퐺 ⊂ (퐻 ∪퐻 ),∀ℎ ∈ 퐻∗,∀ℎ ∈ 퐻∗, we have푥ℎ ∈ 퐻 and 푥ℎ ∈ 퐻 .
Romanian Mathematical Society-Mehedinți Branch 2018
67 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
a. Find an example of a finite group with the property 풫. b. Prove that all the finite groups with the property 풫 are isomorphic. (For the subgroup 퐻 of the group 퐺, we denoted 퐻∗ = 퐻 ∖ {푒}.)
Proposed by Dana Heuberger-Romania
XII.44. We consider the group (퐺,⋅) with 푛 elements, where 푛 ∈ ℕ,푛 ≥ 5. If each time we choose 푘 = + 3 of its elements, at least 2 of them belong to the center of the group, 푍(퐺), then 퐺 is abelian. Proposed by Dana Heuberger-Romania
XII.45. We consider the abelian group (퐺,⋅) with 푛 elements, where 푛 ∈ ℕ,푛 ≥ 4. If each time we choose 푘 = + 1 of its elements, at least two of them have the order ≤ 2 and none of the elements of 퐺 has the order 4, then the group 퐺 has the exponent 2.
Proposed by Dana Heuberger -Romania
XII.46. Let (퐺 ,⋅) be a group with 푛 elements and 푒 its identity element. We say that the subgroup 퐻 of 퐺 has the property (풫) if 퐻 ≠ {푒},퐻 ≠ 퐺 and ∀푥,푦 ∈ 퐺 ∖ 퐻 ⇒ 푥푦 ∈ 퐻. a. If the distinct subgroups 퐻 and 퐻 of 퐺 exist, having the property (풫), with 퐻 ∩ 퐻 = {푒}, prove that the group 퐺 and Klein’s four-group are isomorphic. b. Prove that for 푛 ∈ ℕ,푛 ≥ 2, the group (ℤ , +) has at most a subgroup with the property (풫). Proposed by Dana Heuberger-Romania
XII.47. We consider the group (퐺,⋅) with 푛 elements, where 푛 ∈ ℕ,푛 ≥ 4. We say that the subgroup 퐻 of 퐺 has the property 풫, if 퐻 ≠ 퐺 and ∀푥, 푦 ∈ 퐺 ∖ 퐻 ⇒ 푥푦 ∈ 퐻. a. Find an example of a group 퐺 which has three distinct subgroups 퐻,퐾, 퐿 with the property 풫, such that 퐺 = 퐻 ∪퐾 ∪ 퐿.
b. If 퐻,퐾, 퐿 are distinct subroups of 퐺 which have the property 풫 and 퐺 = 퐻 ∪ 퐾 ∪ 퐿, find |퐻 ∩ 퐾 ∩ 퐿|. Proposed by Dana Heuberger-Romania
XII.48. We consider 푛 ∈ ℕ,푛 ≥ 4 and group (퐺,⋅) with 푛 elements. We denote by 푒 its identity element. Let 퐻 and 퐻 be two subgroups of 퐺, such that 퐻 ∩ 퐻 = {푒}. If each time we choose 푘 = elements of 퐺, at least 2 of them are in the set 퐻 ∪ 퐻 , prove that 퐻 = 퐺 or 퐻 = 퐺.
Proposed by Dana Heuberger-Romania
XII.49. We consider 푛 ∈ ℕ,푛 ≥ 4 and the group (퐺,⋅) with 푛 elements. We denote by 푒 its neutral element. Let 퐻 and 퐻 be two subgroups of 퐺, such that 퐻 ∩ 퐻 = {푒}. If the set 퐻 ∪퐻 has at least + 2 elements, then 퐻 = 퐺 or 퐻 = 퐺. Proposed by Dana Heuberger-Romania
XII.50. We consider 푎, 푏 ∈ ℤ, with (푎, 푏) = 푑 and (퐺,⋅) a group with 푛 elements.For 푘 ∈ ℤ, we donote 퐺 = {푥 |푥 ∈ 퐺}. a. If (푑,푛) = 1, prove that 퐺 퐺 = 퐺. b. If the group is commutative, then 퐺 퐺 = 퐺 ⇔ (푑 ,푛) = 1. Proposed by Dana Heuberger-Romania
XII.51. Determinați funcțiile : (0, ) (0, )f cu proprietatea că:
( )
2
1 1 cos( ) .( )
f x
x
tf x x dtf x x t
Proposed by Dan Nedeianu-Romania
Romanian Mathematical Society-Mehedinți Branch 2018
68 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
ROMANIAN MATHEMATICAL MAGAZINE-R.M.M.-autumn 2017 JP.076. Let 퐴퐵퐶 be an acute triangle. Prove that: (푎 cot퐴) (푏 cot퐵) (푐 cot 퐶) ≤ (2푟)
where 푎 = 퐵퐶, 푏 = 퐶퐴, 푐 = 퐴퐵, and 푟 is the inradius.
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
JP.077. Let 푎 ,푎 , … ,푎 be non-negative real numbers such that 푎 + 푎 +⋯+ 푎 = 1. Prove that for all 휆 ≥ 4, the following inequality holds:
푎 + 휆 푎 푎 ≤2휆 + 1
3.
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
JP.078. Let 푎, 푏, 푐 be positive real numbers such that 푎 = 푏 + 푐 . Prove that 푎푏+ 푏푐 + 푐푎 + √2− 1 ≤ 2푎 .
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
JP.079. Prove that the inequality holds for all positive real numbers 푎, 푏, 푐:
1푎 + 푏
+1
푏 + 푐+
1푐 + 푎
≥4
2푎 + 3푏 + 3푐+
42푏 + 3푐 + 3푎
+4
2푐 + 3푎 + 3푏
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
JP.080. Prove that in any triangle 퐴퐵퐶: + + ≥ 2√3.
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
JP.081. If 푥,푦, 푧 > 0 then: + + ≤ 3
Proposed by Marin Chirciu – Romania
JP.082. If 푎, 푏, 푐 > 0;푎 + 푏 + 푐 = 3 then: + + ≥
Proposed by Marin Chirciu – Romania
JP.083. In 훥퐴퐵퐶 then following relationship holds:
(푎 + 푏 + 푐 )1
(푎 + 푏) +1
(푏 + 푐) +1
(푐 + 푎) ≥ 3 ⋅ 4 ⋅ 푟 ( );푚,푛 ≥ 1
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu - Romania
JP.084. In 훥퐴퐵퐶 the following relationship holds:
Romanian Mathematical Society-Mehedinți Branch 2018
69 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
(푎 + 푏) tan퐶2
⋅1
tan + tan≥ 3 ⋅ 4 ⋅ 푟 ;푚 ≥ 푛 ≥ 1
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu - Romania
JP.085. Let 퐴퐵퐶 denote a triangle, 퐼 its incenter, 푅 its circumradius, 푟 its inradius and 푥,푦 and 푧 the inradii of triangles 퐼퐵퐶, 퐼퐶퐴, and 퐼퐴퐵 respectively.Prove that:
sin퐴푥
+sin퐵푦
+sin 퐶푧
≤4 + 3√3
2푟+
2푅
Proposed by George Apostolopoulos – Messolonghi – Greece
JP.086. Let 푎, 푏, 푐 be the side lengths of a triangle 퐴퐵퐶 with incentre 퐼, circumradius 푅 and inradius 푟.Prove that:
√퐴퐼푎
+√퐵퐼푏
+√퐶퐼푐
≤√22⋅√푅 + 푟
푟
Proposed by George Apostolopoulos – Messolonghi – Greece
JP.087. Let 퐴퐵퐶 be an acute triangle. Prove that:
√cos퐴 ⋅ sin퐵 ⋅ sin퐶 + √sin퐴 ⋅ cos퐵 ⋅ sin 퐶 + √sin퐴 ⋅ sin퐵 ⋅ cos 퐶 ≤32
32
Proposed by George Apostolopoulos – Messolonghi – Greece
JP.088. Let a, b, c be positive real numbers. Prove that:
a + bc + ab
+b + ca + bc
+c + ab + ca
≥9abc
ab + bc + ca
Proposed by Nguyen Ngoc Tu – Ha Giang –Vietnam
JP.089. Let 푎, 푏, 푐 be positive real numbers, take 푋 = + ,푌 = + ,푍 = + .
Prove that: 푋 + 푌 + 푍 ≥ 2 (푋 + 푌 + 푍 − 3)(푋 + 푌 + 푍 + 3).
Proposed by Nguyen Ngoc Tu – HaGiang –Vietnam
JP.090. Let 푟 and 푠 be the inradius and the semiperimeter of a triangle 퐴퐵퐶 respectively. Prove that: ≥ . Proposed by Martin Lukarevski – Skopje
PROBLEMS FOR SENIORS
SP.076. Let 푎, 푏, 푐 be the side – lengths of an acute triangle with perimeter 1. Prove that:
퐸 ≥ 푎 푏 푐 ≥ 퐸
Romanian Mathematical Society-Mehedinți Branch 2018
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where 퐸 = ( )( )( )( ) ( ) ( ) and 퐸 = ( )( )( ) .
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
SP.077. Prove that in any acute triangle 퐴퐵퐶 the following inequality holds:
푚ℎ
cos퐴 +푚ℎ
cos퐵 +푚ℎ
cos퐶 ≥32
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
SP.078. Let 푎, 푏, 푐 be positive real numbers such that 푎 + 푏 + 푐 = 1. Prove that:
푎 푏 푐 + 푎 푏 푐 + 푎 푏 푐 ≤ 푎 + 푏 + 푐 .
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
SP.079. Prove that for all positive real numbers a, b, c and integer n ≥ 3, the following inequality
holds: + + ≥ + + .
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
SP.080. Prove that for all positive real numbers 푎, 푏, 푐 the following inequality holds:
(푎 + 푏)푎 − 푎푏 + 푏
+(푏 + 푐)
푏 − 푏푐 + 푐+
(푐 + 푎)푐 − 푐푎 + 푎
≥9(푎 푏 + 푏 푐 + 푐 푎 + 푎푏푐)
푎 + 푏 + 푐.
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
SP.081. Let 푎, 푏, 푐 be positive real numbers and 푘 ≥ 2. Prove that:
푏푐(푏 + 푘푎)(푐 + 푘푎) +
푐푎(푐 + 푘푏)(푎 + 푘푏) +
푎푏(푎 + 푘푐)(푏 + 푘푐) ≥
3푘 + 1
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
SP.082. Let 퐴퐵퐶 be an equilateral triangle with side-length 푎 and let 푀 be any point inside the
triangle. Prove that: ≥ 푥푀퐴 + 푦푀퐵 + 푧푀퐶 ≥ 2(푥푦 + 푦푧 + 푧푥)
where 푥,푦, 푧 denote the distances from 푀 to the sides 퐵퐶,퐶퐴,퐴퐵, respectively.
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
SP.083. Let 푚 ,푚 ,푚 be the lengths of the medians of a triangle with circumradius 푅. Prove that:
1 +1푚
⋅ 1 +1푚
⋅ 1 +1푚
≥ 1 +2
3푅.
Proposed by George Apostolopoulos – Messolonghi – Greece
SP.084. Prove that if n ∈ ℕ∗ then:
Romanian Mathematical Society-Mehedinți Branch 2018
71 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
2 arctan(x ) arctan(x ) dx ≤ arctan (x ) dx +1
2n− 1
Proposed by Daniel Sitaru – Romania
SP.085. Prove that if 푎, 푏 ∈ (0,∞);푛 ∈ ℕ∗ then:
푎푏
+푏푎
푎푏
+푏푎
푎푏
+푏푎
푏푎
+푎푏
≥ 8푎푏
+푏푎
Proposed by Daniel Sitaru – Romania
SP.086. Prove that if 푎, 푏, 푐 are the lengths’ sides in triangle 퐴퐵퐶 then:
sin 푎 + sin 푏 + sin 푐 ≥ 4 sin 푠 sin(푠 − 푎) sin(푠 − 푏) sin(푠 − 푐)
Proposed by Daniel Sitaru - Romania
SP.087. Let 푧 , 푧 , 푧 be the affixes of 퐴,퐵 respectively 퐶 in acute-angled 훥퐴퐵퐶. Prove that:
푧 − 푧푧 + 푧
+푧 − 푧푧 + 푧
≥32푠푟
(푠 − (2푅 + 푟) )
Proposed by Daniel Sitaru – Romania
SP.088. Let 푎, 푏, 푐 > 0 such that 푎푏 + 푏푐 + 푐푎 + 푎푏푐 = 4. Prove that:
(푎 + 1) (푏 + 1)(푐 + 1) + (푏 + 1) (푐 + 1)(푎 + 1) + (푐 + 1) (푎 + 1)(푏 + 1) ≥ 푎 + 푏 + 푐 + 9
Proposed by Nguyen Ngoc Tu – Ha Giang –Vietnam
SP.089. Let 푟 , 푟 , 푟 be the exradii of a triangle 퐴퐵퐶,ℎ ,ℎ ,ℎ be altitudes and let 푅, 푟, 푠 denote the circumradius, inradius and semiperimeter respectively. Prove that:
푟ℎ
+푟ℎ
+푟ℎ
≥2푠
31푟−
1푅
Proposed by Martin Lukarevski – Skopje - Macedonia
SP.090. If 푢, 푣 > 0, with 2푢 − 푣 > 0 and 훼,훽, 훾 are the measures of the angles of triangle 퐴퐵퐶, then:
sin 훼푢 sin훽 + 푣 sin 훼 sin 훽
≥3
푢 + 푣.
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
UNDERGRADUATE PROBLEMS
UP.076. Evaluate: 푆 = ∑ .
Romanian Mathematical Society-Mehedinți Branch 2018
72 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
Proposed by Shivam Sharma – New Delhi – India
UP.077. Evaluate: S = ∑ e
Proposed by Shivam Sharma – New Delhi – India
UP.078. Find: 훺 = lim → 푛 (2푛 + 1)‼ − (2푛 − 1)‼ (푛 + 1)! − √푛! Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
UP.079. Find: 훺 = lim →( )
( )‼( )‼−
( )‼( !).
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
UP.080. Let be 푓: (0,∞) → (0,∞) a function such that: lim →( ) = 푎 ∈ (0,∞) and
lim →( )
( ) = 푏 ∈ (0,∞).Find: 훺 = lim → 푓(푥 + 1)− 푓(푥)
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu – Romania
UP.081. If 퐵 (푡) = 푛 ( )
( )!−
√ !, with 푡 > 0, then compute lim → 퐵 (푡).
Proposed by Bătinețu – Giurgiu; Neculai Stanciu – Romania
UP.082. Let n ∈ N. Calculate:
I = sin x cos x cosπ2
sin x + sin x cosπ2
cos x dx.
Proposed by Bătinețu – Giurgiu; Neculai Stanciu – Romania
UP.083. Prove that in any triangle ABC the following relationship holds:
R (b + c − 2a) ≤ 4(R− 2r) a
Proposed by Daniel Sitaru – Romania
UP.084. Evaluate: 퐼 = ∫ ∫( ( ) ( )) 푑푥푑푦
Proposed by Shivam Sharma – New Delhi – India
UP.085. Let 푘 be positive integer. Calculate:
lim→
훤(푥 + 2) − 훤(푥 + 1) 훤(푥 + 1) ,
where 훤(푥) is the Gamma function (or Euler’s second integral)
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu - Romania
Romanian Mathematical Society-Mehedinți Branch 2018
73 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
UP.086. Let 푎 > 0, 푏, 푐 > 1 and 푓,푔:푅 → 푅 be continuous and odd functions. Prove that:
푓(푥) ln 푏 ( ) + 푐 ( ) 푑푥 = (ln(푏푐)) 푓(푥)푔(푥) 푑푥
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu - Romania
UP.087. Let 푎, 푏 ∈ 푅,푎 < 푏 and continue functions 푓,푔,ℎ:푅 → 푅 such that 푓(푎+ 푏 − 푥) = −푓(푥),푔(푎 + 푏 − 푥) = 푔(푥),ℎ(푎 + 푏 − 푥) = −ℎ(푥),∀푥 ∈ 푅.Prove that:
푓(푥) (arctan푔(푥)) ln 1 + 푒 ( ) 푑푥 =12
푓(푥)ℎ(푥) arctan푔(푥) 푑푥
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu - Romania
UP.088. Let 푓:푅 → 푅 be a continuous function such that 푓(푥) = 푓(1 − 푥),∀푥 ∈ 푅. Prove that:
√1 − 푥 + √푥1 + √2푥
푓(푥)푑푥 =√22⋅ 푓(푥) 푑푥
Proposed by D.M. Bătinețu – Giurgiu; Neculai Stanciu - Romania
UP.089. Evaluate: ∫ [ln(푥) ln(1 − 푥) + 퐿푖 (푥)] ( )( ) −
( ) 푑푥
Proposed by Shivam Sharma – New Delhi – India
UP.090. Evaluate: ∫ (ln(훤(푥))))(sin(2푘휋푥))푑푥, 푘 ≥ 1
Proposed by Shivam Sharma – New Delhi – India
ROMANIAN MATHEMATICAL MAGAZINE-R.M.M.-winter 2017
JP.091. Prove that the following inequalities holds for all positive real numbers 푎, 푏, 푐
(a) + + ≥ ⋅
(b) ( )
+ ( ) + ( ) ≥ ⋅
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
JP.092. Prove that the following inequalities hold for all positive real numbers 푎, 푏, 푐
(a) + + ≥ ( )
(b) + + ≥
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
Romanian Mathematical Society-Mehedinți Branch 2018
74 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
JP.093. Let 푎, 푏, 푐 be positive real numbers such that 푎 + 푏 + 푐 = 1. Prove that: (a) + + ≤
(b) √√
+ √√
+ √√
≤√
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
JP.094. Let 푎, 푏, 푐 be positive real numbers such that 푎푏 + 푏푐 + 푐푎 = 1. Prove that:
푏푐 푎 + 2푏푐 + 푐푎 푏 + 2푐푎 + 푎푏 푐 + 2푎푏 ≥ 1
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
JP.095. Prove that for all positive real numbers 푎, 푏, 푐: 푎(푏 + 푐 )2푎 + 푏푐
+푏(푐 + 푎 )2푏 + 푐푎
+푐(푎 + 푏 )2푐 + 푎푏
≥6푎푏푐
푎푏 + 푏푐 + 푐푎
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
JP.096. Let 푎, 푏, 푐 positive numbers such that 푎 + 푏 + 푐 = 3. Prove that:
푎푏
+푏푐
+푐푎
푏푎
+푐푏
+푎푐
≥ 9
Proposed by Nguyen Ngoc Tu - Ha Giang – Vietnam
JP.097. Let 푎, 푏, 푐 > 0 such that (푎 + 푏)(푏 + 푐)(푐 + 푎) = 8. Prove that:
푎푎 + 1
+2푏푏 + 1
+ 22푐푐 + 1
≤72
Proposed by Nguyen Ngoc Tu - Ha Giang – Vietnam
JP.098. Let 푎, 푏 and 푐 be the side lengths of a triangle 퐴퐵퐶 with incenter 퐼. Prove that:
1퐼퐴
+1퐼퐵
+1퐼퐶
≥ 31푎
+1푏
+1푐
Proposed by George Apostolopoulos – Messolonghi – Greece
JP.099. Find the value the following expression: 퐸 = + + + + +
where 푥 = tan 20° ,푦 = tan 40 °, 푧 = tan 80°.
Proposed by Kevin Soto Palacios – Huarmey – Peru
JP.100. Let in triangle 푤 ,푤 ,푤 be the angle bisectors and 푅, 푟 the circumradius and inradius respectively. Prove the inequality: ≤ + + ≤
Proposed by D.M. Bătinețu – Giurgiu – Romania, Martin Lukarevski – Skopje
Romanian Mathematical Society-Mehedinți Branch 2018
75 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
JP.101. Let 푥,푦, 푧 be positive real numbers with 푥푦푧 = 1. Prove that:
√푥 + 1 + 푦 + 1 + √푧 + 1푥 + 푦 + 푧
≤ √2
Proposed by George Apostolopoulos – Messolonghi – Greece
JP.102. Let 푥,푦, 푧 > 0 be positive real numbers. Then:
1푥 + 푦
+1
푦 + 푧+
1푧 + 푥
≥4 3푥푦푧(푥 + 푦 + 푧)
(푥 + 푦)(푦 + 푧)(푧 + 푥)
Proposed by D.M. Bătinețu – Giurgiu – Romania, Martin Lukarevski – Skopje
JP.103. Let 푥,푦, 푧 > 0 be positive real numbers. Then in triangle 퐴퐵퐶 with semiperimeter 푠 and
inradius: 푟 cot + cot + cot ≥ 18− .
Proposed by D.M. Bătinețu – Giurgiu – Romania, Martin Lukarevski – Skopje
JP.104. Let 푟 , 푟 , 푟 be the exradii, ℎ ,ℎ ,ℎ the altitudes and 푚 ,푚 ,푚 the medians of a triangle 퐴퐵퐶 with semiperimeter 푠, circumradius 푅 and inradius 푟. Then:
푟ℎ 푚
+푟
ℎ 푚+
푟ℎ 푚
≥54푟
푠 − 푟 − 4푅푟
Proposed by D.M. Bătinețu – Giurgiu – Romania, Martin Lukarevski – Skopje
JP.105. Let 푚 > 0 and 퐹 be the area of the triangle 퐴퐵퐶. Then:
푎푏 + 푐
+푏
푐 + 푎+
푐푎 + 푏
≥ 2√3퐹
Proposed by D.M. Bătinețu – Giurgiu – Romania, Martin Lukarevski – Skopje
PROBLEMS FOR SENIORS
SP.091. Prove that for all positive real numbers 푎, 푏, 푐,푑:
푎푎 + 푏 + 푐
+푏
푏 + 푐 + 푑+
푐푐 + 푑 + 푎
+푑
푑 + 푎 + 푏≥푎 + 푏 + 푐 + 푑
3+
4(2푎 + 푏 − 2푐 − 푑)27(푎 + 푏 + 푐 + 푑)
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
SP.092. Prove that for all positive real numbers 푎, 푏, 푐:
(a) + + ≥ + ( )( ).
(b) + + ≥ + ( )( ) .
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
Romanian Mathematical Society-Mehedinți Branch 2018
76 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
SP.093. Prove that in any triangle 퐴퐵퐶 the following inequality holds:
(푏 + 푐)푎푚
+(푐 + 푎)푏푚
+(푎 + 푏)푐푚
≥ 8.
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
SP.094. Prove that in any acute triangle 퐴퐵퐶 the following inequality holds:
cos퐵 cos퐶sin퐴
+cos 퐶 cos퐴
sin퐵+
cos 퐴 cos퐵sin퐶
≤√32.
Proposed by Nguyen Viet Hung – Hanoi – Vietnam
SP.095. Let 푎, 푏, 푐 be the side lengths of a triangle 퐴퐵퐶 with inradius 푟 and circumradius 푅. Prove that:
(푏 + 푐 ) + sin 퐴 + (푐 + 푎 ) sin 퐵 + (푎 + 푏 ) sin 퐶 ≤814
(3푅 − 16푟 )
Proposed by George Apostolopoulos – Messolonghi – Greece
SP.096. Let 퐴퐵퐶 be a triangle and 푤 ,푤 ,푤 are bisectors of 퐴퐵퐶. Prove that:
+ + ≥ , where 푅 is the circumradius of 퐴퐵퐶, 훥 is area of 퐴퐵퐶.
Proposed by Mehmet Șahin - Ankara - Turkey
SP.097. Let ABC be the side lengths of a triangle ABC with incentre I, circumradius R and inradius r. Prove that:
√AIa
+√BI
b+√CI
c≤√22⋅√R + r
r
Proposed by George Apostolopoulos – Messolonghi – Greece
SP.098. Let 퐴퐵퐶 be an acute triangle with orthocenter 퐻. Prove that
퐴퐻 ⋅ 퐵퐻 + 퐵퐻 ⋅ 퐶퐻 + 퐶퐻 ⋅ 퐴퐻 ≤ 6푅푟,
where 푅 and 푟 are the circumradius and inradius respectively of triangle 퐴퐵퐶.
Proposed by George Apostolopoulos – Messolonghi – Greece
SP.099. Let 푎, 푏, 푐 be non-negative such that 푎 + 푏 + 푐 = 3. Prove that:
|(푎 − 푏)(푏 − 푐)(푐 − 푎)| ≤ √ . Equality occurs when?
Proposed by Nguyen Ngoc Tu - Ha Giang – Vietnam
SP.100. Let a, b, c be the lengths of the sides of a triangle with perimeter 3 and inradius r. Prove that:
Romanian Mathematical Society-Mehedinți Branch 2018
77 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
288r ≤(a + b)a + b
+(b + c)b + c
+(c + a)c + a
≤2r
.
Proposed by George Apostolopoulos – Messolonghi – Greece
SP.101. Let 푎, 푏 and 푐 be the side lengths of a triangle with inradius 푟. Prove that:
1푎 + 2푏 푐
+1
푏 + 2푐 푎+
1푐 + 2푎 푏
≤√36푟
Proposed by George Apostolopoulos – Messolonghi – Greece
SP.102. Let 퐴퐵퐶 be a triangle with circumradius 푅 and inradius 푟. Prove that:
4 ≤ sec퐴2
+ sec퐵2
+ sec퐶2≤
2푅푟
.
Proposed by George Apostolopoulos – Messolonghi – Greece
SP.103. Let 푚,푛 be positive real numbers. Prove that:
1푚
+1푛
≤4034− 2015푚푚 + 2017
+4034 − 2015푛푛 + 2017
+푚 + 푛 + 2009
2
Proposed by Iuliana Trașcă – Romania
SP.104. Prove that in any triangle 퐴퐵퐶 the following relationship holds:
푟1
sin+푎푏푐
21
푎푏푠(푠 − 푐)≤ 푅
Proposed by Daniel Sitaru – Romania
SP.105. Let 퐺 be the centroid in 훥퐴퐵퐶. Prove that:
cot 퐺퐵퐴 + cot 퐺퐶퐵 + cot 퐺퐴퐶 > cot퐴 + cot퐵 + cot 퐶 + 3
Proposed by Daniel Sitaru – Romania
UNDERGRADUATE PROBLEMS
UP.091. Let be 푎 ∈ ℝ∗ and the continuous functions 푓,푔,ℎ:ℝ → ℝ where 푓 and 푔 are odd and ℎ is even. Prove that:
푓(푥) ⋅ ln 1 + 푒 ( ) ⋅ arctan ℎ(푥) 푑푥 = 푓(푥)푔(푥) arctan ℎ(푥) 푑푥.
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
UP.092. Calculate:
lim→
n ⋅ (n + 1)!( )
− √n! .
Romanian Mathematical Society-Mehedinți Branch 2018
78 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
UP.093. Let (a ) , (b ) be positive real sequences such that there exists lim →∞ ⋅
and lim →∞(b − u ⋅ a ). Find:
a) lim →∞ b − b ;
b) lim →∞( ) − .
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
UP.094. Let (푠 ) , 푠 = ∑ . Calculate:
lim→∞
푠 ⋅ (푛 + 1)!–휋6⋅ √푛! .
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
UP.095. Let (푠 ) , 푠 = ∑ and let (푎 ) be a positive real sequence such that
lim →∞ = 푎 ∈ ℝ∗ . Calculate:
lim→∞
푠 ⋅ 푎 −휋6⋅ 푎 .
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
UP.096. Let (푠 ) , 푠 = ∑ . Calculate:
푙푖푚→∞
푠 ⋅ (2푛 + 1)‼ −휋6⋅ (2푛 − 1)‼ .
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
UP.097. Evaluate 퐼 = ∫ ( )( )( )
푑푥, where 푛 ∈ ℕ∗.
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
UP.098. Let 푎, 푏 ∈ ℝ,푎 < 푏 and 푓,푔:ℝ → ℝ continuous functions such that
푓(푥)푓(푎 + 푏 − 푥) = 1,푔(푥) = 푔(푎 + 푏 − 푥),∀푥 ∈ ℝ.
Show that:
푔(푥)1 + 푓(푥)푑푥 =
12⋅ 푔(푥)푑푥
Proposed by D.M. Bătinețu – Giurgiu, Neculai Stanciu – Romania
UP.099. In an arbitrary triangle 퐴퐵퐶 denote by 푙 ,푚 ,ℎ respectively the lengths of the internal angle-bisector, the median and the altitude corresponding to the side 푎 = 퐵퐶 of the triangle. Prove that:
Romanian Mathematical Society-Mehedinți Branch 2018
79 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
a) + + ≥ 2 ⋅ ⋅ + 1.
b) + + ≤ 2 ⋅ ⋅ + 1.
c) explain why each of a) and b) are equivalent to the fundamental inequality of the triangle.
Proposed by Vasile Jiglău – Romania
UP.100. In 훥퐴퐵퐶;푚 ,푚 ,푚 - median’s length. Prove that:
3(푎 + 푏 + 푐 ) < 4(푎푚 + 푏푚 + 푐푚 )
Proposed by Daniel Sitaru – Romania
UP.101. Prove that if 푎, 푏, 푐 ∈ (1,∞) then:
3√2 + 푥 sin휋
3푥푑푥 + 푥 sin
휋3푥
푑푥 + 푥 sin휋
3푥푑푥 > 3 + 푎 + 푏 + 푐
Proposed by Daniel Sitaru – Romania
UP.102. Solve for real numbers: 푛 + 푛 +⋯+ 푛 + 푛 =√
Proposed by Daniel Sitaru – Romania
UP.103. Prove that in any triangle 퐴퐵퐶 the following relationship holds:
|cos 퐴| + |cos퐵| + |cos퐶| ≤ |cos퐴 cos퐵| + cos퐶2
sin퐵 − 퐴
2
Proposed by Daniel Sitaru – Romania
UP.104. Prove that if 푥 ∈ (0,∞); 푖 ∈ 1,푛;푛 ∈ ℕ; 푛 ≥ 3;푥 = 푥 ; 푥 , 푥 ⋅ … ⋅ 푥 = 1, then
+ + 1
푥 + 푥 푥 + 푥≥ 푛√3
Proposed by Daniel Sitaru – Romania
UP.105. In ABC; a, b, c – length sides; s - semiperimeter; A, B, C - angle’s measures. Prove that:
Ab
+Bc
+Ca
Ac
+Ba
+Cb
Aa
+Bb
+Cc
≥π
216s
Proposed by Daniel Sitaru – Romania
Romanian Mathematical Society-Mehedinți Branch 2018
80 ROMANIAN MATHEMATICAL MAGAZINE NR. 20
INDEX OF PROPOSERS RMM-20
Nr.crt. Numele și prenumele Nr.crt. Numele și prenumele 1 DANIEL SITARU-ROMANIA 17 GHEORGHE CĂINICEANU-ROMANIA 2 D.M.BĂTINEȚU-GIURGIU-ROMANIA 18 DANA HEUBERGER-ROMANIA 3 IONICĂ CONSTANTIN-ROMANIA 19 CRISTIAN HEUBERGER-ROMANIA 4 CLAUDIA NĂNUȚI-ROMANIA 20 DAN NEDEIANU-ROMANIA 5 NECULAI STANCIU-ROMANIA 21 MARIUS DRĂGAN-ROMANIA 6 DAN NĂNUȚI-ROMANIA 22 IULIANA TRAȘCĂ-ROMANIA 7 CARMEN-VICTORIȚA CHIRFOT-ROMANIA 23 ROXANA MIHAELA STANCIU-ROMANIA 8 MARIN CHIRCIU-ROMANIA 24 MARIAN URSĂRESCU-ROMANIA 9 VASILE JIGLĂU-ROMANIA 25 MARIA ELENA PANAITOPOL-ROMANIA
10 NICOLAE PAPACU-ROMANIA 26 GHEORGHE BOROICA-ROMANIA 11 ANDREI ȘTEFAN MIHALCEA-ROMANIA 27 GEORGE APOSTOLOPOULOS-GREECE 12 HUNG NGUYEN VIET-VIETNAM 28 MARTIN LUKAREVSKI-SKOPJE 13 NGUYEN NGOC TU-VIETNAM 29 VADIM MITROFANOV-UKRAINE 14 HOANG LE NHAT TUNG-VIETNAM 30 REDWANE EL MELLAS-MOROCCO 15 DO QUOC CHINH-VIETNAM 31 SHIVAM SHARMA-INDIA 16 MEHMET SAHIN-TURKEY 32 ADIL ABDULLAYEV-AZERBAIDIAN
NOTĂ: Pentru a publica probleme propuse, articole și note matematice în RMM puteți trimite materialele pe mailul: [email protected] All solutions for proposed problems can be finded on the http//:www.ssmrmh.ro which is the adress of Romanian Mathematical Magazine-Interactive Journal.