Junior problems - AwesomeMath...Junior problems J469. Letaandbbedistinctrealnumbers. Provethat...

Junior problems

J469. Let a and b be distinct real numbers. Prove that

(3a + 1)(3b + 1) = 3a2b2 + 1

if and only if( 3√a + 3

√b)

3= a2b2.

Proposed by Adrian Andreescu, University of Texas at Austin, USA

First solution by by the authorThe first condition is equivalent to a + b − a2b2 + 3ab = 0.

The identity x3 + y3 + z3 − 3xyz = 1

2(x + y + z)[(x − y)2 + (y − z)2 + (z − x)2]

shows that if x, y, z are real numbers, not all equal, then x3 + y3 + z3 −3xyz = 0 if and only if x+ y + z = 0.In our problem, x = 3

√a, y = 3

√b, and z = − 3

√a2b2, so

3√a + 3

√b − 3

√a2b2 = 0,

implying

3√a + 3

√b = 3

√a2b2.

Hence the conclusion.

Second solution by Takuji Imaiida, Fujisawa, Kanagawa, JapanLet x = 3

√a + 3

√b and y = 3

√ab, then it suffices to show that

3y3 + x(x2 − 3y) = y6 (1)

if and only ifx = y2. (2)

Since 3y3 +x(x2 − 3y) = y6⇔ (x− y2)(x2 +xy2 + y4 − 3y) = 0, clearly, (2)⇒ (1). Conversely, assume that (1)holds. Since a and b are distinct, we obtain x2 − 4y > 0. Therefore

x2 + xy2 + y4 − 3y > x2 + xy2 + y4 − 3

4x2

= 1

4x2 + xy2 + y4 = (1

2x2 + y2)2 ≥ 0.

x − y2 = 0, (2) holds and we are done.

Also solved by Ioannis D. Sfikas, Athens, Greece; Arkady Alt, San Jose, CA, USA; Corneliu Mănescu-Avram, Ploies,ti, Romania; G. C. Greubel, Newport News, VA, USA; Santosh Kumar Mvrk, Hyderabad, India;Marin Chirciu, Colegiul Nat,ional "Zinca Golescu", Pites,ti, Romania; Nguyen Viet Hung, Hanoi Universityof Science, Vietnam; Pantelis.N, Athens, Greece; Pradyumna Atreya, Mumbai, India; Saloni Gole, FIITJEEMumbai, India; Albert Stadler, Herrliberg, Switzerland; Titu Zvonaru, Comănes,ti, Romania; Polyahedra,Polk State College, USA; Bryant Hwang, Korea International School, South Korea; Dumitru Barac, Sibiu,Romania; Prajnanaswaroopa S, Amrita University, Coimbatore, India.

Mathematical Reflections 1 (2019) 1

J470. Solve in real numbers the equation

(x3 − 2)3 + (x2 − 2)2 = 0.

Proposed by Nguyen Viet Hung, Hanoi University of Science, Vietnam

Solution by Polyahedra, Polk State College, USADenote by f(x) the left-hand side of the equation. Then f(1) = 0 and we show that f has no other real zero.

If x > 3√

2 then f(x) > 0.If 1 < x ≤ 3

√2, then 0 ≤ 2 − x3 < 2 − x2 < 1, so

(2 − x3)3 ≤ (2 − x3)2 < (2 − x2)2 ,

thus f(x) > 0.Finally, consider x < 1. Then x2 − 2 ≥ x3 − 2 and

(x2 − 2) + (x3 − 2) = (x − 1) (x2 + 2x + 2) − 2 < 0,

thus ∣x2 − 2∣ ≤ 2 − x3. Therefore,(x2 − 2)2 ≤ (2 − x3)2 < (2 − x3)3 ,

that is, f(x) < 0.

Also solved by Takuji Imaiida, Fujisawa, Kanagawa, Japan; Bryant Hwang, Korea International School,South Korea; Albert Stadler, Herrliberg, Switzerland; G. C. Greubel, Newport News, VA, USA; IoannisD. Sfikas, Athens, Greece; Nicusor Zlota ‚”Traian Vuia” Technical College, Focsani, Romania; Pantelis.N,Athens, Greece; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Rome, Italy; Akash Singha Roy,Chennai Mathematical Institute, India; Titu Zvonaru, Comănes,ti, Romania.


J471. Find all real numbers a for which the equation

( x

x − 1)2

+ ( x

x + 1)2

= a

has four distinct real roots.


Solution by the author

Clearly, a ≥ 0. Completing the square givesx

x − 1+ ( x

x + 1)2

− 2x2

x2 − 1= a, which rewrites

( 2x2

x2 − 1)2

− 2x2

x2 − 1= a.

With the substitution2x2

x2 − 1= t, this is equivalent to t2 − t + 1

4= a + 1

4, that is,

t − 1

2= b or t − 1

2= −b,

where b =√a + 1

4.

It follows that (2b − 3)x2 = 2b + 1, implying b > 3

2, or (2b + 3)x2 = 2b − 1, which has two distinct solutions

for b > 1

2(which is always true for a diff from 0). We cannot have b < 1

2because that would imply a < 0, a

contradiction. It follows that a + 1

4> 9

4, and so the answer is a > 2. The four solutions are distinct as b ≠ 0

implies2b + 1

2b − 3≠ 2b − 1

2b + 3.

Also solved by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Rome, Italy; Takuji Imaiida,Fujisawa, Kanagawa, Japan; Polyahedra, Polk State College, USA; Dumitru Barac, Sibiu, Romania; AkashSingha Roy, Chennai Mathematical Institute, India; Albert Stadler, Herrliberg, Switzerland; Ángel Plaza,University of Las Palmas de Gran Canaria, Spain; G. C. Greubel, Newport News, VA, USA; Ioannis D. Sfikas,Athens, Greece; Konstantinos Kritharidis, American College of Greece - Pierce, Athens, Greece; Lukas Seier,Charters Sixth Form, Sunningdale, England, UK; Marin Chirciu, Colegiul Nat,ional "Zinca Golescu", Pites,ti,Romania; Nicusor Zlota ‚”Traian Vuia” Technical College, Focsani, Romania; Oana Prajitura, College atBrockport, SUNY, NY, USA; Santosh Kumar Mvrk, Hyderabad, India; Titu Zvonaru, Comănes,ti, Romania;Arkady Alt, San Jose, CA, USA; Sebastian Foulger, Charters Sixth Form, Sunningdale, England, UK.


J472. Let a, b, c be positive numbers such that ab + bc + ca = 1. Prove that

a√b2 + 1 + b

√c2 + 1 + c

√a2 + 1 ≥ 2.

Proposed by An Zhenping, Xianyang Normal University, China

First solution by Polyahedra, Polk State College, USABy the Cauchy-Schwarz inequality,

√(b2 + 1) (c2 + 1) ≥ bc + 1, etc. Hence,

(a√b2 + 1 + b

√c2 + 1 + c

√a2 + 1)

2= a2 (b2 + 1) + b2 (c2 + 1) + c2 (a2 + 1)

+2ab√

(b2 + 1) (c2 + 1) + 2bc√

(c2 + 1) (a2 + 1) + 2ca√

(a2 + 1) (b2 + 1)≥ (ab)2 + (bc)2 + (ca)2 + a2 + b2 + c2 + 2ab(bc + 1) + 2bc(ca + 1) + 2ca(ab + 1)= (ab + bc + ca)2 + (a + b + c)2 ≥ 1 + 3(ab + bc + ca) = 4,

completing the proof.

Second solution by Polyahedra, Polk State College, USAAs a branch of the hyperbola y2 − x2 = 1, the function f(x) =

√x2 + 1 is convex. By Jensen’s inequality,

a√b2 + 1 + b

√c2 + 1 + c

√a2 + 1 ≥ (a + b + c)

¿ÁÁÀ(ab + bc + ca

a + b + c )2

+ 1

=√

1 + (a + b + c)2 ≥√

1 + 3(ab + bc + ca) =√

4 = 2.

Also solved by Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Bryant Hwang, Korea In-ternational School, South Korea; Dumitru Barac, Sibiu, Romania; Prajnanaswaroopa S, Amrita University,Coimbatore, India; Akash Singha Roy, Chennai Mathematical Institute, India; Jamal Gadirov, Ishik Univer-sity, Iraq; Marin Chirciu, Colegiul Nat,ional "Zinca Golescu", Pites,ti, Romania; Nguyen Viet Hung, HanoiUniversity of Science, Vietnam; Nicusor Zlota ‚”Traian Vuia” Technical College, Focsani, Romania; PaoloPerfetti, Università degli studi di Tor Vergata Roma, Rome, Italy; Sarah B. Seales, Prescott, AZ, USA;Ioannis D. Sfikas, Athens, Greece; Soumyadeep Paul, D.A.V. Public School, Haldia, India; Albert Stad-ler, Herrliberg, Switzerland; Titu Zvonaru, Comănes,ti, Romania; Loreta Arzumanyan, "Quantum" College,Armenia; Arkady Alt, San Jose, CA, USA.


J473. Let a, b, c be distinct real numbers. Prove that

( a

b − a)2

+ ( b

c − b)2

+ ( c

a − c)2

≥ 1.

Proposed by Anish Ray, Institute of Mathematics, Bhubaneswar, India

Solution by Nguyen Viet Hung, Hanoi University of Science, VietnamLet

a

b − a = x, b

c − b = y,c

a − c = z

then easily to see thatxyz = (x + 1)(y + 1)(z + 1).

This impliesxy + yz + zx + x + y + z + 1 = 0.

Using this relation we get

x2 + y2 + z2 = (x + y + z)2 − 2(xy + yz + zx)= (x + y + z)2 + 2(x + y + z + 1)= (x + y + z + 1)2 + 1

≥ 1

and we are done.

Also solved by Polyahedra, Polk State College, USA; Dumitru Barac, Sibiu, Romania; Akash SinghaRoy, Chennai Mathematical Institute, India; Albert Stadler, Herrliberg, Switzerland; Ángel Plaza, Universityof Las Palmas de Gran Canaria, Spain; Ioannis D. Sfikas, Athens, Greece; Nicusor Zlota ‚”Traian Vuia”Technical College, Focsani, Romania; Pantelis.N, Athens, Greece; Paolo Perfetti, Università degli studi diTor Vergata Roma, Rome, Italy; Titu Zvonaru, Comănes,ti, Romania; Arkady Alt, San Jose, CA, USA.


J474. Let k be a positive integer. Suppose x and y are positive integers such that for every positive integern, n > k,

xn−k + yn ∣ xn + yn+k.

Prove that x = y.

Proposed by Valentio Iverson, Medan, North Sumatra, Indonesia

Solution by Polyahedra, Polk State College, USAFirst, consider x > y. Since xn + yn+k = (xn−k + yn)xk + (yk − xk) yn, xn−k + yn must divide (yk − xk) yn forall n > k. But as n→∞,

0 >(yk − xk) yn

xn−k + yn = yk − xk1xk

(xy )n+ 1→ 0,

thus the ratio cannot be an integer for sufficiently large n.Next, consider x < y. Since xn+yn+k = (xn−k + yn) yk+(xk − yk)xn−k, xn−k+yn must divide (xk − yk)xn−k

for all n > k. But as n→∞,

0 >(xk − yk)xn−k

xn−k + yn = xk − yk

1 + xk ( yx)n → 0,

thus the ratio cannot be an integer for sufficiently large n.

Also solved by Akash Singha Roy, Chennai Mathematical Institute, India; Sarah B. Seales, Prescott, AZ,USA; Albert Stadler, Herrliberg, Switzerland.


Senior problems

S469. Let ABCD be a kite with ∠A = 5∠C and AB ⋅BC = BD2. Find ∠B.

Proposed by Titu Andreescu, University of Texas at Dallas, USA

Solution by Miguel Amengual Covas, Cala Figuera, Mallorca, SpainIf ∠D = x, then ∠A = 5x, and since the angles of △ABD add up to 180○, we have 5x < 180○, implyingx < 36○.

From isosceles triangles ABD and BCD, we have

BD = 2 ⋅AB ⋅ sin 5x

2and BD = 2 ⋅BC ⋅ sin x

2,

respectively.Hence

BD2 = 4 ⋅AB ⋅BC ⋅ sin 5x

2sin

x

2.

Dividing both sides by BD2 = AB ⋅BC gives

sin5x

2sin

x

2= 1

4,

which is equivalent to

cos 3x − cos 2x = −1

2.

This, in turn, can be written as a cubic in cosx

8 cos3 x − 4 cos2 x − 6 cosx + 3 = 0

which factors immediately so that its solutions (cosx = 12 ,√

32 ,−

√

32 ) can be read off.

From x < 36○, the only admissible solution is cosx =√

32 and x = 30○. Thus

∠D = 30○

and∠A = 150○.

Since the angles of kite ABCD add up to 360○ and ∠B =∠C, we conclude that ∠B = 90○.

Also solved by Santosh Kumar Mvrk, Hyderabad, India; Marin Chirciu, Colegiul Nat,ional "Zinca Go-lescu", Pites,ti, Romania; Nicusor Zlota ‚”Traian Vuia” Technical College, Focsani, Romania; PradyumnaAtreya, Mumbai, India; Ioannis D. Sfikas, Athens, Greece; Albert Stadler, Herrliberg, Switzerland; TituZvonaru, Comănes,ti, Romania; Kevin Soto Palacios, Huarmey, Perú.


S470. Let x, y, z be positive real numbers such that xyz(x + y + z) = 4. Prove that

(x + y)2 + 3(y + z)2 + (z + x)2 ≥ 8√

7.


First solution by Nguyen Viet Hung, Hanoi University of Science, VietnamWe know that in any triangle ABC and for all real numbers u, v,w such that uv + vw +wu ≥ 0

ua2 + vb2 +wc2 ≥ 4S√uv + vw +wu

where S denotes the are of the triangle. Using Ravi’s substitutions a = y+z, b = z+x, c = x+y with x, y, z > 0then the above result becomes

u(y + z)2 + v(z + x)2 +w(x + y)2 ≥ 4√xyz(x + y + z)(uv + vw +wu).

Now we apply this result for (u, v,w) = (3,1,1) and note that the condition xyz(x + y + z) = 4 to obtain

(x + y)2 + 3(y + z)2 + (z + x)2 ≥ 8√

7

as desired.

Second solution by Arkady Alt, San Jose, CA, USALet p ∶= y + z, q ∶= yz.Then p2 ≥ 4q, 4 = xyz (x + y + z) = qx2 + pqx and, therefore,

(x + y)2 + 3(y + z)2 + (z + x)2 = 2 (x2 + x (y + z) + 4 (y + z)2 − 2yz) =

2 (4p2 + x2 + px − 2q) = 2(4p2 + qx2 + pqxq

− 2q) = 2(4p2 + 4

q− 2q) ≥

2(4p2 + 4

p2/4 − 2 ⋅ p2

4) = 2(4p2 + 7

p2) ≥

2 ⋅ 2√

4p2 ⋅ 7

p2= 8

√7.

Also solved by Haosen Chen, Zhejiang, China; Dumitru Barac, Sibiu, Romania; Ioannis D. Sfikas, Athens,Greece; Nicusor Zlota ‚”Traian Vuia” Technical College, Focsani, Romania; Albert Stadler, Herrliberg, Swi-tzerland.


S471. Prove that the following inequality holds for all positive real numbers a, b, c:

1

a+ 1

b+ 1

c+ 9 (a + b + c)ab + bc + ca ≥ 8( a

a2 + bc +b

b2 + ca +c

c2 + ab)


Solution by Arkady Alt, San Jose, CA, USASince for any positive a, b, c holds inequality(Vasile Cirtoaje, Algebraic Inequalities, Old and New Methods, Inequality 59,p.13.)

∑cyc

a

a2 + bc ≤∑cyc1

b + c (1)

remains to prove inequality1

a+ 1

b+ 1

c+ 9 (a + b + c)ab + bc + ca ≥ 8 ∑

cyc

1

b + c ⇐⇒

ab + bc + caabc

+ 9 (a + b + c)ab + bc + ca ≥ 8∑

cyc

1

b + c . (2)

Let p ∶= ab + bc + ca, q ∶= abc. Also we may assume that a + b + c = 1 (due homogeneity of (2)).

Then ∑cyc

1

b + c =8 (1 + p)p − q and since

3p = 3 (ab + bc + ca) ≤ (a + b + c)2 = 1,3q = 3abc (a + b + c) ≤ (ab + bc + ca)2 = p2

we obtainab + bc + ca

abc+ 9 (a + b + c)ab + bc + ca − 8∑

cyc

1

b + c =p

q+ 9

p− 8 (1 + p)

p − q ≥

12

p− 8 (1 + p)

p − p2

3

= 12 (1 − 3p)p (3 − p) ≥ 0.

Also solved by Dumitru Barac, Sibiu, Romania; Akash Singha Roy, Chennai Mathematical Institute,India; Loreta Arzumanyan, "Quantum" College, Armenia; Albert Stadler, Herrliberg, Switzerland; IoannisD. Sfikas, Athens, Greece; Nicusor Zlota ‚”Traian Vuia” Technical College, Focsani, Romania; Titu Zvonaru,Comănes,ti, Romania.


S472. Let ABC be a triangle with ∠B and ∠C acute and let D be the foot of the altitude from A. Provethat ∠A is right if and only if

BD

AB2+ CD

AC2= 2

BC.


Solution by Albert Stadler, Herrliberg, SwitzerlandLet x = BD,y = CD, b = AB, c = AC. Assume first that ∠A = π/2. Then, b2 = x(x + y) and c2 = y(x + y).Therefore,

x

b2+ y

c2= 2

x + y ,

as required.

Assume next thatx

b2+ y

c2= 2

x + y . Clearly, b2−x2 = c2−y2. Solving these two equations for x and y yields

x = b2√b2 + c2

, y = c2√b2 + c2

.

Then b2 + c2 − (x + y)2 = 0, which proves that △ABC is a right triangle.

Also solved by Dumitru Barac, Sibiu, Romania; Akash Singha Roy, Chennai Mathematical Institute,India; Kevin Soto Palacios, Huarmey, Perú; Corneliu Mănescu-Avram, Ploies,ti, Romania; Marin Chir-ciu, Colegiul Nat,ional "Zinca Golescu", Pites,ti, Romania; Ioannis D. Sfikas, Athens, Greece; TelemachusBaltsavias, Keramies Junior High School, Kefallonia, Greece; Titu Zvonaru, Comănes,ti, Romania.


S473. Let a, b, c be positive real numbers. Prove that

(a − b)4 + (b − c)4 + (c − a)4 ≤ 6 (a4 + b4 + c4 − abc(a + b + c)) .

Proposed by Nicus,or Zlota, Focs,ani, Romania

Solution by Nguyen Viet Hung, Hanoi University of Science, VietnamThe desired inequality is equivalent to

∑cyc

(a4 − 4a3b + 6a2b2 − 4ab3 + b4) ≤ 6(a4 + b4 + c4 − abc(a + b + c)),

3(a2b2 + b2c2 + c2a2) + 3abc(a + b + c) ≤ 2(a4 + b4 + c4) + 2ab(a2 + b2) + bc(b2 + c2) + ca(c2 + a2).

We have

2(a4 + b4 + c4) +∑cyc

2ab(a2 + b2) ≥ 2(a2b2 + b2c2 + c2a2) +∑cyc

4a2b2

= 6(a2b2 + b2c2 + c2a2)≥ 3(a2b2 + b2c2 + c2a2) + 3abc(a + b + c).

The proof is completed.

Also solved by Dumitru Barac, Sibiu, Romania; Akash Singha Roy, Chennai Mathematical Institute,India; Loreta Arzumanyan, "Quantum" College, Armenia; Albert Stadler, Herrliberg, Switzerland; HaosenChen, Zhejiang, China; Ioannis D. Sfikas, Athens, Greece; Arkady Alt, San Jose, CA, USA; Paolo Perfetti,Università degli studi di Tor Vergata Roma, Rome, Italy; Titu Zvonaru, Comănes,ti, Romania.


S474. Let a, b, c, d be real numbers such that a2 + b2 + c2 + d2 = 12. Prove that

a3 + b3 + c3 + d3 + 9(a + b + c + d) ≤ 84

Proposed by Marius Stănean, Zalău, Romania

Solution by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Rome, Italy

∑cyc

(a3 + 9a) ≤ 84

a3 + 9a is convex, therefore, the maximum of the above sum occurs when at least three variables are equal.Set b = c = d = x. The inequality becomes

a3 + 3x3 + 9(a + 3x) ≤ 84

provided a2 + 3x2 = 12. a =√

12 − 3x2, 0 ≤ x ≤ 2. The inequality becomes√

12 − 3x2 + 3x3 + 9√

12 − 3x2 + 27x − 84 ≤ 0

if and only if(√

12 − 3x2 + 9√

12 − 3x2)2− (84 − 3x3 − 27x)2 ≤ 0

The inequality becomes36(x4 + 2x3 − 6x2 − 28x + 49)(x − 1)2 ≥ 0

Let x = 2t/(1 + t), t ≥ 0, the quantity x4 + 2x3 − 6x2 − 28x + 49 becomes

t4 − 4t3 + 102t2 + 140t + 49

(1 + t)4 ≥ 0

becauset4 + 4t2 ≥ 4t3

and the result follows.

Also solved by Haosen Chen, Zhejiang, China; Ioannis D. Sfikas, Athens, Greece; Nicusor Zlota ‚”TraianVuia” Technical College, Focsani, Romania; Albert Stadler, Herrliberg, Switzerland.


Undergraduate problems

U469. Let x > y > z > t > 1 be real numbers. Prove that

(x − 1)(z − 1) ln y ln t > (y − 1)(t − 1) lnx ln z.

Proposed by Ángel Plaza, University of Las Palmas de Gran Canaria, Spain

Solution by Oana Prajitura, College at Brockport, SUNY, NY, USAAll quantities in the inequality are positive. Therefore the inequality is equivalent to

x − 1

lnx⋅ z − 1

ln z> y − 1

ln y⋅ t − 1

ln t

Let f ∶ (1,∞)→ R given by

f(x) = x − 1

lnx.

To prove the inequality it suffices to show that f is a strictly increasing function.

f ′(x) =lnx − x−1

x

(lnx)2 = xlnx − x + 1

x(lnx)2

The denominator of the last fraction is strictly greater than 0. I need to show that the same is true forthe numerator.

Let g ∶ (0,∞)→ R, given by g(x) = x lnx − x + 1. Then g′(x) = lnx + 1 − 1 = lnx > 0 on (1,∞). Thereforefor x ∈ (1,∞)

g(x) > limx→1

g(x) = g(1) = 0

which completes the proof.

Also solved by Dumitru Barac, Sibiu, Romania; Akash Singha Roy, Chennai Mathematical Institute,India; Ioannis D. Sfikas, Athens, Greece; Thiago Landim de Souza Leão, Federal University of Pernambuco,Recife, Brazil; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Rome, Italy; Albert Stadler,Herrliberg, Switzerland; Arkady Alt, San Jose, CA, USA.


U470. Let n be a positive integer. Evaluate

limx→0

1 − cosn x cosnx

x2


Solution by Thiago Landim de Souza Leão, Federal University of Pernambuco, Recife, BrazilSince cosx = 1 − x2

2 +O(x4) and since (1 + x)n = 1 + nx +O(x2),

cosnx = 1 − n2x2

2+O(x4)

and

cosn x = (1 − x2

2+O(x4))n = 1 − nx

2

2+O(x4).

Therefore,

limx→0

1 − cosn x cosnx

x2= limx→0

1 − (1 − nx22 +O(x4)(1 − n2x2

2 +O(x2)x2

= limx→0

1 − 1 + n(n+1)2 x2 +O(x4)x2

= n(n + 1)2

.

Also solved by Daniel López-Aguayo, Instituto Tecnológico y de Estudios Superiores de Monterrey, CampusMonterrey, Mexico; Dumitru Barac, Sibiu, Romania; Prajnanaswaroopa S, Amrita University, Coimbatore,India; Akash Singha Roy, Chennai Mathematical Institute, India; Lukas Seier, Charters Sixth Form, Sun-ningdale, England, UK; Albert Stadler, Herrliberg, Switzerland; Aenakshee Roy, FIITJEE Chembur, Mumbai,India; Ángel Plaza, University of Las Palmas de Gran Canaria, Spain; Corneliu Mănescu-Avram, Ploies,ti,Romania; G. C. Greubel, Newport News, VA, USA; Ioannis D. Sfikas, Athens, Greece; Marin Chirciu, Co-legiul Nat,ional "Zinca Golescu", Pites,ti, Romania; Matthew Too, College at Brockport, SUNY, NY, USA;Nicusor Zlota ‚”Traian Vuia” Technical College, Focsani, Romania; Oana Prajitura, College at Brockport,SUNY, NY, USA; Paolo Perfetti, Università degli studi di Tor Vergata Roma, Rome, Italy; PradyumnaAtreya, Mumbai, India; Santosh Kumar Mvrk, Hyderabad, India; Sebastian Foulger, Charters Sixth Form,Sunningdale, England, UK.


U471. Let f(x) = ax2 + bx + c, where a < 0 < b and b 3√c ≥ 3

8 . Prove that

f ( 1

∆2) ≥ 0

where ∆ = b2 − 4ac


First solution by the authorSo b, c, and ∆ are positive and it suffices to prove that

a + b(∆2) + c(∆2)2 >= 0.

Let g(x) = cx2 + bx+ a, having the same discriminant ∆ and roots x1 and x2, x1 ≤ x2. It suffices to provethat

∆2 ≥ x2 =−b +

√∆

2c.

This inequality rewrites

∆2 + b

6c+ b

6c+ b

6c≥

√∆

2c

and follows by the AM-GM Inequality and the condition b 3√c ≥ 3

8.

Second solution by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Rome, Italy

f ( 1

∆2) = c∆

4 + b∆2 + a∆4

≥ 0 ⇐⇒ c∆4 + b∆2 + a

that is

∆2 ≤ −b −√b2 − 4ac

cor ∆2 ≥ −b +

√b2 − 4ac

c

Since c > 0 by b > 0 and b 3√c ≥ 3

8 , ∆2 ≤ −b−√

b2−4acc is impossible. Instead

∆2 ≥ −b +√b2 − 4ac

c⇐⇒ 2c∆2 + b ≥

√∆

2c∆2 + b = 2c∆2 + b3+ b

3+ b

3≥®

AGM

4

334

(2c∆2b3)14 ≥

√∆

if and only if

294 c

14 b

34

1

334

≥ 1 ⇐⇒ 23c13b

3≥ 1

that is the result.

Also solved by Ioannis D. Sfikas, Athens, Greece; Albert Stadler, Herrliberg, Switzerland.


U472. If f, g, h ∶ R → R are derivatives, find whether or not the function max{f, g, h} is the derivative of afunction.

Proposed by Dorin Andrica, Babes,-Bolyai University, Cluj-Napoca, Romania

Solution by Albert Stadler, Herrliberg, SwitzerlandWe claim that the function max{f, g, h} is not necessarily the derivative of a function. We will construct acounterexample as follows:

Let F (x) = x2 sin1

x2, if x ≠ 0, and F (0) = 0. F is a differentiable function, since F ′(x) = 2x sin

1

x2− 2

xcos

1

x2,

if x ≠ 0, and F ′(0) = limt→0F (t) − F (0)

t= limt→0 tsin

1

t2= 0.

Put f(x) = F ′(x), g(x) = −F ′(x), h(x) = 0. Then

max{f, g, h} = ∣f ∣

.Suppose that ∣f ∣ is the derivative of a (differentiable) function V . Then V (x) is given (up to a constant)

byV (x) = ∫

x

1∣f(t)∣dt,

since ∣f(x)∣ is continuous for x ≠ 0. We will prove that V (0) = ∫ 01 ∣f(t)∣dt = −∞ which cannot hold if V is

throughout differentiable. Indeed,

∣f(x)∣ ≥ 2

∣x∣ ∣cos1

x2∣ − 2∣x∣,

and therefore,

∫1

0∣f(t)∣dt ≥

∞

∑k=1∫

1√

πk−π41

√

πk+π4

∣f(t)∣dt ≥ −1 +√

2∞

∑k=1∫

1√

πk−π41

√

πk+π4

dt

t= −1 +

√2

2

∞

∑k=1

ln(πk + π

4

πk − π4

) =

−1 +√

2

2

∞

∑k=1

ln

⎛⎜⎜⎝

1 + 1

2k − 1

2

⎞⎟⎟⎠=

√2

2

∞

∑k=1

1

2k − 1

2

+O(1) =∞.

Also solved by Ioannis D. Sfikas, Athens, Greece.


U473. For each continuous function f ∶ [0,1]↦ [0,∞), let

If = ∫1

0(2f(x) + 3x)f(x)dx

andJf = ∫

1

0(4f(x) + x)

√xf(x)dx.

Find the minimum of If − Jf over all such functions f .


Solution by Albert Stadler, Herrliberg, SwitzerlandSet g(x) =

√f(x). We need to find the minimum of

∫1

0(2g4(x) + 3xg2(x) − 4

√xg3(x) − x

√xg(x))dx

over all continuous functions g ∶ [0,1]↦ [0,∞). This is a variational problem whose solution is derived fromthe Euler-Lagrange-equation:

0 = ∂

∂g(2g4(x) + 3xg2(x) − 4

√xg3(x) − x

√xg(x)) =

8g3(x) + 6xg(x) − 12√xg2(x) − x

√x = (2g(x) −

√x)3.

We conclude that g(x) = 1

2

√x and finally f(x) = x

4.

Also solved by Ioannis D. Sfikas, Athens, Greece.


U474. Let f ∶ [0,1]→ R be a differentiable function such that f(1) = 0 and

∫1

0xnf(x)dx = 1

Prove that

∫1

0(f ′(x))2 dx ≥ (2n + 3)(n + 1)2

When does the equality occur?


Solution by Paolo Perfetti, Università degli studi di Tor Vergata Roma, Rome, ItalyIntegrating by parts

1 = ∫1

0xnf(x)dx = xn+1

n + 1f(x)∣

1

0−∫

1

0

xn+1

n + 1f ′(x)dx = −∫

1

0

xn+1

n + 1f ′(x)dx

Cauchy–Schwarz yields

1 = −∫1

0

xn+1

n + 1f ′(x)dx ≤ ∫

1

0( x2n+2

(n + 1)2dx)12

(∫1

0(f ′(x))2 dx)

12

whence

∫1

0(f ′(x))2 dx ≥ (2n + 3)(n + 1)2

Also solved by Stroe Octavian and Marin Chirciu, Colegiul Nat,ional "Zinca Golescu", Pites,ti, Roma-nia; Dumitru Barac, Sibiu, Romania; Akash Singha Roy, Chennai Mathematical Institute, India; AlbertStadler, Herrliberg, Switzerland; Jamal Gadirov, Ishik University, Iraq; Ioannis D. Sfikas, Athens, Greece;Joshua Siktar, Carnegie Mellon University, PA, USA; Thiago Landim de Souza Leão, Federal University ofPernambuco, Recife, Brazil; Nicusor Zlota ‚”Traian Vuia” Technical College, Focsani, Romania.


Olympiad problems

O469. Find the greatest constant k such that the following inequality holds for all positive real numbers aand b

1

a3+ 1

b3+ k

a3 + b3 ≥ 16 + 4k

(a + b)3


Solution by Arkady Alt, San Jose, CA, USAAssuming a + b = 1 (due homogeneity) and denoting t ∶= ab ∈ (0,1/4] we obtain

1

a3+ 1

b3+ k

a3 + b3 ≥ 16 + 4k

(a + b)3⇐⇒ 1 − 3t

t3+ k

1 − 3t≥ 16 + 4k ⇐⇒

1 − 3t

t3− 16 − 4k + k

1 − 3t≥ 0 ⇐⇒

(1 − 4t) (4t2 + t + 1)t3

− 3k ⋅ 1 − 4t

1 − 3t≥ 0 ⇐⇒

(1 − 4t) ((1 − 3t) (4t2 + t + 1) − 3kt3)(1 − 3t) t3 ≥ 0

Since1 − 4t

t3 (1 − 3t) > 0 for any t ∈ (0,1

4) then

k ≤(1 − 3t) (4t2 + t + 1)

3t3,∀t ∈ (0,1/4) ⇐⇒ k ≤ inf

t∈(0,1/4)

(1 − 3t) (4t2 + t + 1)3t3

= 8

because(1 − 3t) (4t2 + t + 1)

3t3= 1

3t(1

t− 1)

2

− 4 ≤ 1

3 ⋅ 1/4 ( 1

1/4 − 1)2

− 4 = 8.

Thus greatest constant k = 8 is maximal value of constant k such that inequality1

a3+ 1

b3+ k

a3 + b3 ≥ 16 + 4k

(a + b)3holds for all positive a, b.

Also solved by Santosh Kumar Mvrk, Hyderabad, India; Dumitru Barac, Sibiu, Romania; Akash SinghaRoy, Chennai Mathematical Institute, India; Albert Stadler, Herrliberg, Switzerland; Corneliu Mănescu-Avram, Ploies,ti, Romania; Ioannis D. Sfikas, Athens, Greece; M.A,Prasad, Mumbai, India; Marin Chir-ciu, Colegiul Nat,ional "Zinca Golescu", Pites,ti, Romania; Nicusor Zlota‚ ”Traian Vuia” Technical College,Focsani, Romania; Titu Zvonaru, Comănes,ti, Romania.


O470. Let a, b, c, x, y, z be nonnegative real numbers such that a ≥ b ≥ c, x ≥ y ≥ z and

a + b + c + x + y + z = 6.

Prove that(a + x)(b + y)(c + z) ≤ 6 + abc + xyz.

Proposed by Marius Stănean, Zalău, Romania

Solution by the authorRewrite the above inequality as

abz + bcx + cay + xyc + yza + zxb ≤ 6.

I will prove the following inequality

[(a + b)z + (b + c)x + (c + a)y]2 ≥ 4(abz + bcx + cay)(x + y + z). (1)

This is equivalent to

(a − b)2z2 + (b − c)2x2 + (c − a)2y2 ≥ 2(a − b)(b − c)zx + 2(b − c)(c − a)xy + 2(c − a)(a − b)yz,

or[(a − b)z − (b − c)x + (c − a)y]2 ≥ 4(c − a)(a − b)yz

which is obviously true.Similarly, if in (1) we take (a, b, c, x, y, z)←→ (x, y, z, a, b, c), we get the following inequality

[(x + y)c + (y + z)a + (z + x)b]2 ≥ 4(xyc + yza + zxb)(a + b + c). (2)

By (1) and (2) it follows that

abz + bcx + cay + xyc + yza + zxb ≤

≤[(a + b)z + (b + c)x + (c + a)y]2

4(x + y + z) + [(x + y)c + (y + z)a + (z + x)b]2

4(a + b + c)

=[(a + b)z + (b + c)x + (c + a)y]2 (a + b + c + x + y + z)4(x + y + z)(a + b + c)

=6 [(a + b)z + (b + c)x + (c + a)y]2

4(x + y + z)(a + b + c) .

But,3 [(a + b)z + (b + c)x + (c + a)y] ≤ 2(a + b + c)(x + y + z)

because it reduces to(a + b − 2c)z + (b + c − 2a)x + (c + a − 2b)y ≤ 0,

or(b − c)z − (c − a)z + (c − a)x − (a − b)x + (a − b)y − (b − c)y ≤ 0,

or(a − b)(x − y) + (b − c)(y − z) + (c − a)(z − x) ≥ 0,

obviously true.


Hence, using this result and AM-GM Inequality,

abz + bcx + cay + xyc + yza + zxb ≤2(a + b + c)(x + y + z)3

≤2(a + b + c + x + y + z)212

=6.

Equality holds if and only if a = b = c and x = y = z.

Also solved by M.A,Prasad, Mumbai, India; Albert Stadler, Herrliberg, Switzerland; Titu Zvonaru, Comănes,ti,Romania.


O471. Let a, b, c be positive real numbers such that a2+ b2+ c2+abc = 4. Prove that for all real numbers x, y, zthe following inequality holds

ayz + bzx + cxy ≤ x2 + y2 + z2.


Solution by Nguyen Viet Hung, Hanoi University of Science, VietnamFrom the relation

a2 + b2 + c2 + abc = 4

we deduce that there exist an acute triangle ABC such that

a = 2 cosA, b = 2 cosB, c = 2 cosC.

Then the inequality becomes

2yz cosA + 2zx cosB + 2xy cosC ≤ x2 + y2 + z2.

This is equivalent to(z − y cosA − x cosB)2 + (y sinA − x sinB)2 ≥ 0

which is obvious and we are done.

Also solved by Dumitru Barac, Sibiu, Romania; Akash Singha Roy, Chennai Mathematical Institute,India; Ioannis D. Sfikas, Athens, Greece; M.A,Prasad, Mumbai, India; Marin Chirciu, Colegiul Nat,ional"Zinca Golescu", Pites,ti, Romania; Nicusor Zlota ‚”Traian Vuia” Technical College, Focsani, Romania;Albert Stadler, Herrliberg, Switzerland; Titu Zvonaru, Comănes,ti, Romania.


O472. Let △ABC be an acute triangle and A1,B1,C1 the tangency points between BC,AC,AB and ABCincircle. Circumcircles of △BB1C1,△CB1C1 cut BC in A2, respectively A3, analogously, circumcirclesof △AB1A1,△BA1B1 cut AB in C2,C3 and circumcircles of △AC1A1,△CC1A1 cut AC in C2,C3. IfA2B1 ∩ A3C1 = (A′} ,B2A1 ∩ B3C1 = {B′} ,C2A1 ∩ C3B1 = {C ′}, show that A1A

′, B1B′, C1C

′ areconcurrent lines.

Proposed by Mihaela Berindeanu, Bucharest, Romania

Solution by the authorDenote by I the triange ABC incentre

IA1 ⊥ BC, IB1 ⊥ AC, IC1 ⊥ AB ⇒ IB1AC1 is inscribed in the circle with AI diameter.

AB1 ≡ AC1 (tangents from A to ABC incircle)⇒ △AC1B1 is an isosceles triangle ⇒ ∡AC1B1 ≡∡AB1C1 =

180○ −∡A2

= 90○ − ∡A2


• Show thatA′A2A3 is an isisceles triangle. (Analogously△B′B2B3 and△C ′C2C3 are isosceles triangles).

B,A2,B1,C1 = concyclic points⇒∡A3A2B1 ≡∡B1C1A(suplement∡B1C1B)C,B1,C1,A3 = concyclic points⇒∡C1A2A3 ≡∡C1B1A(suplement∡C1B1C) }⇒

⇒∡A′A3A2 ≡∡A′A2A3 = 90○ − ∡BAC2

⇒∡A3A′A2 ≡∡BAC

So, A′,A,C1, I,B1 are concyclic points on the circle with AI diameter. ⇒ IC1 = IB1 ⇒ ∡IA′C1 =∡IA′B1

• Show that A′, I,A1 are collinear points (analogously B′, I,B1 and C ′, I,C1 )In the isosceles triangle A3A

′A2, A′I is angle bisector ⇒ A′I ⊥ BC

From A′I ⊥ BC and IA1 ⊥ BC ⇒ A′, I,A1 = collinear points.

• Show that A1A′ ∩B1B

′ ∩C1C′ = {I}

I ∈ A1A′

I ∈ B1B′

I ∈ C1C′

⎫⎪⎪⎪⎬⎪⎪⎪⎭⇒ A1A

′ ∩B1B′ ∩C1C

′ = {I}

Also solved by Shuborno Das, Ryan International School, Bangalore, India; Loreta Arzumanyan, "Quan-tum" College, Armenia.


O473. Let x, y, z be positive real numbers such that x6 + y6 + z6 = 3. Prove that

x + y + z + 12 ≥ 5(x6y6 + y6z6 + z6x6).

Proposed by Hoan Le Nhat Tung, Hanoi, Vietnam

Solution by Nguyen Viet Hung, Hanoi University of Science, VietnamWe rewrite the inequality as

x + y + z + 12 ≥ 5

2[(x6 + y6 + z6)2 − (x12 + y12 + z12)]

or equivalently5(x12 + y12 + z12) + 2(x + y + z) ≥ 21.

Now we use the AM-GM inequality to obtain

x12 +⋯ + x12´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

15

+x +⋯ + x´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

6

+1 +⋯ + 1´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

10

≥ 31 31√

(x12)15x6 = 31x6.

Similarly15y12 + 6y + 10 ≥ 31y6,

15z12 + 6z + 10 ≥ 31z6.

Adding these three inequalities we get

15(x12 + y12 + z12) + 6(x + y + z) + 30 ≥ 31(x6 + y6 + z6) = 93.

This yields5(x12 + y12 + z12) + 2(x + y + z) ≥ 21.

The proof is completed. Equality occurs if and only if x = y = z = 1.

Also solved by Dumitru Barac, Sibiu, Romania; Loreta Arzumanyan, "Quantum" College, Armenia;Lukas Seier, Charters Sixth Form, Sunningdale, England, UK; M.A,Prasad, Mumbai, India; Nicusor Zlota‚”Traian Vuia” Technical College, Focsani, Romania; Albert Stadler, Herrliberg, Switzerland.


O474. Let P (x) = adxd + ad−1xd−1 + ⋅ ⋅ ⋅ + a2x2 + a0 be a polynomial with positive integer coefficients of degreed ≥ 2. We define the sequence (bn)n≥1, where b1 = a0 and bn = P (bn−1), for all n ≥ 2. Prove that for alln ≥ 2, there is a prime p such that p ∣ bn and p does not divide b1 . . . bn−1.

Proposed by Navid Safaei, Sharif University of Technology, Tehran, Iran

Solution by the authorWe shall argue by contradiction. Let there be a positive integer n ≥ 2 that appears in prime decompositionof b1 . . . bn−1. Let q be an arbitrary prime divisor of bn, then bn = qrl where gcd(l, q) = 1. Now, one can findthat

bn+1 = P (bn) = ad(qrl)d + ad−1(qrl)d−1 + ⋅ ⋅ ⋅ + a2(qrl)2 + a0 ≡ a0 = b1 (mod q1+r)

Hence, one can inductively prove that for all i we have bn+i ≡ bi (mod q1+r). That is,

bn+i+1 = P (bn+i) ≡ P (bi) = bi+1 (mod q1+r).

Now, we find that bn ≡ b2n ≡ ⋅ ⋅ ⋅ ≡ bkn (mod q1+r). Since vp(bn) = r, we find that:

vp(bn) = vp(b2n) = ⋅ ⋅ ⋅ = vp(bkn) = r.

By our assumption, there must exist an index i,1 ≤ i ≤ n − 1 such that bi must be divisible by q. Repeatingthe same procedure yields to

vp(bi) = vp(b2i) = . . .

This implies that vp(bi) = vp(bni) = vp(bn) = r. Then for any prime divisor of bn, the exponent of it would bethe same as exponent of p in prime decomposition of bi for some 1 ≤ i ≤ n − 1. Hence,

bn ∣ b1 . . . bn−1.

Therefore, bn ≤ b1 . . . bn−1. But bn = P (bn−1) > b2n−1, thus, bn−1 <√bn. Now,

bn−k <√bn−k+1 < 4

√bn−k+2 < ⋅ ⋅ ⋅ < bn

1

2k .

Yielding 0 < b1 . . . bn−1 < bn1

2n−1 bn1

2n−2 . . . bn12 = bn

12+⋅⋅⋅+

12n−1 < bn. Contradiction!

Note that the last inequality is true, because1

2+ ⋅ ⋅ ⋅ + 1

2n−1< 1.

Also solved by Prajnanaswaroopa S, Amrita University, Coimbatore, India; M.A,Prasad, Mumbai, India.


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