FYHS-Kulai by Chtan 1

Chapter 28

INEQUALITIES

不等式

What will be taught in this chapter?

1. Some fundamental properties of inequalities.

2. Logarithmic function inequalities .

3. absolute function inequalities .

4. Use the To determine relationship between coeff. and roots.

FYHS-Kulai by Chtan 3

Some properties of inequalities

𝟏 .𝒂>𝒃⇔𝒃<𝒂

2

3

4

5

6

7

9.

10.

11. 𝒏∈𝒁 ,𝒏>𝟏

12.

𝒏∈𝒁 ,𝒏>𝟏

13. have same sign.

have opposite sign.

Some common

inequalities formulae

𝟏 .𝒂𝟐≥𝟎 , (𝒂∈𝑹)Equality holds when

2

Equality holds when

3

Equality holds when

4

Equality holds when

AM-GM inequality

5

Equality holds when

6

Equality holds when

FYHS-

Kulai by Chtan

7

Equality holds when

Equality holds when

e.g.1

Consider the function

𝒇 (𝒙 , 𝒚 ,𝒛 )= 𝒙𝒚

+√ 𝒚𝒛 +𝟑√ 𝒛𝒙𝑓𝑜𝑟 𝑥 , 𝑦 ,𝑧 𝑎𝑟𝑒𝑎𝑙𝑙𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒𝑟𝑒𝑎𝑙𝑛𝑢𝑚𝑏𝑒𝑟𝑠 .

𝑭𝒊𝒏𝒅 𝒕𝒉𝒆𝒎𝒊𝒏𝒊𝒎𝒖𝒎𝒗𝒂𝒍𝒖𝒆𝒐𝒇 𝒕𝒉𝒊𝒔 𝒇𝒖𝒏𝒄𝒕𝒊𝒐𝒏 .

Soln :𝒇 (𝒙 , 𝒚 ,𝒛 )= 𝒙

𝒚+√ 𝒚𝒛 +𝟑√ 𝒛𝒙

¿𝒙𝒚

+𝟏𝟐 √ 𝒚𝒛 +

𝟏𝟐 √ 𝒚𝒛 +

𝟏𝟑

𝟑√ 𝒛𝒙 +𝟏𝟑

𝟑√ 𝒛𝒙 +𝟏𝟑

𝟑√ 𝒛𝒙¿𝟔 ∙

𝟏𝟔

[𝒙𝒚

+𝟏𝟐 √ 𝒚𝒛 +

𝟏𝟐 √ 𝒚𝒛 +

𝟏𝟑

𝟑√ 𝒛𝒙 +𝟏𝟑

𝟑√ 𝒛𝒙 +𝟏𝟑

𝟑√ 𝒛𝒙 ]

≥𝟔 ∙𝟔√ 𝒙𝒚 ∙

𝟏𝟐 √ 𝒚𝒛 ∙

𝟏𝟐 √ 𝒚𝒛 ∙

𝟏𝟑

𝟑√ 𝒛𝒙 ∙𝟏𝟑

𝟑√ 𝒛𝒙 ∙𝟏𝟑

𝟑√ 𝒛𝒙

¿𝟔 ∙𝟔√ 𝒙𝒚 ∙

𝟏𝟐

∙𝟏𝟐𝒚𝒛

∙𝟏𝟑

∙𝟏𝟑

∙𝟏𝟑

∙𝒛𝒙

¿𝟔 ∙𝟔√ 𝟏𝟐 ∙

𝟏𝟐

∙𝟏𝟑

∙𝟏𝟑

∙𝟏𝟑

9

10

1

13

(𝒂 ,𝒃∈𝑹)

14(𝒂 ,𝒃∈𝑹)

𝟏𝟓 .|𝒂𝟏+𝒂𝟐+⋯+𝒂𝒏|≤|𝒂𝟏|+|𝒂𝟐|+⋯+|𝒂𝒏|

𝒂𝟏 ,𝒂𝟐 ,⋯ ,𝒂𝒏∈𝑹

Exponential inequalities

𝑮𝒊𝒗𝒆𝒏𝒂𝒇 (𝒙 )>𝒂𝒈 (𝒙 ) ,(𝒂>𝟏)

To solve this inequality, it is equivalent to solve :

𝒇 (𝒙 )>𝒈 (𝒙 )

𝑮𝒊𝒗𝒆𝒏𝒂𝒇 (𝒙 )>𝒂𝒈 (𝒙 ) ,(𝟎<𝒂<𝟏)

To solve this inequality, it is equivalent to solve :

𝒇 (𝒙 )<𝒈 (𝒙 )

logarithmic inequalities

𝑰𝒇 𝐥𝐨𝐠𝒂 𝒇 (𝒙 )>𝐥𝐨𝐠𝒂𝒈 (𝒙 ) ,𝒂>𝟏

It is equivalent to solve :

𝒇 (𝒙 )>𝟎𝒈 (𝒙 )>𝟎

𝒇 (𝒙 )>𝒈 (𝒙 )

𝑰𝒇 𝐥𝐨𝐠𝒂 𝒇 (𝒙 )>𝐥𝐨𝐠𝒂𝒈 (𝒙 ) ,𝟎<𝒂<𝟏

It is equivalent to solve :

𝒇 (𝒙 )>𝟎𝒈 (𝒙 )>𝟎

𝒇 (𝒙 )<𝒈 (𝒙 )

e.g.2

𝑭𝒐𝒓 𝒘𝒉𝒂𝒕 𝒗𝒂𝒍𝒖𝒆𝒐𝒇 𝒙𝟐 𝒙−𝟑>𝒙+𝟏?

e.g.3

:

𝒙 (𝒙 −𝟏 )<𝟎

e.g.4

Find the range of values of x for which :

|𝟏−𝟑 𝒙|>𝟕

e.g.5

Find the range of values of x for which :

|𝟒 𝒙+𝟏|>𝟑

e.g.6

:

−𝟏<𝟐𝒙 −𝟏<𝟏

e.g.7

Express in the modulus form :

−𝟑<𝟐𝒙<𝟓

e.g.8

Express in the modulus form :

−𝟒<𝒙<𝟎

e.g.9

For what values of x is :

|𝟐 𝒙−𝟓|>𝟔

e.g.10

:

𝒎𝟐−𝟕𝒎+𝟏𝟐<𝟎

e.g.11

:

𝒎𝟐−𝟕𝒎+𝟏𝟐 ≥𝟎

e.g.12

:

(𝒎+𝟓 ) (𝒎𝟐−𝟕𝒎+𝟏𝟐) ≥𝟎

e.g.13

:

(𝒎+𝟓 ) (𝒎𝟐−𝟕𝒎+𝟏𝟐) ≤𝟎

e.g.14

e.g.15

𝑺𝒉𝒐𝒘𝒕𝒉𝒂𝒕 𝒙𝟑−𝟐>𝒙𝟐+𝒙 .

e.g.16

For what values of x is :

(𝒙+𝟐 ) (𝒙−𝟐 )(𝒙+𝟏 ) (𝒙−𝟏 )

positive .

e.g.17

Find the range of values of x which satisfy the inequality :

𝒙+𝟏<𝟔𝒙

e.g.18

Find the range of values of x which satisfy the inequality :

𝒙𝟐−𝟒 𝒙+𝟑𝒙𝟐+𝟏

<𝟏

e.g.19

For what values of x is :

𝒙𝟑+𝟏≥ 𝒙𝟐+𝒙

e.g.20

Solve the inequality :

𝒙𝟑+𝟒 𝒙𝟐+𝟒 𝒙 ≥𝟎

e.g.21

For what values of x is :

𝟓 𝒙−𝟏𝟏𝒙𝟐−𝟒 𝒙+𝟑

≥𝟏

Harder examples

e.g.22

𝒂𝒃

+𝒃𝒄

+𝒄𝒂

≥𝒂+𝒃+𝒄

Soln :

Using AM-GM inequality, consider the 3 numbers :

𝒂𝒃

+𝒂𝒃

+𝒃𝒄

=𝟐𝒂𝒃

+𝒃𝒄

≥𝟑𝟑√𝒂𝟐𝒃𝒃𝟐𝒄

𝟐𝒂𝒃

+𝒃𝒄

≥𝟑𝟑√ 𝒂𝟐𝒃𝒃𝟐𝒄=𝟑𝟑√ 𝒂𝟐

𝒃𝒄=𝟑𝟑√ 𝒂

𝟐

𝟏𝒂

=𝟑𝒂

Similarly,𝟐𝒃𝒄

+𝒄𝒂

≥𝟑𝒃 and𝟐𝒄𝒂

+𝒂𝒃

≥𝟑𝒄

Adding the 3 inequalities,

𝟐𝒂𝒃

+𝒃𝒄

+𝟐𝒃𝒄

+𝒄𝒂

+𝟐𝒄𝒂

+𝒂𝒃

≥𝟑𝒂+𝟑𝒃+𝟑𝒄

𝟑𝒂𝒃

+𝟑𝒃𝒄

+𝟑𝒄𝒂

≥𝟑𝒂+𝟑𝒃+𝟑𝒄

∴ 𝒂𝒃

+𝒃𝒄

+𝒄𝒂

≥ 𝒂+𝒃+𝒄

e.g.23

𝒂𝟐+𝒃𝟐+𝒄𝟐+𝟐𝒂𝒃𝒄=𝟏

Soln :

(i) Given 𝒂𝟐+𝒃𝟐+𝒄𝟐+𝟐𝒂𝒃𝒄=𝟏Using AM-GM inequality, consider the 4 numbers : :

𝑎2+𝑏2+𝑐2+2𝑎𝑏𝑐4

≥4√𝑎2𝑏2𝑐2 (2𝑎𝑏𝑐 )

¿ 4√2𝑎3𝑏3𝑐3

∴ 𝑎2+𝑏2+𝑐2+2𝑎𝑏𝑐4

≥4√2𝑎3𝑏3𝑐3

𝑎2+𝑏2+𝑐2+2𝑎𝑏𝑐 ≥ 44√2𝑎3𝑏3𝑐3

1 ≥ 44√2𝑎3𝑏3𝑐3

4√2𝑎3𝑏3𝑐3 ≤14

2𝑎3𝑏3𝑐3 ≤1

44

𝑎3𝑏3𝑐3≤1

2 ∙ 44 =1

2 ∙28 =1

29

∴𝑎𝑏𝑐≤( 129 )

13= 1

23 =18

(ii) 𝐿𝑒𝑡𝑎 ,𝑏 ,𝑐∈ (0,1 ) ,𝑎𝑛𝑑 𝑠=𝑎+𝑏+𝑐

𝑮𝒊𝒗𝒆𝒏𝒂𝟐+𝒃𝟐+𝒄𝟐+𝟐𝒂𝒃𝒄=𝟏

=2

¿𝟐 (𝟏−𝒂−𝒃+𝒂𝒃−𝒄+𝒂𝒄+𝒃𝒄−𝒂𝒃𝒄 )

¿𝟐−𝟐𝒂−𝟐𝒃+𝟐𝒂𝒃−𝟐𝒄+𝟐𝒂𝒄+𝟐𝒃𝒄−𝟐𝒂𝒃𝒄

¿𝟏−𝟐 (𝒂+𝒃+𝒄 )+𝟐𝒂𝒃+𝟐𝒂𝒄+𝟐𝒃𝒄+𝟏−𝟐𝒂𝒃𝒄

¿𝟏−𝟐 (𝒂+𝒃+𝒄 )+𝟐𝒂𝒃+𝟐𝒂𝒄+𝟐𝒃𝒄+𝒂𝟐+𝒃𝟐+𝒄𝟐

¿𝟏−𝟐 (𝒔 )+(𝒔 )𝟐

∴𝟐 (𝟏−𝒂) (𝟏−𝒃 ) (𝟏−𝒄 )=𝒔𝟐−𝟐𝒔+𝟏

Since

𝒃𝒚 𝑨𝑴 −𝑮𝑴𝒊𝒏𝒆𝒒𝒖𝒂𝒍𝒊𝒕𝒚 ,

𝟏−𝒂+𝟏−𝒃+𝟏−𝒄𝟑

≥𝟑√ (𝟏−𝒂) (𝟏−𝒃 ) (𝟏−𝒄 )

𝟑− 𝒔𝟑

≥𝟑√ (𝟏−𝒂) (𝟏−𝒃) (𝟏−𝒄 )

(𝟑−𝒔𝟑 )

𝟑

≥ (𝟏−𝒂) (𝟏−𝒃 ) (𝟏−𝒄 )

𝟐(𝟑−𝒔𝟑 )

𝟑

≥𝟐 (𝟏−𝒂) (𝟏−𝒃 ) (𝟏−𝒄 )

𝑩𝒆𝒄𝒂𝒖𝒔𝒆 ,𝟐 (𝟏−𝒂 ) (𝟏−𝒃) (𝟏−𝒄 )=𝒔𝟐−𝟐 𝒔+𝟏

∴𝒔𝟐−𝟐 𝒔+𝟏≤𝟐(𝟑− 𝒔𝟑 )

𝟑

𝟐𝟕 𝒔𝟐−𝟓𝟒 𝒔+𝟐𝟕≤𝟐 (𝟐𝟕−𝟐𝟕 𝒔+𝟗 𝒔𝟐−𝒔𝟑 )

𝟐 𝒔𝟑+𝟗 𝒔𝟐−𝟐𝟕≤𝟎 2+9+0-27 -3

2

-6

3

-9

-9

27

0 -3

2

-6-3

+90

(𝒔+𝟑 )𝟐 (𝟐 𝒔−𝟑 )≤𝟎

Q:𝑮𝒊𝒗𝒆𝒏𝒕𝒉𝒂𝒕 𝒙∈𝑹 ,𝒑𝒓𝒐𝒗𝒆𝒕𝒉𝒂𝒕

.

(𝒔+𝟑 )𝟐 (𝟐 𝒔−𝟑 )≤𝟎

∴𝟐 𝒔−𝟑≤𝟎

∴𝒔 ≤𝟑𝟐

𝒂+𝒃+𝒄≤𝟑𝟐

𝒂𝒙𝟐+𝒃𝒙+𝒄=𝟎

𝒃𝟐−𝟒𝒂𝒄¿ ,=,> ,≤ ,≥𝟎

The notorious cases :

When you will use this

?

1. Find the maximum and minimum values of :

𝒙𝟐+𝟑 𝒙+𝟑𝟐 𝒙+𝟑

2. Find the range of the function :

𝒇 (𝒙 )=𝒙𝟐+𝟑 𝒙+𝟑𝟐 𝒙+𝟑

3. Find the possible range of the values of

𝑘𝑥2+(𝑘− 2 )𝑥−3=0

so that the equation has equal roots, no real roots or 2 real roots.

4. Show that, if is real, the expression

(𝟐𝒙 −𝟓 ) (𝒙+𝟏 )𝒙−𝟏

can take all real values.

5. Many other cases …

The end