100 Inequality Problems

• Address: Cao Minh Quang, Mathematics teacher, Nguyen Binh Khiem specialized High School, Vinh Long town, Vinh Long, Vietnam... The inequality is equivalent to... If a=b, the inequality

• Address: Cao Minh Quang, Mathematics teacher, Nguyen Binh Khiem specialized High School, Vinh Long town, Vinh Long, Vietnam

• Email: kt13quang@yahoo.com

(a 1 b 1− )( − ≥ ⇔) 0 ab 1 a b+ ≥ + ,

(ab 1 c 1− )( − ≥ ⇔) 0 abc 1 ab c+ ≥ + Adding these two inequalities, we obtain

abc 2+ ≥ + +a b c Thus,

We set a x , b= 3 =y , c z3 = and observe that ab3 + bc+ ca 1=

The inequality is equivalent to

Without loss of generality, we can assume that a≥ ≥b c, we set x= +b c, y= +c a, z a b= +

We observe that x≤ ≤ , therefore, y z

We set 1 x− i =a , i 1, 2, ., ni ( = ) and observe that 0 a≤ < , i 1 (i 1, 2, ., n= )

The inequality is equivalent to

If a=b, the inequality is true

If a≠b, the inequality is equivalent to

We set x= +a b, y= ab and observe that a2+b2 =x2−2y2

The inequality is equivalent to

( 2+ 2)−( + ) ( 2+ 2)≥ ⇔( − )3 ≥ ⇔ ≥ ⇔ + ≥

12 (x∈ ),

3 3sin x sin 2x sin 3x

( )( ) ( )( ) n

1 3 2 4 n 1 1 n 2

1 3 2 4 n 1 1 n 2

n n

Adding these four inequalities, we obtain

2n 1 2n 11

,2n

2n 1 2n 11

,2n

2n 1 2n 11

Adding these three inequalities, we obtain

⇔tg x sin x cot g x cos xn 2 + n 2 ≥cot g x sin x tg x cos xn 2 + n 2

1 2 2

Applying the AM – GM Inequality we get

3 6

=4096 cos x 256 256 2564 + + + ≥4 4096 cos x.2564 4 3 =2048 cos x ≥2048 cos x

The inequality is equivalent to

44 (α β γ ∈, , , sinα +sinβ +sinγ ≥2),

cosα+cosβ+cosγ ≤ 5

16 16

Since a, b, c 0> , thus there exist A, B, C ∈( )0,π such that

A B C+ + = π and a tan= A, tan1 = B, c tan= C

It is easy to show that ( ) 1

3 3

( + + )= +( + +) ( + )≥ ( )( + ) ⇒⎛⎜ ⎞⎟ ≥

3 2

Since a, b, c > 1, thus log c 0, log bb > c > 0

Applying the AM – GM Inequality we get

Without loss of generality, we can assume that x=max x, y{ }

Applying the AM – GM Inequality we get

{ } ( )2 ( ) 2 ( 2) ( )2 1( )( )3 1 3 3y 1 y 1 y 1 y 4 27

( ) ( ) ( )( )

≤ 2 a b a a b b( + ) ( + )= 2 a a b b( + ) Thus,

( ) ( )

2

4 4

5 4

2c a 3 3 3 31

53

13 3 10

3 4 16

.32

Applying the AM – GM Inequality we get

Let α >i 0,i 1, 2, ,5= Applying the AM – GM Inequality we get

2 2 1

sin7

π+

π

Therefore,

2 2 2 2 2 2 1

2 cos7

a27