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Introduction to Inequalities Roman Kvasov

Idea #1 :

Perform equivalent transformations of the initial inequality to bring it to the inequality that is obvious or known to be true

Problem Prove the inequality: a 2 + b 2 ≥ 2ab

Solution Start with the initial inequality and perform equivalent transformations passing all terms to one side and completing the square of the obtained expression: 2 a 2 + b 2 ≥ 2ab ⇔ a 2 + b 2 − 2ab ≥ 0 ⇔ ( a − b )( a − b ) ≥ 0 ⇔ ( a − b ) ≥ 0 Since the last inequality is obviously true then the initial inequality is true as well. 2 Note that the equality is reached only when ( a − b ) = 0 or equivalently when a = b .

Practice Problems 1. a 2 + 9b 2 ≥ 6ab , for any a, b 2. 2a 2 + b 2 + c 2 ≥ 2a ( b + c ) , for any a, b, c 3. a 2 + b 2 ≥ 2 ( a + b − 1) , for any a, b

4. a 2 + 6ab + 10b 2 ≥ 0 , for any a, b 5. a + b 2

2

( a + b) ≥

2

, for any a, b 2 1 1 6. a 2 + 2 ≥ a + , for any a a a

7. ( a 2 − b 2 ) ≥ 4ab ( a − b ) , for any a, b 2

2

8. ( a 3 − b3 ) ( a − b ) ≥ 3ab ( a − b ) , for any a, b 2

9. a 4 + b 4 ≥ a 3b + b3a , for any a, b

10. ( a 2 + b 2 )( c 2 + d 2 ) ≥ ( ac + bd ) , for any a, b, c, d 2

11.

2

≤ ab ≤

a+b a 2 + b2 ≤ , for a, b > 0 2 2

1 1 + a b 12. ( a + c )( b + d ) ≥ ab + cd , for any a, b, c, d > 0


Apply the AM-GM Inequality to the terms of the initial inequality

Idea #2 :

Problem Prove the inequality:

a b + ≥ 2 , for a, b > 0 b a

Solution Apply the AM-GM Inequality to the numbers a b + b a ≥ a⋅b 2 b a

a b and : b a

a b + b a≥ 1 ⇔ 2

a b + ≥2 b a a b Note that the equality is reached only when = or equivalently when a = b . b a ⇔

Practice Problems 1 1 1. ( a + b )  +  ≥ 4 , for a, b > 0 a b 3 2. ( a + b )( a + b3 ) ≥ 4a 2b 2 , for a, b > 0

3. ( a + b )( ab + 9 ) ≥ 12ab , for a, b > 0

4. b ( a 2 + 1) + a ( b 2 + 1) ≥ 4ab , for a, b > 0

1 1 + 2 ≥ 2 , for a, b > 0 a +1 b +1  a 2   b2   c2  6. 1 +  1 +   1 +  ≥ 8 , for a, b, c > 0  bc   ac   ab  5. a 2 + b 2 +

2

7. a 4 + b 4 + 2c 2 ≥ 4abc , for a, b, c > 0 a b c 8. + + ≥ 3 , for a, b, c > 0 b c a a+b b+c a+c 9. + + ≥ 6 , for a, b, c > 0 c a b 10. ( a + b + c ) ( a 2 + b 2 + c 2 ) ≥ 9abc , for a, b, c > 0  bc  cd   ad   ab  11. 1 +  1 +   1 +   1 +  ≥ 16 , for a, b, c, d > 0  ad  ab   bc   cd  12. a + b + c ≥ ab + bc + ac , for a, b, c > 0 13. a 3 + b3 + c3 ≥ a 2 bc + b 2 ca + c 2 ab , for a, b, c > 0 14. a 4 + b 4 + c 4 ≥ abc ( a + b + c ) , for a, b, c > 0


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