The Art Of Inequality Vol 1

Ho Chi Minh, 5/12/2017 Vol. 1

The art of Inequality Pham Quoc Sang1 , Le Minh Cuong2 * Preface If a, b, c are positive real numbers such that ab + bc + ac = 3, then 1 1 3 1 + 2 + 2 ≥ (a + b)(b + c)(c + a) 2 a b c 8 *Corresponding author: [email protected] - [email protected]

Contents 1 Problem 1

1

2 Problem 2

3

3 Problem 3

4

4 Problem 4

4

5 Problem 5

5

References

5

Problem 1. Problem 1 If a, b, c be positive real number such that a + b + c = 3 then 1 1 1 3 + 2 + 2 > (a + b) (b + c) (c + a) 2 a b c 8 Proposed by Pham Quoc Sang Solution 1 by Diego Alvariz, Kolkata, West Bengal, India. By AM − GM we have   2 (a + b + c) 3 (a + b) (b + c) (c + a) 6 =8 3 On the other hand, we get r 1 1 1 1 3 1 3 1 + 2+ 2 >3 . 2 . 2 > 3 > (a + b) (b + c) (c + a) 2 2 a b c a b c 8

The art of Inequality — 2/5

Equality occurs if and only if a = b = c = 1. Solution 2 by Sanong Huayreai, Nakon Pathom, Thailand We have   1 1 1 1 1 1 2 27 1 + 2+ 2> + + =3 > 2 a b c 3 a b c (a + b + c)2 On the other hand, we have 

2 (a + b + c) (a + b) (b + c) (c + a) 6 3

3 =8

Equality occurs if and only if a = b = c = 1. Solution 3 by professor Leo Giugiuc, Romania By AM − GM we have 1 1 1 1 1 3 9(a + b + c) 27 1 + + ≥ + + = ≥ = a2 b2 c2 ab bc ca abc (ab + bc + ca)2 (ab + bc + ca)2 And (ab + bc + ca)2 (a2 + b2 + c2 ) ≤ So

(a2 + b2 + c2 + 2(ab + bc + ca))3 = 27 27

1 1 1 + 2 + 2 > a2 + b2 + c2 (1) 2 a b c

On the other hand, we have
8 a2 + b2 + c2 (a + b + c) > 9 (a + b) (b + c) (c + a) " # " # ⇔8

∑ a3 + ∑ a2 (b + c) cyc

cyc

3

>9

∑ a2 (b + c) + 2abc cyc

2

⇔ 8 ∑ a > ∑ a (b + c) + 18abc cyc

cyc

We have

∑ cyc


a3 + b3 > ∑ ab (a + b) ⇔ 2 ∑ a3 > ∑ a2 (b + c) cyc

cyc

cyc

And applying AM − GM we get
6 a3 + b3 + c3 > 18abc So 9 (a + b) (b + c) (c + a) 6 8 (a + b + c) a2 + b2 + c2 Since (1) and (2) we have 1 1 3 1 + 2 + 2 > (a + b) (b + c) (c + a) 2 a b c 8 Equality occurs if and only if a = b = c = 1. The following are some of the extensions to the problem.




(2)

The art of Inequality — 3/5

2. Problem 2 If a, b, c are positive real number such that ab + bc + ca = 3 then 1 1 1 3 + + > (a + b) (b + c) (c + a) a2 b2 c2 8 Proposed by Le Minh Cuong Solution 1 by proposer. 1 1 3 1 + 2 + 2 > (a + b)(b + c)(c + a) 2 a b c 8  3 1 1 1 ⇔ abc 2 + 2 + 2 > abc(a + b)(b + c)(c + a) a b c 8 bc ac ab 3 ⇔ + + > (ac + bc)(ba + ca)(cb + ab) a b c 8 On the other hand, we get (ab + bc + ac)2 bc ac ab a2 b2 + b2 c2 + c2 a2 3 + + = ≥ s  =3 a b c abc ab + bc + ac 3 3 and

  3 3 (ab + bc) + (ba + ca) + (cb + ab) 3 (ac + bc)(ba + ca)(cb + ab) ≤ =3 8 8 3 Therefore (∗) is true. The equality holds for a = b = c = 1. Solution 2 by Pham Quoc Sang. We have 1 1 3 1 + 2 + 2 > (a + b) (b + c) (c + a) 2 a b c 8 1 1 1 3 ⇔ 2 + 2 + 2 > [(a + b + c) (ab + bc + ca) − abc] a b c 8  1 1 1 ⇔ 3abc + 8 2 + 2 + 2 > 9 (a + b + c) a b c
⇔ 3a3 b3 c3 + 8 a2 b2 + b2 c2 + c2 a2 > 9a2 b2 c2 (a + b + c) On the other hand, we have 9a2 b2 c2 (a + b + c) = 3abc.3abc (a + b + c) 6 3abc(ab + bc + ca)2 = 27abc And

1 a2 b2 + b2 c2 + c2 a2 > (ab + bc + ca)2 = 3 3

So we need prove that 3a3 b3 c3 + 24 > 27abc ⇔ a3 b3 c3 + 8 − 9abc > 0 ⇔ a3 b3 c3 + 1 + 1 + 6 − 9abc > 3abc + 6 − 9abc = 6 (1 − abc) > 0

The art of Inequality — 4/5

Equality occurs if and only if a = b = c = 1.

3. Problem 3 If a, b, c be positive real number such that (ab + bc + ca)(a + b + c) = 9 then 

 9 1 1 1 + 2 + 2 (a + b + c) > (a + b) (b + c) (c + a) 2 a b c 8 Proposed by Pham Quoc Sang

Solution by Le Minh Cuong. We have 1 9 = (a + b + c) (ab + bc + ca) 6 (a + b + c) . (a + b + c)2 ⇒ a + b + c > 3 3 So ab + bc + ca =

9 63 a+b+c

The inequality is equivalent to   ab bc ca 9 + + (a + b + c) > (ab + bc) (bc + ca) (cb + ab) c a b 8 By AM − GM we get   9 2 (ab + bc + ca) 3 9 (ab + bc) (bc + ca) (cb + ab) 6 69 8 8 3 And 

 ab bc ca + + (a + b + c) > (a + b + c) (a + b + c) > 9 c a b

So 

 1 1 1 9 + 2 + 2 (a + b + c) > (a + b) (b + c) (c + a) 2 a b c 8

Equality occurs if and only if a = b = c = 1.

4. Problem 4 If a, b, c be positive real number such that a 6 1 6 b 6 c and a + b + c = 3 then 1 1 1 3 + + > (a + b) (b + c) (c + a) a b c 8 Proposed by Pham Quoc Sang

The art of Inequality — 5/5

5. Problem 5 If a, b, c be positive real number such that a 6 1 6 b 6 c and a + b + c =

1 1 1 + + then a b c 

 1 1 1 9 1 + 2 + 2 > abc a + b + c − 2 a b c 8 a+b+c

Proposed by Pham Quoc Sang Note. This is a problem in Romania’s 2006 TST test. If a, b, c be positive real number such that a + b + c = 3 then 1 1 1 + 2 + 2 > a2 + b2 + c2 2 a b c On the other hand, we have some interesting results below. • If a, b, c be positive real number such that a 6 1 6 b 6 c and a + b + c = 3 then 1 1 1 1 1 1 + + > 2+ 2+ 2 a b c a b c Proposed by Vasile Cirtoaje • If a, b, c be positive real number such that a 6 1 6 b 6 c and a + b + c =

1 1 1 + + then a b c

1 1 1 + 2 + 2 > a2 + b2 + c2 2 a b c Proposed by Vasile Cirtoaje

References [1]

A. J. AAFigueredo and P. S. A. Wolf. Assortative pairing and life history strategy - a cross-cultural study. Human Nature, 20:317–330, 2009.

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