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28 Jul 2008 12:46:38 IST

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Improvement on AM-GM

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We all know the AM-GM Inequality predicts that for positive reals a, b we have

$\frac{a+b}{2} - \sqrt {ab} \geq 0$

Can you improve the inequality to

$\frac{a+b}{2} - \sqrt {ab} \geq \frac{(a-b)^2(a+3b)(3a+b)}{8(a+b)(a^2+6ab+b^2)} \\ \\ $

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Comments (8)

Rajat Sen

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3 Aug 2008 13:13:50 IST

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Sir, can you please give the solution of this.

student9_iit

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3 Aug 2008 13:21:33 IST

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i think we should give it a try first..

Rajat Sen

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3 Aug 2008 17:05:05 IST

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I tried a lot. This question is there for a long time.

Dipanjan

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3 Aug 2008 17:19:36 IST

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Re:Improvement on AM-GM

Dipanjan

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3 Aug 2008 17:27:58 IST

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The first step used the fact that

if x>y y then the exp [(x+c)/(y+c)]<x/c {x,y,c>0}

the second step uses the fact that

(x+y)²<2(x²+y²) (cauchy,schwartz ineq)

the next one is the same solution; only it is clearer

Dipanjan

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3 Aug 2008 17:45:15 IST

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Re:Improvement on AM-GM

Dipanjan

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3 Aug 2008 17:54:19 IST

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this one is supposed to be clearer

and

the first exp actually uses the fact that

if x>y then [(x+c)/(y+c)]<=(x)/(y)

Hari Shankar

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3 Aug 2008 19:37:30 IST

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I had done this on another forum which is now sadly defunct. I liked this because it is a much sharper inequality than simple AM-GM and that makes it really exciting. And kudos to dipanjan for solving this one.

$\text{Since the inequality is homogenous, we can let a+b = 1} \\ \\ \text{So, the inequality becomes} \\ \\ \frac{1}{2} - \sqrt{a(a-1)} \geq \frac {(2a-1)^2(3-2a)(2a+1)}{8(1+4a(1-a))} \\ \\ \text{Now let} \ \sqrt{4a(a-1)} = y \ \text{So} \ y \leq 1 \\ \\ \text{The inequality becomes} \\ \\ \frac{1-y}{2} \geq \frac{(1-y^2)(3+y^2)}{8(1+y^2)} \\ \\ \text{or} \ 4(1+y^2) \geq (1+y)(3+y^2) \\ \\ \text{or} 1-3y+3y^2-y^3 \geq 0 \\ \\ \Rightarrow (1-y)^3 \geq 0 \\ \\ \text{which is true as} \ y = 4a(1-a) \leq 1$

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