Inequality 15 « Mathsmaster’s Blog

Inequality 15

Problem: let $a,b,c$ be positive reel numbers; Prove that:

$\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\geq \frac{9}{4(a+b+c)}$

Solution: assuming that, $a \geq b \geq c$ we get ,

$\frac{1}{(b+c)^2} \geq \frac{1}{(c+a)^2} \geq \frac{1}{(a+b)^2}$ then by Chebychev,

$S=\sum \frac{a}{(b+c)^2}\geq \frac{1}{3}(a+b+c)\left(\sum\frac{1}{(a+b)^2}\right)$

and by Cauchy-Scwarz tow times,

$\sum \frac{1}{(a+b)^2}\geq \frac{1}{3}\left(\frac{1}{a+b}\right)^2\geq \frac{1}{3}\left(\frac{9}{2(a+b+c)}\right)^2$

then $S\geq \frac{1}{3}(a+b+c)\left(\frac{1}{3}\left(\frac{9}{2(a+b+c)}\right)^2\right)=\frac{9}{4(a+b+c)}$

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on avril 7, 2009 at 7:10 Laisser un commentaire

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