Problem 58: Iran 1996 inequality

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by henderson, Jan 6, 2017, 3:21 PM

${\color{red}\bf{Problem \ 58}}$ Prove that, for all positive real numbers $a,b,c$
$(ab+bc+ca)\left(\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2} \right)\geq \frac{9}{4}$ holds.
(Iran 1996)
${\color{red}\bf{Solution}}$ We may assume that $a\geq b \geq c.$ Firstly, let's show that
$\frac{1}{(b+c)^2}+\frac{1}{(c+a)^2}+\frac{1}{(a+b)^2}\geq \frac{1}{4ab}+\frac{2}{(a+c)(b+c)}.$ This can be rewritten as
$\left(\frac{1}{a+c}-\frac{1}{b+c}\right)^2\geq \frac{(a-b)^2}{4ab(a+b)^2},$ or equivalently $4ab(a+b)^2\geq (a+c)^2(b+c)^2.$ This is obvious, since $4ab\geq (b+c)^2$ and $(a+b)^2\geq (a+c)^2.$
Thus, it remains to prove that
$(ab+bc+ca)\left(\frac{1}{4ab}+\frac{2}{(a+c)(b+c)} \right)\geq \frac{9}{4}.$ Using the identities
$\frac{ab+bc+ca}{4ab}=\frac{1}{4}+\frac{c(a+b)}{4ab}, \quad \frac{2(ab+bc+ca)}{(a+c)(b+c)}=2 -\frac{2c^2}{(a+c)(b+c)},$ this becomes
$\frac{c(a+b)}{4ab}\geq \frac{2c^2}{(a+c)(b+c)}.$ But this is equivalent to
$(a+b)(b+c)(c+a)\geq 8abc,$ which is a simple consequence of $AM-GM.$ $\quad$ $\square$

This post has been edited 3 times. Last edited by henderson, Jan 6, 2017, 6:07 PM

Inequality

Iran 1996 inequality

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