Posted here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=964550#p964550

$2S$ : perimeter of $ABC$ we know : $\sqrt{S(S-a)(S-b)(S-c)}=S.r$ (*) Let : $a=x+y$ , $b=y+z$ , $c=z+x$ using (*) we have : $\frac{1}{4r^2}=\frac{1}{4xy}+\frac{1}{4yz}+\frac{1}{4zx}$ so we have to prove that : $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \le \frac{1}{4xy}+\frac{1}{4yz}+\frac{1}{4zx}$ hence , we have to prove that : $\frac{1}{(x+y)^2}+\frac{1}{(y+z)^2}+\frac{1}{(z+x)^2} \le \frac{1}{4xy}+\frac{1}{4yz}+\frac{1}{4zx}$ $(x-y)^2 \le 0 \Longrightarrow x^2+y^2 \ge 2xy \Longrightarrow x^2+y^2+2xy \ge 4xy \Longrightarrow (x+y)^2 \ge 4xy$ $\Longrightarrow \frac{1}{(x+y)^2} \le \frac{1}{4xy} \Longrightarrow \sum_{cyc}\frac{1}{(x+y)^2} \le \sum_{cyc}\frac{1}{4xy}$ , and proof has completed

We know that $a,b,c$ are sides of triangle, we can denote that: $a=x+y; b=y+z; c=z+x$, We know $r=2S/(a+b+c) =S/(x+y+z) =\sqrt{\frac{xyz} {x+y+z}}$ $\Rightarrow$ We have to prove $\frac{1} {4r^2}=\frac{x+y+z} {4xyz}\ge \frac{1} {(x+y)^2}+\frac{1} {(y+z)^2}+\frac{1} {z+x)^2}$ By inequality AM-GM $x^2+y^2\ge 2xy$ $y^2+z^2\ge 2yz$ $z^2+x^2\ge 2zx$ Than we have $\frac{1} {(x+y)^2}+\frac{1} {(y+z)}^2+\frac{1} {(z+x)^2}\le \frac{1} {4xy}+\frac{1} {4yz}+\frac{1} {4zx}=\frac{x+y+z} {4xyz}=\frac{1} {4r^2}$