# INEQUALITY'S BLOG

## August 13, 2011

### Problem 240 Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 9:50 am

Let $a, b, c$ be the lengths of the sides of a trangle. Prove that:

$\displaystyle \frac{1}{(a^2+b^2)^2}+\frac{1}{(b^2+c^2)^2}+\frac{1}{(c^2+a^2)^2}\geq \frac{9}{4(a^3b+b^3c+c^3a)}$