# INEQUALITY'S BLOG

## July 12, 2011

### Problem 233 (van khea)

Filed under: Problem by Van Khea — KKKVVV @ 12:10 am

Let $a, b, c$ be the lengths of the sides of a trangle. Prove that: $\displaystyle \frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\geq 6+\frac{1}{2}\biggl[\frac{(a-b)^2}{c^2}+\frac{(b-c)^2}{a^2}+\frac{(c-a)^2}{b^2}\biggl]$