Problem

Source: Bosnia and Herzegovina Junior Balkan Mathematical Olympiad TST 2013

Tags: Inequality, positive real, algebra, inequalities



Let $a$, $b$ and $c$ be positive real numbers such that $a^2+b^2+c^2=3$. Prove the following inequality: $\frac{a}{3c(a^2-ab+b^2)} + \frac{b}{3a(b^2-bc+c^2)} + \frac{c}{3b(c^2-ca+a^2)} \leq \frac{1}{abc}$