Mathematical Reflections 2012, Issue 6 - Problem J251

Problem:
Let $a,b,c$ be positive real numbers such that $a \geq b \geq c$ and $b^2>ac$. Prove that
$$\dfrac{1}{a^2-bc}+\dfrac{1}{b^2-ca}+\dfrac{1}{c^2-ab}>0.$$

Proposed by Titu Andreescu.

Solution:
By AM-GM Inequality,
$$\dfrac{1}{a^2-bc}+\dfrac{1}{b^2-ca}+\dfrac{1}{c^2-ab} \geq \dfrac{3}{a^2+b^2+c^2-ab-bc-ca} > 0,$$
where the last inequality follows from the fact that $a,b,c$ cannot all be equal.