Let $a, b, c$ be positive real numbers such that $a^3 + b^3 = c^3$. Prove that $a^2 + b^2 - c^2 \gt 6(c - a)(c - b)$ . 
Let $a$, $b$ and $c$ be positive real numbers such that $a^3 + b^3 = c^3$. Prove that $$a^2 + b^2 - c^2 \gt 6(c - a)(c - b).$$ 

This problem is from the Indian MO 2009, however, I do not know how to approach it, any help? 

We need to prove that
$$a^2+b^2-\frac{a^3+b^3}{c}>6(c-a)(c-b)$$ or
$$\frac{a^2(c-a)}{c}+\frac{b^2(c-b)}{c}>6(c-a)(c-b)$$ or
$$\frac{(c^3-b^3)(c-a)}{ac}+\frac{(c^3-a^3)(c-b)}{bc}>6(c-a)(c-b)$$ or
$$\frac{c^2+bc+b^2}{ac}+\frac{c^2+ac+c^2}{bc}>6.$$ 
Now, by AM-GM
$$\frac{c^2+bc+b^2}{ac}+\frac{c^2+ac+c^2}{bc}\geq2\sqrt{\frac{c^2+bc+b^2}{ac}\cdot\frac{c^2+ac+c^2}{bc}}\geq$$
$$\geq2\sqrt{\frac{3\sqrt[3]{b^3c^3}}{ac}\cdot\frac{3\sqrt[3]{a^3c^3}}{bc}}=6.$$
In our using AM-GM the equality occurs for $a=b=c$, which is impossible.
Thus, $$a^2 + b^2 - c^2 \gt 6(c - a)(c - b)$$ and we are done!
