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

Question

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?

Answer 1

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!

Answer 2

Hint: Consider the quadratic equation $f(t) = 7t^2-6t(a+b)+6ab-a^2-b^2.$ You are to show that $f(\sqrt[3]{a^3+b^3})<0.$ But instead of plugging it in, it easier to show that the number $c = \sqrt[3]{a^3+b^3}$ lies strictly between the two roots of the quadratic $f(t).$