If $a^3+b^3=c^3,$ prove that

Question

If $a,b,c\in\mathbb R^{+},$ such that $a^3+b^3=c^3,$

Prove that $$a^2+b^2-c^2>6(c-a)(c-b).$$

Edit: This question comes from the Indian Mathematical Olympiad, and I have posted below the proof I found.

Answer 1

$$a^3=c^3-b^3$$ $$\frac{a^2}{(c-b)}=\frac{c^2+cb+b^2}{a}.......(1)$$

Similarly, $$b^3=c^3-a^3$$ $$\frac{b^2}{(c-a)}=\frac{c^2+ca+a^2}{b}.......(2)$$

We can observe that $$\frac{a^2}{(c-b)}+\frac{b^2}{(c-a)}=\frac{c(a^2+b^2)-a^3-b^3}{(c-b)(c-a)}=\frac{c(a^2+b^2-c^2)}{(c-b)(c-a)}........(3)$$

Adding $(1)$ and $(2),$ $$\frac{a^2}{(c-b)}+\frac{b^2}{(c-a)}=\frac{c^2+cb+b^2}{a}+\frac{c^2+ca+a^2}{b}$$

By AM-GM inequality, $b^2+c^2>2bc$ and $a^2+c^2>2ac$

So,

$$\frac{b^2+c^2+bc}{ac}>\frac{3b}{a}$$

$$\frac{a^2+c^2+ac}{bc}>\frac{3a}{b}$$

Adding,

$$\frac{c(a^2+b^2-c^2)}{(c-b)(c-a)}>3(\frac{a}{b}+\frac{b}{a})..........[from(3)]$$ Again by AM-GM inequality, $a^2+b^2>2ab$ $$\frac{c(a^2+b^2-c^2)}{(c-b)(c-a)}>3(2)$$ $$a^2+b^2-c^2>6(c-b)(c-a)$$