# An easy inequality from Nordic mathematical contests !?

Reference: Nordic Mathematical Contest, 1987-2009, R. Todev.

Question:

Let a, b, and c be real numbers different from 0  and $a \geq b \geq c$. Prove that inequality

$\frac{a^{3}-c^{3}}{3} \geq abc(\frac{a-b}{c} + \frac{b-c}{a})$

holds. When does the equality hold?

Proof:

We know that a, b and c are real, distinct and also non-zero and also that $a \geq b \geq c$.

Hence, $c-b \leq 0 \leq a-b$, we have $(a-b)^{3}\geq (c-b)^{3}$, or

$a^{3}-3a^{a}b+3ab^{2}-b^{3} \geq c^{3}-3bc^{2}+3b^{2}c-b^{3}$

On simplifying this, we immediately have

$\frac{1}{3}{(a^{3}-c^{3})} \geq a^{2}b-ab^{2}+b^{2}c-bc^{2}=abc(\frac{a-b}{c}+\frac{b-c}{a})$.

A sufficient condition for equality is $a=c$. If $a>c$, then $(a-b)^{3}>(c-b)^{3}$. which makes the proved inequality a strict one. So, $a=c$ is a necessary condition for equality too.

-Nalin Pithwa.