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.

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