Question

If a, b, c $>0$, then prove that $\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\ge 3$.

Answer 1

We know that for a given number of real terms

A.M$\ge$G.M...(1)

consider the terms to be $\frac{a^3}{b^3},\frac{b^3}{c^3},\frac{c^3}{a^3}$

then A.M $=\frac{\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}}{3}$

G.M $=\sqrt[3]{3\frac{a^3}{b^3}\times \frac{b^3}{c^3}\times \frac{c^3}{a^3}}=1$

applying (1) $\frac{a^3}{b^3}+\frac{b^3}{c^3}+\frac{c^3}{a^3}\ge 3$

Hence proved