Question

If a, b, c $>0$ then prove $\left(abc\right){(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}^3\ge 27$.

Answer 1

We know that for a finite number of real terms,

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

Consider the terms to be $\frac{1}{a},\frac{1}{b},\frac{1}{c}$

A.M$=\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}$

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

Applying (1)

$\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\ge \sqrt[3]{3\frac{1}{abc}}$

$\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\ge \sqrt[3]{3\frac{1}{abc}}$

$\frac{{(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}^3}{27}\ge \frac{1}{abc}(\because a,b,c>0)$

$\left(abc\right){(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})}^3\ge 27$

Hence proved