Question

Prove $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}+\frac{1}{\sqrt{ab}}$

Answer 1

$\frac{1}{a}+\frac{1}{b}\ge 2\sqrt{\frac{1}{ab}}$
$\ge 2\frac{1}{\sqrt{ab}}$...........(1)
$\frac{1}{b}+\frac{1}{c}\ge 2\frac{1}{\sqrt{bc}}$............(2)
$\frac{1}{c}+\frac{1}{a}\ge 2\frac{1}{\sqrt{ca}}$.............(3)
Adding (1) + (2) + (3)
$2[\frac{1}{a}+\frac{1}{b}+\frac{1}{c}]\ge 2[\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}]$
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}$