Question

If $a,b,c$ are positive and unequal, then prove that:
$\frac{bc}{b+c}+\frac{ca}{c+a}+\frac{ab}{a+b}<\frac{1}{2}(a+b+c)$

Answer 1

A.M.>H.M.

$\therefore \frac{a+b}{2}>\frac{2}{\frac{1}{a}+\frac{1}{b}}$

$\therefore \frac{b+c}{2}>\frac{2}{\frac{1}{b}+\frac{1}{c}}$

$\therefore \frac{c+a}{2}>\frac{2}{\frac{1}{c}+\frac{1}{a}}$

Adding all the equations, we get,

$\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}>\frac{2}{\frac{1}{a}+\frac{1}{b}}+\frac{2}{\frac{1}{b}+\frac{1}{c}}+\frac{2}{\frac{1}{c}+\frac{1}{a}}$

$a+b+c>2(\frac{ab}{a+b}+\frac{bc}{c+b}+\frac{ac}{a+c})$

$\therefore \frac{ab}{a+b}+\frac{bc}{c+b}+\frac{ac}{a+c}<\frac{1}{2}(a+b+c)$