# work 17 – CRUX

Prove that for $a,b,c$ sides of triangle the following inequality hold good-

$3 \ge \sum \frac{a(b+c)}{a^2+bc}$

Proof

$3 - \sum \frac{a(b+c)}{a^2+bc} = \sum \frac{(a-b)(a-c)}{a^2+bc}$

So we have to prove that –

$\sum \frac{(a-b)(a-c)}{a^2+bc} \ge 0$

Note that –

$b^2+ac \ge c^2+ab$ as a,b,c are sidelengths of a triangle

Therefore.

$\frac{1}{c^2+ab} \ge \frac{1}{b^2+ac}$

thus by Vornicu-Schur the inequality is true