# work 016 – An useful result

Problem

Prove that for positive reals $a,b,c$

$\sum \frac{a}{b+c} + \frac{4abc}{(a+b)(b+c)(c+a)} \ge 2$

Proof

Clear the denominators and observe the above inequality is equivalent to –

$2\sum a^3 + 6abc \ge 2 \sum a^2(b+c)$

which is the schur’s inequality for degree 1 .

Hence proved with equality for $a=b=c$ or $a=b , c=0$