# work 027 – mathlinks user CSS-MU

Problem

Let $a,b,c$ be nonnegative numbers, no two of which are zero. Prove that:
$\frac {ab + bc - ca}{c^2 + a^2} + \frac {bc + ca - ab}{a^2 + b^2} + \frac {ca + ab - bc}{b^2 + c^2} \geqslant \frac {3}{2}$

Proof

It is equivalent to –
$\sum (a - b)^2 \cdot (\frac {\sum ab - a^2)}{(c^2 + a^2)(c^2 + b^2)} \ge 0$

Therefore ,
$S_a = (\sum ab - a^2)(b^2 + c^2)$
$S_b = (\sum ab - b^2)(a^2 + c^2)$
$S_c = (\sum ab - c^2)(a^2 + b^2)$

Let , $a\ge b\ge c$
So, $S_c \ge S_b \ge S_a$
and $S_c , S_b \ge 0$

also note that –
$S_a + S_b = (\sum ab - a^2)(b^2 + c^2) + (\sum ab - b^2)(a^2 + c^2)$

$= b^2(\sum ab - a^2) + a^2(\sum ab - b^2) + c^2(2ab - a^2 - b^2 + 2c(a + b))$

$= (ab - c^2)(a - b)^2 + c(a + b)(a^2 + b^2) + 2c^3(a + b) \ge 0$
thus the inequality is proved.