2016/099) Let $a\,b$ and $c$ be the sides of a triangle. Prove that $\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge 3$

if we put $x = b+c-a, y = a+c-b, z = a + b -c$ we get
given expression
$= \frac{1}{2}(\frac{y+z}{x} +\frac{z+x}{y} + \frac{x+y}{z})$
$= \frac{1}{2}((\frac{y}{x}+ \frac{x}{y})+( \frac{z}{y} + \frac{y}{z}) + ( \frac{z}{x} + \frac{x}{z}))$
$= \frac{1}{2}(((\sqrt{\frac{y}{x}}- \sqrt{\frac{x}{y}})^2+ 2) +((\sqrt{\frac{y}{z}}- \sqrt{\frac{z}{y}})^2+ 2) + ((\sqrt{\frac{z}{x}}- \sqrt{\frac{x}{z}})^2+ 2)$
$>= \frac{1}{2}(2+2+2)\,or\,3$