If $a, b, c$ are sides of a triangle, prove that $\frac{a}{b+c-a} + \frac{b}{a+c-b} + \frac{c}{a+b-c} \ge 3$
Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$.