# Inequality in triangle.

If $a,b,c$ are sides of a triangle prove that-

$$\frac a{c+a-b}+\frac b{a+b-c}+\frac c{b+c-a}\geq3$$

I am having problem in approaching the problem as the sides are not mentioned as integers.How do I approach the problem?

Thanks for any help!!..

By using the Ravi transformation let $a=x+y$, $b=y+z$ and $c=z+x$, then \begin{align} \frac a{c+a-b}+\frac b{a+b-c}+\frac c{b+c-a}&=\frac{x+y}{2x}+\frac{y+z}{2y}+\frac{z+x}{2z}\\ &=\frac{3}{2}+\frac{1}{2}\left(\frac{y}{x}+\frac{z}{y}+\frac{x}{z}\right) \end{align} Now use AM-GM inequality.