$\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b} \geq 0$ I recently met the inequality $\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b} \geq 0$ , where a , b , c are all positive real numbers.                                                            I wanted to prove it but had some difficult time , seeing no connection to any known standard inequalities I began to simplify the expression multiplied by $(a+b)(b+c)(c+a)$ , after some simple and elementary but tedious calculations I obtained:-
$$(a+b)(b+c)(c+a) \left(\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b}\right) \\=a(a-c)^2 + b(b-a)^2 + c(c-b)^2$$ , which is obviously non-negative thus proving the inequality.  But I am wondering , is there any other way(s) to prove the inequality? Also, what is the shortest way of deriving the above said identity ?

To address the first question, let $x = b+c$, $y = c+a$ and $z = a+b$.  Then $2c=x+y-z$, so the inequality rewrites as 
$$ \frac{y-(x+y-z)}{x} + \frac{z-(y+z-x)}{y} + \frac{x-(x+y-z)}{z} \ge 0 $$
or $$\frac{z}{x} + \frac{x}{y} + \frac{y}{z} \ge 3,$$ which follows immediately from the AM-GM inequality.
