$\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b} \geq 0$ 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 ?
 A: We can set $a+b=C$ and so on, then minimize
$$ f(A,B,C)=\sum_{cyc}\frac{2C-B}{A}. $$
If we regard $f$ as a function of $B$ and $C$ only, we have that $\frac{\partial f}{\partial B}=0$ implies $(B^2-AC)(2A-C)=0$, and $\frac{\partial f}{\partial C}=0$ implies $(C^2-AB)(2B-A)=0$. So we have four stationary points:
$$ (B/A,C/A)\in\left\{(1,1),(2,4),(2,1/2),(1/2,1/4)\right\}.$$
Without loss of generality we can additionally assume $A\leq B\leq C$, having stationary points for
$$(A,B,C) = (\lambda,\lambda,\lambda)\quad\mbox{or}\quad(\lambda,2\lambda,4\lambda).$$
We can now substitute these values into $f$ to prove the inequality.
A: 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.
A: $$\sum_{cyc}\frac{b+c-2a}{c+a}=\sum_{cyc}\frac{c-a-(a-b)}{c+a}=\sum_{cyc}(a-b)\left(\frac{1}{a+b}-\frac{1}{a+c}\right)=$$
$$=\sum_{cyc}\frac{(a-b)(c-b)}{(a+b)(a+c)}=\sum_{cyc}\frac{(a-b)(c-a+a-b)}{(a+b)(a+c)}=$$
$$=\sum_{cyc}\frac{(a-b)^2}{(a+b)(a+c)}-\sum_{cyc}\frac{(a-b)(a-c)}{(a+b)(a+c)}=$$
$$=\sum_{cyc}\frac{(a-b)^2}{(a+b)(a+c)}-\sum_{cyc}\frac{(a-b)(a-c)(b+c)}{\prod\limits_{cyc}(a+b)}=$$
$$=\sum_{cyc}\frac{(a-b)^2}{(a+b)(a+c)}-\sum_{cyc}\frac{a^2b+a^2c-2abc}{\prod\limits_{cyc}(a+b)}=$$
$$=\sum_{cyc}\frac{(a-b)^2(b+c)}{\prod\limits_{cyc}(a+b)}-\sum_{cyc}\frac{c(a-b)^2}{\prod\limits_{cyc}(a+b)}=\sum_{cyc}\frac{(a-b)^2b}{\prod\limits_{cyc}(a+b)}\geq0.$$
Done!
