$\frac{a+b-2c}{b+c} + \frac{b+c-2a}{c+a} + \frac{c+a-2b}{a+b} \geq 0$ 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 ?

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.
