3-variable non-symmetric inequality Prove that for $a,b,c > 0$
$$\frac{a+b-2c}{b+c}+\frac {b+c-2a}{c+a}+\frac {c+a-2b}{a+b} >0$$
What I did is this:- Let $f(a,b,c)=\frac{a+b-2c}{b+c}+\frac {b+c-2a}{c+a}+\frac {c+a-2b}{a+b}$. Therefore
$$\begin{align}f(a,b,c)&=\frac{a+b-2c}{b+c}+\frac {b+c-2a}{c+a}+\frac {c+a-2b}{a+b}\\&=(a+b+c)\left(\frac{1}{b+c}+\frac {1}{c+a}+\frac {1}{a+b}\right)-3\left(\frac{c}{b+c}+\frac {a}{c+a}+\frac {b}{a+b}\right)\end{align}$$
The first term is greater than $9/2$ by the Cauchy-Schwarz inequality. How do I prove that the term inside the second bracket is less than $3/2$? Can I use the rearrangement inequality here??

your inequality is equivalent to
$${\frac {{a}^{3}+{a}^{2}b-2\,{a}^{2}c-2\,a{b}^{2}+a{c}^{2}+{b}^{3}+{b}^
{2}c-2\,b{c}^{2}+{c}^{3}}{ \left( b+c \right)  \left( c+a \right) 
 \left( a+b \right) }}
>0$$ and the numerator can written as
$$a(a-c)^2+b(b-a)^2+c(b-c)^2\geq 0$$ and the equal sign holds if $$a=b=c$$
