<s>[INST] Prove that $\frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$ Prove that $\frac{a}{b+c} +\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}$
My steps
A= $\frac{a}{b+c}$,B= $\frac{b}{c+a}$ &C=$\frac{c}{a+b}$
A.M$\ge$H.M
$\frac{A+B+C}{3} \ge \frac{3ABC}{AB+BC+AC}$
I am struck after this step
${A+B+C} \ge \frac{9}{\frac{a}{c}+\frac{c}{a}+\frac{b}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c}}$
 [/INST] set $$x=b+c,y=c+a$$ and $$z=a+b$$ then we get
$$\frac{-x+y+z}{2x}+\frac{x+z-y}{2y}+\frac{x+y-z}{2z}\geq \frac{3}{2}$$
and this is 
$$\frac{x}{y}+\frac{y}{x}+\frac{x}{z}+\frac{z}{x}+\frac{y}{z}+\frac{z}{y}-3\geq 3
$$
and then use $$\frac{A}{B}+\frac{B}{A}\geq 2$$ for $$A,B>0$$
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