<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] Hint: 
\begin{eqnarray*}
\sum_{cyc} (a+b)(a-b)^2 \geq 0
\end{eqnarray*}
This can be rearranged to
\begin{eqnarray*}
2(a(a+b)(a+c)+b(b+c)(b+a)+c(c+a)(c+b))-3(a+b)(a+c)(b+c) \geq 0
\end{eqnarray*}
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