<s>[INST] How to prove the inequality $a/(b+c)+b/(a+c)+c/(a+b) \ge 3/2$ Suppose $a>0, b>0, c>0$. 
Prove that: 
$$a+b+c \ge \frac{3}{2}\cdot [(a+b)(a+c)(b+c)]^{\frac{1}{3}}$$

Hence or otherwise prove: 
  $$\color{blue}{\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}\ge \frac{3}{2}}$$

 [/INST] Using AM-GM inequality:$$\frac{x_1+\dots+x_n}{n}\geq (x_1\dots x_n)^{1/n}$$
let $n=3$ and $x_1=a+b,\; x_2=b+c,\; x_3=c+a$:
$$\frac{2(a+b+c)}{3}\geq [(a+b)(b+c)(c+a)]^{1/3}$$
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