<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] Hint: Substitute $$b+c=x,a+c=y,a+b=z$$ so $$a=\frac{-x+y+z}{2}$$
$$b=\frac{x-y+z}{2}$$
$$c=\frac{x+y-z}{2}$$
And we get
$$\frac{-x+y+z}{2x}+\frac{x-y+z}{2y}+\frac{x+y-z}{2z}\geq \frac{3}{2}$$
Can you finish?
And we get $$\frac{x}{y}+\frac{y}{x}+\frac{y}{z}+\frac{z}{y}+\frac{x}{z}+\frac{z}{x}\geq 6$$
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