Ta có:
$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq \frac{3}{2}$
$\Leftrightarrow (1+\frac{a}{b+c})+(1+\frac{b}{c+a})+(1+\frac{c}{a+b}) \geq \frac{9}{2}$
$\Leftrightarrow 2(a+b+c)(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}) \geq 9$
$\Leftrightarrow [(a+b)+(b+c)+(c+a)](\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}) \geq 9$ $(\bigstar)$
Theo bất đẳng thức $\mathbb {AM - GM}$ thì $(\bigstar)$ đúng $\Rightarrow \mathbb {DPCM..}$