Theo cô si, ta có:
$a+b+c\geq \sqrt{a(b+c)}\Rightarrow \frac{1}{a+b+c}\leq \frac{1}{\sqrt{a(b+c)}}$
$\Rightarrow \frac{2}{a+b+c}\leq \frac{a}{\sqrt{a(b+c)}}\Rightarrow\frac{2a}{a+b+c}\leq \sqrt{\frac{a^{2}}{a(b+c)}}\Rightarrow \frac{2a}{a+b+c}\leq \sqrt{\frac{a}{b+c}}$tương tự
$\Rightarrow \frac{2b}{a+b+c}\leq \sqrt{\frac{b}{a+c}}$
$\Rightarrow \frac{2c}{a+b+c}\leq \sqrt{\frac{c}{a+b}}$
$\Rightarrow \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}} \geq \frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}$
có $\frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}\geq 2$
$\Rightarrow \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}\geq 2$
Do dấu "=" không thể xảy ra nên
$\Rightarrow \sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{a+b}}> 2$