Giải thích các bước giải:
Ta có:
$\sqrt{\dfrac{a}{b+c}}=\dfrac{a}{\sqrt{a(b+c)}}\ge \dfrac{a}{\dfrac12(a+(b+c))}=\dfrac{2a}{a+b+c}$
Tương tự
$\sqrt{\dfrac{b}{c+a}}\ge \dfrac{2b}{a+b+c}$
$\to \sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2a}{a+b+c}+ \dfrac{2b}{a+b+c}$
$\to \sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2(a+b)}{a+b+c}$