Ta có:
\(\sqrt{\dfrac{a}{b+c}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\dfrac{2a}{a+b+c}\)
Tương tự ta có: \(\left\{{}\begin{matrix}\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c}\\\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\end{matrix}\right.\)
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Dễ thấy dấu = không thể xảy ra nên
\(\Rightarrow\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}>2\)