Ta có \(\Sigma\sqrt{\dfrac{a}{b+c}}=\Sigma\dfrac{a}{\sqrt{a\left(b+c\right)}}\)
Theo AM-GM ta có
\(\Sigma\dfrac{a}{\sqrt{a\left(b+c\right)}}\ge\Sigma\dfrac{a}{\dfrac{a+b+c}{2}}=\Sigma\dfrac{2a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
dấu bằng xảy ra khi \(\left\{{}\begin{matrix}a=b+c\\b=a+c\\c=a+b\end{matrix}\right.\Rightarrow a+b+c=2\left(a+b+c\right)\Rightarrow1=2\) (vô lí)
nên\(\Sigma\sqrt{\dfrac{a}{b+c}}>2\)
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}\ge2\sqrt{1+\dfrac{abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
