Áp dụng BĐT cosi:
\(\sqrt{\dfrac{b+c}{a}}\le\dfrac{\dfrac{b+c}{a}+1}{2}=\dfrac{\dfrac{a+b+c}{a}}{2}=\dfrac{a+b+c}{2a}\\ \Leftrightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Cmtt \(\sqrt{\dfrac{b}{c+a}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{c+a}}\ge\dfrac{2c}{a+b+c}\)
Cộng vế theo vế 3 BĐT trên:
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}a=b+c\\b=c+a\\c=a+b\end{matrix}\right.\Leftrightarrow a+b+c=2\left(a+b+c\right)\)
\(\Leftrightarrow a+b+c=0\) (vô lí vì \(a,b,c>0\))
Do đó dấu "=" ko xảy ra hay \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}>2\)
