Áp dụng BĐT cosi với 2 số dương:
\(\dfrac{\dfrac{b+c}{a}+1}{2}\ge\sqrt{\dfrac{b+c}{a}}\\ \Leftrightarrow\dfrac{a+b+c}{2a}\ge\sqrt{\dfrac{b+c}{a}}\\ \Leftrightarrow\sqrt{\dfrac{a}{b+c}}\ge\dfrac{2a}{a+b+c}\)
Chứng minh tương tự, ta được
\(\sqrt{\dfrac{b}{a+c}}\ge\dfrac{2b}{a+b+c};\sqrt{\dfrac{c}{a+b}}\ge\dfrac{2c}{a+b+c}\)
Cộng vế theo vế 3 BĐT trên, ta được:
\(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}\\ \ge\dfrac{2a+2b+2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{b+c}{a}}=1\\\sqrt{\dfrac{a+c}{b}}=1\\\sqrt{\dfrac{a+b}{c}}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b+c=a\\a+c=b\\a+b=c\end{matrix}\right.\)
\(\Leftrightarrow2\left(a+b+c\right)=a+b+c\Leftrightarrow a+b+c=0\) (vô lí vì \(a,b,c>0\))
Nên dấu \("="\) không xảy ra
Vậy \(\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{a+c}}+\sqrt{\dfrac{c}{a+b}}>2\)
