Ta có \(\sqrt{bc\left(1+a^2\right)}=\sqrt{bc+a^2bc}=\sqrt{bc+a\left(a+b+c\right)}\)
\(=\sqrt{\left(a+b\right)\left(a+c\right)}\)
Đặt BT đề cho là P
\(\Leftrightarrow P=\sum\dfrac{a}{\sqrt{bc\left(1+a^2\right)}}=\sum\sqrt{\dfrac{a}{a+b}\cdot\dfrac{a}{a+c}}\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{a}{a+c}+\dfrac{b}{b+c}+\dfrac{b}{b+a}+\dfrac{c}{c+a}+\dfrac{c}{c+b}\right)\\ \Leftrightarrow P\le\dfrac{1}{2}\left(\dfrac{a+b}{a+b}+\dfrac{b+c}{b+c}+\dfrac{c+a}{c+a}\right)=\dfrac{1}{2}\cdot3=\dfrac{3}{2}\)
Dấu \("="\Leftrightarrow a=b=c=\sqrt{3}\)