Ta có: \(3\ge a+b+c\Leftrightarrow9\ge\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow3\ge ab+bc+ca\)
Khi đó:
\(A=\Sigma\left(\frac{bc}{\sqrt{a^2+3}}\right)\le\Sigma\left(\frac{bc}{\sqrt{a^2+ab+bc+ca}}\right)=\Sigma\left(\frac{bc}{\sqrt{\left(a+b\right)\left(c+a\right)}}\right)=\Sigma\left(\sqrt{\frac{bc}{a+b}\cdot\frac{bc}{c+a}}\right)\)
\(\le\Sigma\left[\frac{1}{2}\cdot\left(\frac{bc}{a+b}+\frac{bc}{c+a}\right)\right]=\frac{1}{2}\cdot\left(\frac{bc}{a+b}+\frac{bc}{c+a}+\frac{ca}{b+a}+\frac{ca}{b+c}+\frac{ab}{b+c}+\frac{ab}{c+a}\right)\)
\(=\frac{1}{2}\cdot\left(\frac{c\left(a+b\right)}{a+b}+\frac{b\left(c+a\right)}{c+a}+\frac{a\left(b+c\right)}{b+c}\right)=\frac{1}{2}\cdot\left(a+b+c\right)\le\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
