$A=\Sigma \frac{bc}{\sqrt{3a+bc}}$$A=\Sigma \frac{bc}{\sqrt{a^2+ab+ac+bc}}$
$A=\Sigma \frac{bc}{\sqrt{(a+b)(a+c)}}$
$A=\Sigma \frac{b}{\sqrt{a+b}} . \frac{c}{\sqrt{a+c}}$
$\Rightarrow 2A = \Sigma bc.2.(\frac{1}{\sqrt{a+b}}.\frac{1}{\sqrt{a+c}})$
$2A \leq \Sigma bc.(\frac{1}{a+b}+\frac{1}{a+c})$
$2A \leq \Sigma \frac{bc}{a+b}+\frac{bc}{a+c}$
$2A \leq \frac{bc+ac}{a+b} + \frac{bc+ab}{a+c} + \frac{ab+ac}{b+c}$
$2A \leq a+b+c=3$
$A \leq 1,5$
Dấu = xảy ra $\Leftrightarrow$ $a=b=c=1$