2. Ta có $3(ab+bc+ca) \le (a+b+c)^2\Rightarrow ab+bc+ca \le 3$. Do đó
$\sum \frac{bc}{\sqrt{a^2+3}} \le \sum \frac{bc}{\sqrt{a^2+ab+bc+ca}} = \sum \frac{bc}{\sqrt{(a+b)(a+c)}} \le \frac{1}{2}bc.\sum\left (\frac{1}{a+b} +\frac{1}{a+c}\right )=\frac{1}{2}\sum\left (\frac{bc}{a+b} +\frac{bc}{a+c}\right ) =\frac{1}{2}(a+b+c)=\frac{3}{2}$.