cách này dùng cô-si này:
$b+c\leq \sqrt{2(b^{2}+c^{2})}$ ; $c+a\leq \sqrt{2(c^{2}+a^{2})}$ ; $a+b\leq \sqrt{2(a^{2}+b^{2})}$
$\rightarrow$ $\sum \frac{a^{2}}{b+c}\geq \sum \frac{a^{2}}{\sqrt{2(b^{2}+c^{2})}}$
Đặt $\sqrt{b^{2}+c^{^{2}}}=x$; $\sqrt{c^{2}+a^{2}}=y; \sqrt{a^{2}+b^{2}}=z$
$\rightarrow a^{2}=\frac{y^{2}+z^{2}-x^{2}}{2}; b^{2}=\frac{x^{2}+z^{2}-y^{2}}{2}; c^{2}=\frac{x^{^{2}}+y^{2}-z^{2}}{2}$
$\rightarrow \sum \frac{a^{2}}{b+c}\geq \frac{1}{2\sqrt{2}}.\left [ \right \sum (\frac{y^{2}+z^{2}}{x}+2x)-3.\sum x\left \right ]$
Có $\frac{y^{2}+z^{2}}{x}+2x\geq \frac{(y+z)^{2}}{2x}+2x\geq 2(y+z)$
Tương tự,...
$\rightarrow \sum \frac{a^{2}}{b+c}\geq \frac{1}{2\sqrt{2}}.(x+y+z)=\frac{1}{2}\sqrt{\frac{2011}{2}}$