Ta có $\sum_{}\frac{a^{2}+2ab}{(a+\sqrt{bc}+c)^{2}}\geqslant \frac{1}{3}\sum_{}\frac{a^{2}+2ab}{a^{2}+bc+c^{2}}=\frac{2}{3}\sum{}\frac{a^{2}+2ab}{2a^{2}+2c^{2}+2bc}\geqslant $
$\frac{2}{3}\sum{}\frac{a^{2}+2ab}{2a^{2}+2c^{2}+b^{2}+c^{2}}=\frac{2}{3}\sum{}\frac{a^{2}+2ab}{2a^{2}+b^{2}+3c^{2}} =1-\frac{1}{3}\sum\frac{3c^{2}-4ab+b^{2}}{2a^{2}+b^{2}+3c^{2}}\geqslant $
$1-\frac{1}{3}\sum\frac{3c^{2}-2a^{2}-2b^{2}+b^{2}}{2a^{2}+b^{2}+3c^{2}}=1-\frac{1}{3}\sum\frac{3c^{2}-2a^{2}-b^{2}}{2a^{2}+b^{2}+3c^{2}}=\frac{2}{3}\sum\frac{2a^{2}+b^{2}}{2a^{2}+b^{2}+3c^{2}}=$
$\frac{2}{3}\sum\frac{2x+y}{2x+y+3z}$ $(x=a^{2},y=b^{2},z=c^{2})$
$=\frac{4}{3}\sum\frac{x}{2x+y+3z}+\frac{2}{3}\sum\frac{y}{2x+y+3z}=\frac{4}{3}\sum\frac{x^{2}}{2x^{2}+xy+3zx}+\frac{2}{3}\sum\frac{y^{2}}{2xy+y^{2}+3yz} \geqslant $
$\frac{4(x+y+z)^{2}}{3.2(x+y+z)^{2}}+\frac{2(x+y+z)^{2}}{3[(x+y+z)^{2}+3(xy+yz+zx)]}\geqslant \frac{2}{3}+\frac{2(x+y+z)^{2}}{6(x+y+z)^{2})}=\frac{2}{3}+\frac{1}{3}=1$
Vậy BĐT đã đc Cm
Dấu = xảy ra $\Leftrightarrow a=b=c$