VT=$\frac{(a+b)^{2}}{c^{2}+4c(a+b)+4ab}\geq \frac{(a+b)^{2}}{c^{2}+4c(a+b)+(a+b)^{2}}$
$= \frac{(\frac{a}{c}+\frac{b}{c})^{2}}{1+4(\frac{a}{c}+\frac{b}{c})+(\frac{a}{c}+\frac{b}{c})^{2}}$
Đặt $\frac{a}{c}+\frac{b}{c}=t$
ta cần CM
$\frac{t^{2}}{1+4t+t^{2}}\geq \frac{1}{6}$
$\Leftrightarrow 5t^{2}-4t-1\geq 0\Leftrightarrow (t-1)(5t+1)\geq0$
Dấu "=" $\Leftrightarrow a=b=1;c=2$