mk nghĩ $c\in \left[1 {;}2 \right]$ P $\geq \frac{(a+b)^{2}}{c^{2}+4(a+b)c+(a+b^{2}}=\frac{(\frac{a}{c}+\frac{b}{c})^{2}}{1+4(\frac{a}{c}+\frac{b}{c})+(\frac{a}{c}+\frac{b}{c})^{2}}$
đặt $t=\frac{a+b}{c}$ vs $t\in \left[1 {;} 4\right]$
$\Rightarrow P\geq \frac{t^{2}}{1+4t+t^{2}}$ ta cần cm $P\geq \frac{1}{6}$
$5t^{2}-4t-1\geq0$ (lđ vs $t\in \left[1 {;}4 \right]$
Dấu "=" $\Leftrightarrow a=b=1;c=2$