Áp dụng BDT quen thuộc: $\frac{1}{a^2}+\frac{1}{b^2}\ge \frac{8}{(a+b)^2}$.
Cm: $\frac{1}{a^2}+\frac{1}{b^2}\ge \frac{2}{ab}\ge \frac{8}{(a+b)^2}$.
Khi đó ta có: $P=\frac{1}{(a+1)^2}+\frac{1}{(\frac{b}{2}+1)^2}\ge \frac{8}{(a+\frac{b}{2}+2)^2}$.
Ta có: $2a+b+4=2a+4b+4-3b\le (a^2+1)+(b^2+4)+4-3b=8(\text{do gt})$
$\implies a+\frac{b}{2}+2\le 4\implies P\ge \frac{8}{16}=\frac{1}{12}$.
Dấu $=$ xảy ra khi $a=1;b=2$