Có $(16+a^4)(16+1)\geq (16+a^2)^2;(16+16b^4)(16+1)\geq (16+4b^2)^2$ (bđt B.C.S)$\Rightarrow \sqrt{17(16+a^4)}+4\sqrt{17(b^4+1)}\geq 16+a^2+16+4b^2$
$\Leftrightarrow \sqrt{17}P\geq (1+a^2)+(1+4b^2)+30\geq 2a+4b+30$
$=2(a+2)+4(b+1)+22\geq 2\sqrt{8(a+2)(b+1)}+22=34$
$\Rightarrow P\geq 2\sqrt{17}$
Dấu $=$ có khi $a=1;b=\frac{1}{2}$