Giả sử $b \ge a \ge 0$
$P+\frac 12=\left[\frac{1}{2(a+b)^2}+\frac{1}{2(a+b)^2}+\frac{8}{c^4}+\frac 12 \right]+\frac{8}{c^4}+\frac{abc}{2}$$\ge \frac{4}{(a+b).c}+\frac{8}{2^4}+\frac{0bc}{2} $
$\Leftrightarrow P \ge \frac{4}{2-ac} \ge 2$
$\Rightarrow \min P=2$ khi $(a,b,c)=\{(0,1,2);(1,0,2) \}$