$1-a\geq 0,1-b\geq 0\Rightarrow (1-a)(1-b)\geqslant 0\Rightarrow 1-(a+b)+ab\geq 0\Rightarrow ab\geq a+b-1$
$a^2+b^2=(a+b)^2-2ab\leq (a+b)^2-2(a+b-1)$
$a^4+b^4\geq \frac{(a^2+b^2)^2}{2}\geq \frac{(\frac{(a+b)^2}{2})^2}{4}=\frac{(a+b)^4}{8}$
Do đó: $P\geq \frac{(a+b-1)(a+b)^4}{8}+\frac{6}{(a+b)^2-2(a+b-1)}-3(a+b)$
Đặt $t=a+b\Rightarrow P\geq f(t)=\frac{(t-1).t^4}{8}+\frac{6}{t^2-2(t-1)}-3t$
Có $ab\geq a+b-1\geq 0\Rightarrow a+b\geq 1\Rightarrow t\geq 1$
Có $a\leq 1,b\leq 1\Rightarrow a+b\leq 2\Rightarrow t\leq 2$
Khhs $f(t)$, với đk: $1\leq t\leq 2$
Tìm đc. $f_{min}=-1$ khi $t=2\Rightarrow a=b=1$