$\Rightarrow \frac{1}{a+1}\geq (1-\frac{1}{b+1})+(1-\frac{1}{c+1)}= \frac{b}{b+1}+\frac{c}{c+1}\geq 2\sqrt{\frac{bc}{(b+1)(c+1)}}$
Tương tự $...\Rightarrow\prod_{cyc}^{}\frac{1}{a+1}\geq 8\frac{abc}{(a+1)(b+1)(c+1)} $
$\Rightarrow 1\geq 8abc\Rightarrow abc\leq \frac{1}{8}$
vậy $P_{min}=\frac{1}{8}$ tại $a=b=c$