Dễ chứng minh BĐT $ (x+y-z)(y+z-x)(x+z-y) \le xyz, \quad \forall a,b,c \ge 0$
Suy ra
$(1-2x)(1-2y)(1-2z) \le xyz$
$\Leftrightarrow 1 -2(x+y+z) +4(xy+yz+zx) -8xyz \le xyz$
$\Leftrightarrow -1 +4(xy+yz+zx) -8xyz \le xyz$
$\Leftrightarrow xy+yz+zx -2xyz \le \frac{1}{4}(1+xyz)$
$\Rightarrow P \le \frac{1}{4}\left ( 1+\left ( \frac{x+y+z}{3} \right )^3 \right )=\frac{7}{27}$
Vậy $\max P=\frac{7}{27}\Leftrightarrow x=y=z=\frac{1}{3}$.