$P = 3 - \frac{1}{x+1} - \frac{1}{y+1}-\frac{1}{z-1}$
áp dụng bdt cosi cho 2 bộ 3 số $x+1,y+1,z+1,$và $ \frac{1}{x+1},\frac{1}{y+1}\frac{1}{z+1}$
ta có $(x+1+y+1+z+1)(\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1})\geqslant 9$
$<=>\frac{1}{x+1} + \frac{1}{y+1} + \frac{1}{z+1} \geqslant \frac{9}{4} (x+y+z=1)$
$=>P\leqslant 3 - \frac{9}{4}=\frac{3}{4}$
vậy min$P=\frac{3}{4}<=>x=y=z=\frac{1}{3}$