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Ta có $\displaystyle \frac{1}{1+x} +\frac{1}{1+y} +\frac{1}{1+z} \geqslant 2$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Leftrightarrow \frac{1}{1+x} \geqslant \left( 1-\frac{1}{1+y}\right) +\left( 1-\frac{1}{1+z}\right)\\
\Leftrightarrow \frac{1}{1+x} \geqslant \frac{y}{1+y} +\frac{z}{1+z}
\end{array}$
Do $\displaystyle y,z >0$ nên $\displaystyle \frac{y}{1+y}  >0;\frac{z}{1+z}  >0$ 

Áp dụng bất đẳng thức Côsi ta có 
$ \displaystyle \frac{y}{1+y} +\frac{z}{1+z} \geqslant 2\sqrt{\frac{yz}{( 1+y)( 1+z)}}$
Do đó $\displaystyle \frac{1}{1+x} \geqslant 2\sqrt{\frac{yz}{( 1+y)( 1+z)}}$  $\displaystyle ( 1)$ 

Chứng minh tương tự ta có 
$\displaystyle \frac{1}{1+y} \geqslant 2\sqrt{\frac{xz}{( 1+x)( 1+z)}}$ $\displaystyle ( 2)$ và $\displaystyle \frac{1}{1+z} \geqslant 2\sqrt{\frac{xy}{( 1+x)( 1+y)}}$  $\displaystyle ( 3)$
Từ $\displaystyle ( 1) ,( 2) ,( 3)$ ta có $\displaystyle \frac{1}{( 1+x)( 1+y)( 1+z)} \geqslant 8\sqrt{\frac{x^{2} y^{2} z^{2}}{( 1+x)^{2}( 1+y)^{2}( 1+z)^{2}}}$
$\displaystyle  \begin{array}{{>{\displaystyle}l}}
\Leftrightarrow \frac{1}{( 1+x)( 1+y)( 1+z)} \geqslant 8\frac{xyz}{( 1+x)( 1+y)( 1+z)}\\
\Leftrightarrow xyz\leqslant \frac{1}{8} \Leftrightarrow P\leqslant \frac{1}{8}
\end{array}$
Dấu $\displaystyle =$ xảy ra khi $\displaystyle \frac{x}{1+x} =\frac{y}{1+y} =\frac{z}{1+z} \Leftrightarrow x=y=z=\frac{1}{2}$
Vậy GTLN của P là $\displaystyle \frac{1}{8}$ khi $\displaystyle x=y=z=\frac{1}{2}$