\(\frac{x}{1+x}+\frac{2y}{1+y}=1\Leftrightarrow1-\frac{1}{1+x}+\frac{2y}{1+y}=1\)
\(\Leftrightarrow\frac{1}{1+x}=\frac{2y}{1+y}\Leftrightarrow x+1=\frac{y+1}{2y}\Rightarrow x=\frac{1-y}{2y}\)
\(\Rightarrow P=\frac{\left(1-y\right)y^2}{2y}=\frac{y\left(1-y\right)}{2}\le\frac{1}{8}\left(y+1-y\right)^2=\frac{1}{8}\)
\(\Rightarrow P_{max}=\frac{1}{8}\) khi \(x=y=\frac{1}{2}\)