\(P=\frac{x}{y+\sqrt{2}}\Rightarrow P.y+P\sqrt{2}=x\Rightarrow x-P.y=P\sqrt{2}\)
\(\Rightarrow2P^2=\left(x-P.y\right)^2\le\left(1+P^2\right)\left(x^2+y^2\right)=1+P^2\)
\(\Rightarrow P^2\le1\Rightarrow P_{max}=1\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\frac{\sqrt{2}}{2}\\y=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)