\(A=\frac{1}{\sqrt{2}}.\sqrt{2}x.y.y\le\frac{1}{27\sqrt{2}}\left(\sqrt{2}x+2y\right)^3\)
\(A\le\le\frac{1}{27\sqrt{2}}\left(\sqrt{\left(2+4\right)\left(x^2+y^2\right)}\right)^3=\frac{4\sqrt{6}}{9}\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}x=\frac{\sqrt{6}}{3}\\y=\frac{2\sqrt{2}}{3}\end{matrix}\right.\) \(\Rightarrow x+y^2=\frac{4+\sqrt{6}}{3}\)
\(\Rightarrow P=61\)