\(2=2\left(x^2+y^2\right)\ge\left(x+y\right)^2\Rightarrow\hept{\begin{cases}\left(x+y\right)^2\le2\\x+y\le\sqrt{2}\end{cases}.}\)
Dấu ''='' xảy ra khi \(\hept{\begin{cases}x=y\\x^2+y^2=1\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{\sqrt{2}}\\y=\frac{1}{\sqrt{2}}\end{cases}}}\)
\(P=x+y+2\left(x+y\right)^2\le\sqrt{2}+2.2=4+\sqrt{2}\)