\(\frac{3}{2}\left(x^2+y^2\right)=1+\frac{1}{2}\left(x+y\right)^2\ge1\Rightarrow x^2+y^2\ge\frac{2}{3}\)
\(P\ge x^4+y^4-\frac{x^4+y^4}{2}=\frac{x^4+y^4}{2}\ge\frac{\left(\frac{x^2+y^2}{2}\right)^2}{2}=\frac{1}{18}\)
\(P_{min}=\frac{1}{18}\) khi \(\left(x;y\right)=\left(\frac{\sqrt{3}}{3};-\frac{\sqrt{3}}{3}\right)\) và hoán vị
Bình phương 2 vế giả thiết:
\(x^4+y^4+2x^2y^2=x^2y^2+2xy+1\)
\(\Rightarrow x^4+y^4=-x^2y^2+2xy+1\)
\(\Rightarrow P=-2x^2y^2+2xy+1=-\frac{1}{2}\left(2xy-1\right)^2+\frac{3}{2}\le\frac{3}{2}\)
\(P_{max}=\frac{3}{2}\) khi \(\left\{{}\begin{matrix}xy=\frac{1}{2}\\x^2+y^2=\frac{3}{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=...\\y=...\end{matrix}\right.\)