Đặt \(x+y=z\ge3\Rightarrow y=z-x\)
\(P=x^2+\left(z-x\right)^2+\frac{1}{x}+\frac{1}{z}=2x^2-2zx+z^2+\frac{1}{x}+\frac{1}{z}\)
\(P=\frac{1}{6}\left(9x^2-12zx+4z^2\right)+\frac{1}{2}x^2+\frac{1}{x}+\frac{1}{3}z^2+\frac{1}{z}\)
\(P=\frac{1}{6}\left(3x-2z\right)^2+\frac{x^2}{16}+\frac{1}{2x}+\frac{1}{2x}+\frac{z^2}{54}+\frac{1}{2z}+\frac{1}{2z}+\frac{7}{16}x^2+\frac{17}{54}z^2\)
\(P\ge3\sqrt[3]{\frac{x^2}{64x^2}}+3\sqrt[3]{\frac{z^2}{216z^2}}+\frac{7}{16}.2^2+\frac{17}{54}.3^2=\frac{35}{6}\)
\(P_{min}=\frac{35}{6}\) khi \(\left\{{}\begin{matrix}x=2\\z=3\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)
