\(P=\frac{x^2}{y}+\frac{y^2}{x}\ge\frac{\left(x+y\right)^2}{x+y}=x+y\)
Đặt \(\left(\sqrt{x}+1;\sqrt{y}+1\right)=\left(a;b\right)\Rightarrow\left\{{}\begin{matrix}a;b>1\\ab\ge4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(a-1\right)^2\\y=\left(b-1\right)^2\end{matrix}\right.\)
\(\Rightarrow P\ge\left(a-1\right)^2+\left(b-1\right)^2\ge\frac{1}{2}\left(a+b-2\right)^2\)
\(\Rightarrow P\ge\frac{1}{2}\left(2\sqrt{ab}-2\right)^2\ge\frac{1}{2}\left(2\sqrt{4}-2\right)^2=2\)
Dấu "=" xảy ra khi \(a=b=2\) hay \(x=y=1\)
