\(A=\frac{x^2+y^2}{x-y}=\frac{x^2+y^2-2xy+2xy}{x-y}\)
\(=\frac{\left(x-y\right)^2+2}{x-y}=x-y+\frac{2}{x-y}\)
Áp dụng bất đẳng thức Cô-si :
\(A\ge2\sqrt{\frac{2\left(x-y\right)}{x-y}}=2\sqrt{2}\)
Dấu "=" xảy ra
\(\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{2}{x-y}\\xy=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)^2=2\\xy=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+y^2+2xy-4xy=2\\xy=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=6\\xy=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+y=\sqrt{6}\\xy=1\end{matrix}\right.\)
Theo Vi-ét : \(\left\{{}\begin{matrix}x=\frac{\sqrt{6}+\sqrt{2}}{2}\\y=\frac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)( thỏa )
Vậy \(minA=2\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x=\frac{\sqrt{6}+\sqrt{2}}{2}\\y=\frac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)
