\(\frac{a^2+b^2}{a-b}=\frac{a^2+b^2-2ab+2ab}{a-b}=\frac{\left(a-b\right)^2}{a-b}+\frac{2}{a-b}=a-b+\frac{2}{a-b}\ge2\sqrt{\frac{2\left(a-b\right)}{a-b}}=2\sqrt{2}\)
Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}ab=1\\a-b=\sqrt{2}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=\frac{\sqrt{6}+\sqrt{2}}{2}\\b=\frac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)