Question

Cho \(a;b\) là 2 số thỏa mãn \(a>b>0\) và \(ab=1\)

Chứng minh: \(\frac{a^2+b^2}{a-b}\ge2\sqrt{2}\)

Answer 1

\(\frac{a^2+b^2}{a-b}=\frac{\left(a-b\right)^2+2ab}{a-b}=\frac{\left(a-b\right)^2+2}{a-b}=a-b+\frac{2}{a-b}\ge2\sqrt{\frac{2\left(a-b\right)}{a-b}}=2\sqrt{2}\)

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