Question

Cho a > b > 0 và ab=1. CMR:

\(\dfrac{\left(a+b\right)^2-2}{\left(a+1\right)\left(1-b\right)}\ge2\sqrt{2}\)

Answer 1

\(\dfrac{\left(a+b\right)^2-2}{\left(a+1\right)\left(1-b\right)}=\dfrac{\left(a+b\right)^2-2ab}{a\left(b+1\right)\left(1-b\right)}=\dfrac{a^2+b^2}{a\left(1-b^2\right)}=\dfrac{a^2+b^2}{ab\left(a-b\right)}=\dfrac{a^2+b^2}{a-b}\)

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

Dấu "=" khi \(\left\{{}\begin{matrix}ab=1\\\left(a-b\right)^2=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\b=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)