Question

Cho a,b>0 và \(\sqrt{ab}\left(a-b\right)=a+b\).Tìm Min B= a+b

Answer 1

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

\(\Leftrightarrow\left(a+b\right)\left(2-\frac{a+b}{2}\right)\le0\Leftrightarrow\frac{a+b}{2}\ge2\Leftrightarrow a+b\ge4\)(vì \(a,b>0\))

Dấu \(=\)khi \(\hept{\begin{cases}2\sqrt{ab}=a-b\\a+b=4\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2+\sqrt{2}\\b=2-\sqrt{2}\end{cases}}\).