Áp dụng bđt Cô-si ta có:
\(\frac{\sqrt{a}+\sqrt{b}}{\sqrt{ab}}=\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\ge2\sqrt{\frac{1}{\sqrt{ab}}}=\frac{2}{\sqrt{\sqrt{ab}}}\)
\(\Rightarrow\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\frac{\sqrt{\sqrt{ab}}}{2}\Leftrightarrow\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le2\frac{\sqrt{\sqrt{ab}}}{2}\)'
\(\Leftrightarrow\frac{2\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\le\sqrt{\sqrt{ab}}\)
Dấu '=' xảy ra <=> a=b
Thanks bạn rất nhiều