\(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}=\frac{\left(\sqrt{a}\right)^2}{\sqrt{b}}+\frac{\left(\sqrt{b}\right)^2}{\sqrt{a}}\ge\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}=\sqrt{a}+\sqrt{b}\left(dpcm\right)\)
Theo bđt Cauchy :
\(\frac{a}{\sqrt{b}}+\sqrt{b}\ge2\sqrt{\frac{a}{\sqrt{b}}\cdot\sqrt{b}}=2\sqrt{a}\)
Dấu "=" \(\Leftrightarrow\frac{a}{\sqrt{b}}=\sqrt{b}\Leftrightarrow a=b\)
+ Tươ tự ta cm đc : \(\frac{b}{\sqrt{a}}+\sqrt{a}\ge2\sqrt{b}\)
Dấu "=" <=> a = b
Do đó : \(\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}+\sqrt{a}+\sqrt{b}\ge2\left(\sqrt{a}+\sqrt{b}\right)\)
=> đpcm
Dấu "=" <=> a = b
