Question

Cho a,b>0. CM:

\(\sqrt{\frac{a^2}{b}}+\sqrt{\frac{b^2}{a}}>=\sqrt{a}+\sqrt{b}\)

Answer 1

\(VT=\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}\ge\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}=\sqrt{a}+\sqrt{b}\)

Dấu "=" xảy ra khi a=b 

Answer 2

Ta có:\(\sqrt{\frac{a^2}{b}}+\sqrt{\frac{b^2}{a}}=\frac{a}{\sqrt{b}}+\frac{b}{\sqrt{a}}=\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{ab}}\) (1)

Mặt khác:  \(\sqrt{a}+\sqrt{b}=\frac{\sqrt{a}+\sqrt{b}}{\sqrt{ab}}\) (2)

Từ (1) và (2) \(\Rightarrow\sqrt{\frac{a^2}{b}}+\sqrt{\frac{b^2}{a}}\ge\sqrt{a}+\sqrt{b}\left(đpcm\right)\)