Question

a,b>0 c/m \(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\)≥\(\sqrt{a}+\sqrt{b}\)

Answer 1

Ta cần chứng minh: \(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}\ge\sqrt{a}+\sqrt{b}\) với a,b>0

\(\Leftrightarrow a\sqrt{a}+b\sqrt{b}\ge\left(\sqrt{a}+\sqrt{b}\right)\sqrt{ab}=a\sqrt{b}+b\sqrt{a}\) 

\(\Leftrightarrow a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{a}\ge0\Leftrightarrow a\left(\sqrt{a}-\sqrt{b}\right)-b\left(\sqrt{a}-\sqrt{b}\right)\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)\ge0\)

\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2\left(\sqrt{a}+\sqrt{b}\right)\ge0\) luôn đúng với mọi a,b>0

Vậy...

Answer 2

\(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}=\left(\dfrac{a}{\sqrt{b}}+\sqrt{b}\right)+\left(\dfrac{b}{\sqrt{a}}+\sqrt{a}\right)-\left(\sqrt{a}+\sqrt{b}\right)\)

Áp dụng bất AM-GM: \(\ge2\sqrt{a}+2\sqrt{b}-\left(\sqrt{a}+\sqrt{b}\right)=\sqrt{a}+\sqrt{b}\left(đpcm\right)\)