Ta cần chứng minh:
\(\sqrt{x}\left(\sqrt{\dfrac{x}{y}}-1\right)\ge\sqrt{y}\left(1-\sqrt{\dfrac{y}{x}}\right)\)
\(\sqrt{\dfrac{x^2}{y}}-\sqrt{x}\ge\sqrt{y}-\sqrt{\dfrac{y^2}{x}}\)
\(\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\ge\sqrt{x}+\sqrt{y}\left(x,y>0\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2}=x\\\sqrt{y^2}=y\end{matrix}\right.\right)\)(1)
Áp dụng BĐT Cô si cho 2 số dương, ta có:
\(\left\{{}\begin{matrix}\dfrac{x}{\sqrt{y}}+\sqrt{y}\ge2\sqrt{x}\\\dfrac{y}{\sqrt{x}}+\sqrt{x}\ge2\sqrt{y}\end{matrix}\right.\)
\(\Rightarrow\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}\ge\sqrt{x}+\sqrt{y}\) (2)
Từ \(\left(1\right)và\left(2\right)\Rightarrowđpcm\)
Dấu "=" xảy ra \(\Leftrightarrow x=y\)
