ta có : \(\dfrac{\sqrt{x}+\sqrt{y}}{2}\ge\sqrt{\sqrt{xy}}\) \(\Leftrightarrow\dfrac{\left(\sqrt{x}+\sqrt{y}\right)^2}{4}\ge\sqrt{xy}\)
\(\Leftrightarrow\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)^2}{4}\le-\sqrt{xy}\) \(\Leftrightarrow-\sqrt{xy}\ge\dfrac{-\left(4\right)^2}{4}=-4\)
vậy min của \(-\sqrt{xy}\) là \(-4\) dấu "=" xảy ra khi \(\sqrt{x}=\sqrt{y}=2\Leftrightarrow x=y=4\)