\(\dfrac{x}{\sqrt{y}}+\dfrac{y}{\sqrt{x}}-\sqrt{x}-\sqrt{y}\)
=\(\dfrac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{xy}}-\left(\sqrt{x}+\sqrt{y}\right)\)
=\(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{xy}}-\left(\sqrt{x}+\sqrt{y}\right)\)
=\(\left(\sqrt{x}+\sqrt{y}\right)\left(\dfrac{x-\sqrt{xy}+y}{\sqrt{xy}}-1\right)\)
=\(\left(\sqrt{x}+\sqrt{y}\right)\left(\dfrac{\sqrt{x}}{\sqrt{y}}-1+\dfrac{\sqrt{y}}{\sqrt{x}}-1\right)\)
Mà \(\dfrac{\sqrt{x}}{\sqrt{y}}+\dfrac{\sqrt{y}}{\sqrt{x}}\ge2\) ( Vì x,y > 0 )
Vậy GTNN của biểu thức:\(\left(\sqrt{x}+\sqrt{y}\right)\left(\dfrac{\sqrt{x}}{\sqrt{y}}+\dfrac{\sqrt{y}}{\sqrt{x}}-2\right)\ge\left(\sqrt{x}+\sqrt{y}\right)\left(2-2\right)=0\)
Dấu bằng xảy ra khi \(\sqrt{x}=\sqrt{y}\) hay x=y (Tích đúng cho mình nha)
