If $0<x<y$, then prove that $\sqrt{x} <\sqrt{y}$ and $x <\sqrt{xy} <y$
In terms of Inequalities, we have covered till AM-GM inequality and Cauchy-Schwarz Inequality in class. I was thinking about applying AM-GM inequality but not sure if its right.
$0<x<y$.
$(y-x)=$
$(√y-√x)(√y+√x)>0.$
Since $√y+√x >0$, we get
$√y -√x>0$, or $√y>√x$.
$x=√x√x <√x√y <√y√y =y$
Because $$\sqrt{y}-\sqrt{x}=\frac{y-x}{\sqrt{y}+\sqrt{x}}>0,$$ $$y-\sqrt{xy}=\sqrt{y}(\sqrt{y}-\sqrt{x})>0$$ and $$\sqrt{xy}-x=\sqrt{x}(\sqrt{y}-\sqrt{x})>0.$$