If $0<x<y$, then prove that $\sqrt{x} <\sqrt{y}$ and $x <\sqrt{xy} <y$

Question

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.

Answer 1

$0<x<y$.

$(y-x)=$

$(√y-√x)(√y+√x)>0.$

Since $√y+√x >0$, we get

$√y -√x>0$, or $√y>√x$.

$x=√x√x <√x√y <√y√y =y$

Answer 2

Because $$\sqrt{y}-\sqrt{x}=\frac{y-x}{\sqrt{y}+\sqrt{x}}>0,$$ $$y-\sqrt{xy}=\sqrt{y}(\sqrt{y}-\sqrt{x})>0$$ and $$\sqrt{xy}-x=\sqrt{x}(\sqrt{y}-\sqrt{x})>0.$$

Answer 3

$ \sqrt{x} <\sqrt{y} \implies x = \sqrt{x} \sqrt{x} < \sqrt{x} \sqrt{y} = \sqrt{xy} $

$ \sqrt{x} <\sqrt{y} \implies \sqrt{xy} = \sqrt{x} \sqrt{y} < \sqrt{y} \sqrt{y} = y $