$0I want to show that given that $x,y>0$, we can deduce that $0<x^2<y^2 \Rightarrow x<y$. I am having problems with squarerooting the inequalities here.
$\sqrt{x^2}=|x|$ and $|x|=x$ here since x is positive so can we just squareroot the double inequality and deduce that the implication holds here.
In other words $$0<x^2<y^2 \Rightarrow 0<|x|<|y| \Rightarrow 0<x<y$$ and hence we get our result. Is this valid?

If $x \geq y > 0$, then $x^{2} \geq yx \geq y^{2}$; so $x^{2} < y^{2}$ implies $x<y$. 
For a why you did not do it right, check out the first comment below.
