If $x>\sqrt{xy}>y$, then $x>y>0$.

Question

I am trying to prove the following:

If $x>\sqrt{xy}>y$, then show that $x>y>0$.

My argument is as follows:

We only need to show $y>0$. Suppose $y<0$. Then, for $\sqrt{xy}$ to be defined, we need $x<0$. Thus, $x<\sqrt{xy}$, which is a contradiction. Hence, $y>0$.

Somehow, my proof doesn't sound "rigorous" enough to me. Is there any other method to prove the result?

Answer 1

Yes that's a correct proof by contradiction, as an alternative by a constructive proof we can observe that $xy=0$ is not a solution then we need

$$xy> 0 \implies x>\sqrt{xy}>0 \quad \land \quad y> 0$$

Answer 2

$x$ is positive since $x>\sqrt {\text{something}}$ which means that $y$ is also positive since $xy$ is positive and $y\ne \sqrt {xy}$. Therefore $$x^2>xy>y^2$$which in two ways means that $$x>y>0$$