Why does the proof of $|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|}$ work from the case solution?

Question

There is certain proof:

$|\sqrt{x}-\sqrt{y}| \leq \sqrt{|x-y|}$ where $x, y, \geq 0$

And by doing some rough work (starting with the above equation to see if the proof is true), it ends up to be of the case (i.e. case 1) $x \geq y$:

$\sqrt{xy} \geq y$

However, the question solution states that this is true because:

$\sqrt{xy} > \sqrt{y\cdot y} = y$

I do not seem to get this. An explanation and/or next steps would be appreciated. Thanks.

Answer 1

Multiplying $x \geq y$ with $y$ we get $xy \geq yy = y^2$, because these are positive numbers. Since the square root is a strictly increasing function, it preserves orders and thus $\sqrt{xy}\geq \sqrt{y^2}=y$.