Proving inequality involving absolute value

Question

Question:

Prove that $\left|\sqrt{x}-\sqrt{y}\right|\le \sqrt{\left|x-y\right|} $ where $x ,y \ge 0$

I know this question has been asked many times before, but I just want to confirm if the following way is correct since I couldn't find any answers that did it this way.

Attempt:

$ \left|\sqrt{x}-\sqrt{y}\right|\le \sqrt{\left|x-y\right|} \iff $ $ \left(\sqrt{x}-\sqrt{y}\right)^2\ \le \left|x-y\right| \\\\$ $\iff$ $ x\ -\ 2\sqrt{xy}+y\ \le \left|x-y\right|\le \left|x\right|+\left|y\right|=\ x\ +y$ $ \iff$ $ x\ -\ 2\sqrt{xy}+y\le x+y$ $\iff$ $ 2\sqrt{xy}\ge 0$

Answer 1

Starting from $$2\sqrt{xy} \geq 0$$

We have $$x-2\sqrt{xy}+y \leq x+y$$

Also from triangle inequality, we have $$|x-y| \leq x+y$$

I don't see how do you conclude $$x-2\sqrt{xy}+y \leq |x-y|$$

Answer 2

Wlog, we can suppose that $x>y\ge 0$

we want to prove $$(\sqrt {x}-\sqrt {y})^2\le x-y $$

or

$$x+y-2\sqrt {xy}\le x-y $$ $$\iff $$

$$y\le \sqrt {xy} $$

which is true since $\sqrt {x}>\sqrt {y } $.

Answer 3

Multiply both sides by $|\sqrt{x}+\sqrt{y}|$: $$|x-y|\le \sqrt{|x-y|}\cdot |\sqrt{x}+\sqrt{y}| \Rightarrow$$ $$\sqrt{|x-y|}\le \sqrt{|x|+|y|}\le |\sqrt{x}+\sqrt{y}| \Rightarrow$$ $$x+y\le x+y+2\sqrt{xy} \Rightarrow 0\le 2\sqrt{xy}.$$