Prove inequality $|\sqrt{x}- \sqrt{y}| \le \sqrt{|x-y|}$ How to prove that $|\sqrt{x}- \sqrt{y}| \le \sqrt{|x-y|}$ ? 
Of course, I know that $|\sqrt{x}- \sqrt{y}| \le \sqrt{|x-y|} \Leftrightarrow ... \Leftrightarrow x \le y \quad \text{or} \quad x \ge y$, but for me it isn't a good proof.
 A: Square both sides. Observe that it must be that $\;0\le y\le x\;$ (why?):
$$x+y-2\sqrt{xy}\le x-y\iff y\le\sqrt{xy}$$
But, of course, $\;\sqrt{xy}\ge\sqrt{y^2}=y\;\ldots$
A: This is not the most elegant proof, but it uses a technique that may be useful for other problems.  Assume without loss of generality $0\leq x \leq y$.  Then by the Net Change Theorem, and using the fact that $1/(2\sqrt{t})$ is a decreasing function of $t$, and a change of variables $s = x+t$,
$$0 \leq \sqrt{y}-\sqrt{x} = \int_x^y \frac{d}{dt}\sqrt{t}\,dt = \int_x^y \frac{1}{2{\sqrt{t}}}\,dt  = \int_0^{y-x} \frac{1}{2{\sqrt{s+x}}}\,ds \leq \ldots $$
$$\ldots \leq \int_0^{y-x} \frac{1}{2{\sqrt{s}}}\,ds =\sqrt{y-x}.$$
Another way of stating your question is you want to show $\omega(t) = \sqrt{t}$ is a modulus of continuity for $f(x)=\sqrt{x}$.
A: Let $a=\sqrt{x}, b=\sqrt{y}$, the it suffices to prove that $$(a^2 - b^2)^2 - (a-b)^4\geq0.$$
which is trivial after you simplify the inequality. 
A: Geometric Proof:
Assume WLOG that $y<x$. Consider a right angled triangle with legs having sides: $\sqrt{y},\sqrt{(\sqrt{x})^2-(\sqrt{y})^2}$ and a hypotenuse having a length of: $\sqrt{x}$. Check using Pythagoras theorem that there exists such a right- angled triangle. Now using the triangle inequality, we get:
$$\sqrt{y}+\sqrt{(\sqrt{x})^2-(\sqrt{y})^2}\geq \sqrt{x}$$
Therefore:
$$\sqrt{(\sqrt{x})^2-(\sqrt{y})^2}\geq \sqrt{x}-\sqrt{y}$$

$$\sqrt{|x-y|}=\sqrt{x-y}\geq \sqrt{x}-\sqrt{y}=|\sqrt{x}-\sqrt{y}|$$

