Inequality in difference of square roots I think this inequality is true and I'm trying to prove it: 
For real, non-negative $a$ and $b$
$$\lvert \sqrt{a} - \sqrt{b}\rvert \le \lvert \sqrt{a - b} \rvert$$
Closest thing I've found is equation
$$\sqrt{a} + \sqrt{b} = \sqrt{a + b + \sqrt{4ab}}$$
But this breaks if I set $b$ to $-b$ because then you're taking the root of a negative.
Does anyone have a proof of this, or a more general result that implies this?
 A: As you wrote $\sqrt{a-b}$, I suppose $a\ge b$. Then
$|\sqrt{a}-\sqrt{b}|^2\le|\sqrt{a}-\sqrt{b}||\sqrt{a}+\sqrt{b}|=a-b$. Take square roots on both sides and we are done.
A: This is not true if $a < b$.  If $a < b $ then $a -b < 0$ and $\sqrt{a-b}$ does not exist and $|\sqrt{a-b}|$ does not exist.  And so it is not true that $|\sqrt{a} - \sqrt{b}| \le |\sqrt{a-b}|$.
I think what you meant was to prove $|\sqrt{a} -\sqrt{b}| \le \sqrt{|a - b|}$ which is true.
Notice: $\min (a,b) \le \sqrt{ab}\le \max(a,b)$.  Why? If $c = \min (a,b)$ and $d = \max(a,b)$ then $c =\sqrt{c\cdot c}\le \sqrt{c\cdot d}(=\sqrt{ab})\le \sqrt{d\cdot d} = d$.
So.  Letting $c = \min(a,b)$ and $d=\max(a,b)$ we have
$|\sqrt a-\sqrt b|^2 = (\sqrt{d} - \sqrt{x})^2 = d  -2\sqrt{cd} + c\le d - 2c+c = d-c= |a-b|$.
And as we are dealing with non-negative values it follows 
$\sqrt {|\sqrt{a} - \sqrt{b}|^2}\le \sqrt{|a-b|}$ or
$|\sqrt{a} - \sqrt{b}|^2 \le |a-b|$
.....
Although an easier way to do it is if we don't get bogged down in absolute values; they are only there to assure to take the positive values anyway.
For any $0 \le  b\le a$ then
$\sqrt a - \sqrt b \le \sqrt{a-b} \iff$ 
$(\sqrt a - \sqrt b)^2 \le (\sqrt{a-b})^2 \iff$ 
$a -2\sqrt{ab} + b \le a -b\iff$
$b \le \sqrt{ab}$ and as $b < a$ it follow $b = \sqrt{b^2}\le \sqrt{ab}$.
....
Or directly:
$\sqrt{a - b} = \sqrt{a -2b + b}=\sqrt{a -2\sqrt{b}\sqrt{b} + b} \ge \sqrt{a - 2\sqrt{a}\sqrt{b} + b}=\sqrt{(\sqrt a - \sqrt b)^2} = \sqrt a - \sqrt b$.
A: $a\ge b\ge 0$ then
$$
\\|\sqrt{a}-\sqrt{b}|\le|\sqrt{a-b}| <=>
\\\sqrt{a}\le \sqrt{b}+\sqrt{a-b}<=>
\\a\le b+a-b+2\sqrt{b(a-b)}<=>
\\\sqrt{b(a-b)}\ge0
$$
