Prove the difference of roots is less than or equal to the root of the difference I am doing a larger proof that requires this to be true:
$|\sqrt{a} - \sqrt{b}| \leq \sqrt{|a - b|}$
I can square both sides to get
$a - 2\sqrt{a}\sqrt{b} + b \leq |a - b|$
Note that a and b are > 0.
I also know that
$|c| - |d| \leq |c - d|$
It seems like a mistake that $+b$ is there on the left...
How can I prove this is true
 A: $\sqrt{a}$ and $\sqrt{a}$ are positive, therefore,
$$-\sqrt{b}\leq \sqrt{b}$$ ant it follows that 
$$\sqrt{a}-\sqrt{b}\leq \sqrt{a} +\sqrt{b} $$
by symmetry we have 
$$|\sqrt{a}-\sqrt{b}|\leq |\sqrt{a} +\sqrt{b} |=\sqrt{a} +\sqrt{b}$$
Now muptiply both sides by $|\sqrt{a}-\sqrt{b}|$ to get 
$$|\sqrt{a}-\sqrt{b}|^2\leq |a-b|$$
A: We may suppose that $a\ge b$. So,
$$\begin{align}a-2\sqrt{ab}+b\le|a-b|&\iff a-2\sqrt{ab}+b\le a-b\\&\iff b\le \sqrt {ab}\\&\iff b^2\le ab\\&\iff b(a-b)\ge 0.\end{align}$$
Hence, all we need is to prove that $b(a-b)\ge 0$ holds for any $a\ge b\ge 0$ and it does hold for any $a\ge b\ge 0$. So, $a-2\sqrt{ab}+b\le|a-b|$ also holds for any $a\ge b\ge 0$.
A: Late answer, but we can prove a stronger inequality. Let $a,b\in\mathbb{R}$ and assume, without loss of generality, $|a|\geq |b|$.
$$\begin{align}\big|\sqrt{|a|}-\sqrt{|b|}\big|\leq \sqrt{\big||a|-|b|\big|}&\Longleftrightarrow |a|-2\sqrt{|a||b|}+|b|\leq \big||a|-|b|\big|\\&\Longleftrightarrow |a|-2\sqrt{|a||b|}+|b|\leq |a|-|b|\\&\Longleftrightarrow|b|\leq\sqrt{|a||b|}\\&\Longleftrightarrow|b|^2\leq |a||b|\\&\Longleftrightarrow0\leq |b|(|a|-|b|)\end{align} $$
The third and final inequalities are true since $\big||a|-|b|\big|=|a|-|b|$ if and only if  $|a|-|b|\geq 0$ if and only if $|a|\geq |b|$, as assumed.
Recall that $\big||a|-|b|\big|\leq|a-b|$ so $\sqrt{\big||a|-|b|\big|}\leq \sqrt{|a-b|}$, and finally, $$\big|\sqrt{|a|}-\sqrt{|b|}\big|\leq\sqrt{\big||a|-|b|\big|}\leq \sqrt{|a-b|}$$
A: A (very) late proof, but no assumptions as to the relative sizes for $a$ and $b$, the elimination of the absolute value operators, and the use of simple elementary operations make this very straightforward. (Note that all actions that square or multiply involve strictly positive terms, so the inequality signs do not change.)
$$|\sqrt{a} - \sqrt{b}| \le \sqrt{|a - b|}$$
Square both sides: $$a+b-2\sqrt{ab}\le|a-b|$$
Square both sides again: $$a^2+b^2+6ab-4a\sqrt{ab}-4b\sqrt{ab}\le a^2+b^2-2ab$$
Rearrange terms: $$8ab\le4a\sqrt{ab}+4b\sqrt{ab}$$
Divide by $4\sqrt{ab}$: $$2\sqrt{ab}\le a+b$$
Square both sides one final time: $$4ab\le a^2+b^2+2ab$$
Rearrange terms: $$0\le a^2+b^2-2ab$$
Factor the right side: $$0\le (a-b)^2$$
Since a squared real term is always greater or equal than zero, the proof is complete.
