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

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, $$\big|\sqrt{|a|}-\sqrt{|b|}\big|\leq\sqrt{\big||a|-|b|\big|}\leq \sqrt{|a-b|}$$
