Inequality $ \vert \sqrt{a}-\sqrt{b} \vert \leq \sqrt{ \vert a -b \vert } $ I have the following inequality on my class notes that I haven't been able to prove, I was even wondering if it is actually true:
$$
\forall a,b \in \mathbb{R}^{\ge0} \left( \left| \sqrt{a}-\sqrt{b} \right| \leq \sqrt{ \vert a -b \vert}\right) 
$$
Thanks
 A: This is not going to be a polished argument. Rather, I’m presenting it as I found it, as an illustration of how you might attack such a problem.
Since the absolute value and the square root are non-negative, $|\sqrt{a}-\sqrt{b}|\le\sqrt{|a-b|}$ if and only if $\left(\sqrt{a}-\sqrt{b}\right)^2\le|a-b|$, i.e., if and only if $a-2\sqrt{ab}+b\le|a-b|$. Without loss of generality we may assume that $a\ge b$ and ask whether $a+b-2\sqrt{ab}\le a-b$. Evidently this is the case if and only if $2b\le2\sqrt{ab}$, or $b\le\sqrt{ab}$. This is certainly true if $b\le 0$. Since we’re assuming that $a\ge b$, it’s also true if $b>0$.
A: $$|a-b|=|(\sqrt a-\sqrt b)(\sqrt a+\sqrt b)|=|\sqrt a-\sqrt b|\cdot|\sqrt a+\sqrt b|\ge |\sqrt a-\sqrt b|\cdot|\sqrt a-\sqrt b|$$
A: we assume $\sqrt{a}\geq \sqrt{b}$ and we get $a\geq b$ and $a,b\geq0$ then we have
$\sqrt{a}-\sqrt{b}\le \sqrt{a-b}$ and squaring both sides we get
$a+b-2\sqrt{ab}\le a-b$ 
this is equivalent to
$2b\le 2\sqrt{ab}$ 
squaring again
we get
$b^2\le ab$
for $b=0$ it is true, for $b>0$ we have $b\le a$.
In the the other case $a<b$ the proof is analogously.
A: how about this? 
we will take $0 < b < a.$ then $(\sqrt a - \sqrt b)(\sqrt a + \sqrt b) = (a - b)$ which gives $\sqrt a - \sqrt b =  \frac{(a - b)}{(\sqrt a + \sqrt b)} = \sqrt{a-b} \  \{ \frac{\sqrt{a-b}}{\sqrt a + \sqrt b} \}$ now we will
show $\frac{\sqrt{a -b}}{\sqrt a + \sqrt b} < 1.$ this follows because 
$\frac{\sqrt{a-b}}{\sqrt a + \sqrt b} < \frac{\sqrt a}{\sqrt a + \sqrt b} < 1$ in the second step $b > 0$ was used. and hopefully we are done.
