Bounding the difference between square roots I want to compute the value of $\frac{1}{\sqrt{a + b + c}}$.  Say I can observe a and b, but not c.  Instead, I can observe d which is a good approximation for c in the sense that $P( |c-d| \leq 0.001 )$ is large (say 95%), and both c and d are known to have $|c| \leq 1, |d| \leq 1$ so a difference of 0.001 is actually small.
I want to argue that $\frac{1}{\sqrt{a + b + d}}$ is a good approximation because d is a good approximation for c.  Is there anything I can say about the difference
$\frac{1}{\sqrt{a + b + c}} - \frac{1}{\sqrt{a + b + d}}$?
Maybe I could say something like $P(\frac{1}{\sqrt{a + b + c}} - \frac{1}{\sqrt{a + b + d}} \leq 0.001) = x?$  I'm worried that being under a square root might mess things up.
Would I need to find out the exact distribution of $c - d$, or anything else before I can make such claims?

$\frac{1}{\sqrt{a + b + c}} - \frac{1}{\sqrt{a + b + d}} = \frac{\sqrt{a + b + d}-\sqrt{a + b + c}}{\sqrt{a + b + c}\sqrt{a + b + d}} $
$ =\frac{(\sqrt{a + b + d}+\sqrt{a + b + c})(\sqrt{a + b + d}-\sqrt{a + b + c})}{(\sqrt{a + b + d}+\sqrt{a + b + c})\sqrt{a + b + c}\sqrt{a + b + d}}$
$ =\frac{(d-c)}{(\sqrt{a + b + d}+\sqrt{a + b + c})\sqrt{a + b + c}\sqrt{a + b + d}}$
Denominator is positive.  This difference is smaller when denominator is greater than 1.
