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?
 A: Use a Taylor series (or equivalently the Binomial Theorem) to expand around $c$.  This is valid provided $|d-c| \lt |a+b+c|$:
$$\eqalign{
&\frac{1}{\sqrt{a+b+c}} - \frac{1}{\sqrt{a+b+d}}\\
&= (a+b+c)^{-1/2} - (a+b+c + (d-c))^{-1/2} \\
&= (a+b+c)^{-1/2} - (a+b+c)^{-1/2}\left(1 + \frac{d-c}{a+b+c}\right)^{-1/2} \\
&= (a+b+c)^{-1/2} - (a+b+c)^{-1/2}\sum_{j=1}^{\infty}\binom{-1/2}{j}\left(\frac{d-c}{a+b+c}\right) ^j\\
&= \frac{1}{2}(a+b+c)^{-3/2}(d-c) - \frac{3}{8}(a+b+c)^{-5/2}O(d-c)^2
}
$$
The difference therefore is approximately  $\frac{1}{2}(a+b+c)^{-3/2}$ times $(d-c)$ and the error is (a) negative (because this is an alternating series when $d-c$ is positive and has all negative terms when $d-c$ is negative), (b) proportional to  $\frac{3}{8}(a+b+c)^{-5/2}$, and (c) of second order in $d-c$.  That should be enough to complete your analysis.  (This leads essentially to the delta method.)
A: $\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.
A: New comment:
You problem should be defined as finding a sequence ${x_n}$ so that $f(x_n)=\dfrac{1}{\sqrt{r+x_n}}\rightarrow f(x)=\dfrac{1}{\sqrt{r+x}}$ with given constant $r=a+b$.
Obviously, this problem is easy to be solved by Newton's method.
You can assume that for specific parameter $x_0$, $f(x)$ is equivalent to constant $k$. It means $f(x_0)=\dfrac{1}{\sqrt{r+x_0}}=k$. This is equivalent to finding the solution to $\dfrac{1}{\sqrt{r+x_0}}=k$. The function to use Newton's method is then,
$g(x)=\dfrac{1}{\sqrt{r+x}}$, with derivate $g'(x)=-\dfrac{x}{2\left(r+x\right)^{3/2}}$. With condition that parameter $x$ is $|x|\leq1$, the initial guess of $x$ is in $[-1,1]$. The sequence given by Newton's method is
$x_n=x_{n-1}-\dfrac{g(x_{n-1})}{g'(x_{n-1})}=x_{n-1}+(\dfrac{1}{r+x_{n-1}}-k)\cdot \dfrac{2(r+x_{n-1})^{3/2}}{x_{n-1}}$.
Thus sequence $x_{n}$ is your result.
Here is a R code to generate the sequence $x_n$. It converge to $98.99$.
    y<-rep(0,1000);y[1]<-0.5;r<-1;k<-0.1;for (i in seq(2,1000)) {print(y[i-1]);y[i]<-y[i-1]+(1/sqrt(r+y[i-1])-k)*(2*(r+y[i-1])^(3/2)/y[i-1])}

Old comment:
How do you define the term "good approximation"? Value $c$ and value $d$ are follow normal distribution with mean $\mu_{\{c,d\}}$ and sd $\sigma_{\{c,d\}}$ that
you want to maximize the probability CD where $CD=\dfrac{1}{\sqrt{x+c}}-\dfrac{1}{\sqrt{x+d}}$ with given constant $x=a+b$? If so, maximize the likelihood $L(c,d)=\mathrm{Pr}(CD=r|c,d)$ is proper way.
Anyway, it's better you give a mathematical definition of "good approximation" for us and you to discuss a proper question.
