Does this inequality involving differences between powers hold on a particular range? Let 
$$f(x)=\left(1-\frac{2x}{x+c}\right)^{-n}-\left(1+\frac{x}{c}\right)^{2n}$$ and 
$$g(x)=\left(1-\frac{2x}{c}\right)^{-n}-\left(1-\frac{x}{c}\right)^{-2n}$$ 
where $c>0$ and $n>0$ are constants.
I am wondering if $f(x)\leq g(x)$ for $0\leq x <c/2$ (I am actually interested in small positive $x$).  
Clearly, this holds with equality when $x=0$ and I think the inequality holds for $0< x <c/2$.  This is based on the numerical evaluations as well as the series expansion in Mathematica of $h(x)=g(x)-f(x)$ which yields: 
$$h(x)=\frac{2 n x^3}{c^3}+\frac{3 \left(2 n^2+n\right) x^4}{c^4}+\frac{2 \left(4 n^3+5 n^2+3 n\right) x^5}{c^5}+\ldots$$
However, I am having trouble actually proving this.  Any help?
 A: Let's normalise things a bit first.
$$\left(1 - \frac{2x}{x+c}\right)^{-n} = \left(\frac{c-x}{c+x}\right)^{-n} = \left(\frac{1+\frac{x}{c}}{1-\frac{x}{c}}\right)^n.$$
So we have everything clearly as a function of $\frac{x}{c}$, whence we can without loss of generality assume $c = 1$. Then the inequality we want to show becomes
$$\begin{align}
\left(\frac{1+x}{1-x}\right)^n - (1+x)^{2n} &\leqslant \frac{1}{(1-2x)^n} - \frac{1}{(1-x)^{2n}}\\
\iff (1-x^2)^n - (1-x^2)^{2n} &\leqslant \left(\frac{(1-x)^2}{1-2x}\right)^n - 1\\
\iff (1-x^2)^n\left(1 - (1-x^2)^{n}\right) &\leqslant \left(1+\frac{x^2}{1-2x}\right)^n - 1
\end{align}$$
Now, for $0 \leqslant x < \frac12$, we have $x^2 \leqslant \frac{x^2}{1-2x}$, and by Bernoulli's inequality, we have
$$(1+x^2)^n - 1 \geqslant n\cdot x^2.$$
On the other hand, also by Bernoulli's inequality, we have
$$1 - (1-x^2)^n \leqslant n\cdot x^2,$$
and since that is multiplied with $(1-x^2)^n \leqslant 1$, we conclude
$$\begin{align}(1-x^2)^n\left(1 - (1-x^2)^{n}\right) &\leqslant (1-x^2)^n\cdot n\cdot x^2\\
&\leqslant n\cdot x^2\\
&\leqslant (1 + x^2)^n - 1\\
&\leqslant \left(1+\frac{x^2}{1-2x}\right)^n - 1.
\end{align}$$
