Proving $(1-\frac{1}{x-c})^{\epsilon x-c}\geq(1-\frac{1}{x})^{\epsilon x}$ How do I formally prove that $(1-\frac{1}{x-c})^{\epsilon x-c}\geq(1-\frac{1}{x})^{\epsilon x}$ which can be seen graphically. Here $\epsilon$ is a small positive constant $\ll1$, $x\gg0$ and $c>0$.
 A: Suppose that $x$ is very large and use Taylor expansion
$$\log \left(1-\frac{1}{x-c}\right)=-\frac{1}{x}-\frac{c+\frac{1}{2}}{x^2}+O\left(\frac{1}{x^3}\right)$$
$$(\epsilon x-c)\log \left(1-\frac{1}{x-c}\right)=-\epsilon +\frac{-c \epsilon +c-\frac{\epsilon }{2}}{x}+O\left(\frac{1}{x^2}\right)$$
$$\text{lhs}=\left(1-\frac{1}{x-c}\right)^{\epsilon x-c}=\exp\left((\epsilon x-c)\log \left(1-\frac{1}{x-c}\right) \right)$$
$$\text{lhs}=e^{-\epsilon }+\frac{e^{-\epsilon } \left(\left(-c-\frac{1}{2}\right) \epsilon
   +c\right)}{x}+O\left(\frac{1}{x^2}\right)$$
Do the same for the rhs to get
$$\text{rhs}=e^{-\epsilon }-\frac{e^{-\epsilon } \epsilon }{2
   x}+O\left(\frac{1}{x^2}\right)$$
$$\text{lhs}-\text{rhs}=\frac{c e^{-\epsilon } (1-\epsilon )}{x}+O\left(\frac{1}{x^2}\right)$$
A: Let us prove the inequality in a more precise manner.
The inequality is written as
$$(\epsilon x - c)\ln \big(1 - \frac{1}{x-c}\big) \ge  \epsilon x \ln \big(1 - \frac{1}{x}\big)$$ 
or
$$\epsilon \le \frac{-c \ln \big(1 - \frac{1}{x-c}\big)}{x \ln \big(1 - \frac{1}{x}\big) - x \ln \big(1 - \frac{1}{x-c}\big) }.\tag{1}$$
To proceed, we need the following results. Their proof is given later.
Claim 1: Let $c > 0$ and $x - c > 1$. Then
$$\frac{-c \ln \big(1 - \frac{1}{x-c}\big)}{x \ln \big(1 - \frac{1}{x}\big) - x \ln \big(1 - \frac{1}{x-c}\big) }
\ge -(x-c-1)\ln \big(1 - \frac{1}{x-c}\big) .$$
Claim 2: $-(x-c-1)\ln \big(1 - \frac{1}{x-c}\big)$
is strictly increasing with $x-c$.
Let us proceed. From Claims 1 and 2, we can obtain some sufficient conditions for the inequality in (1) to be true: if $x-c \ge 2$ and $\epsilon \le \ln 2$, the inequality in (1) is true;
or if $x- c\ge \frac{11}{10}$ and $\epsilon \le \frac{\ln 11}{10} \approx 0.2397$, the inequality in (1) is true; etc.
$\phantom{2}$
Proof of Claim 1: It suffices to prove that
$$f(c) = \frac{c}{x-c-1} - x \ln \Big(1 - \frac{1}{x}\Big) + x \ln \Big(1 - \frac{1}{x-c}\Big)\ge 0.$$
Note that $\lim_{c\to 0} f(c) = 0$ and $f'(c) = \frac{c}{(x-c)(x-c-1)^2} > 0$. The desired result follows.
Proof of Claim 2: Let $g(u) = -(u-1)\ln \big(1 - \frac{1}{u}\big)$. 
We have $g'(u) = -\frac{1}{u} - \ln \big(1 - \frac{1}{u}\big) > 0$ for $u > 1$
where we have used $\ln (1-v) < -v$ for $v\in (0, 1)$. We are done.
