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 $c>0$.

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
$$\text{rhs}=e^{-\epsilon }-\frac{e^{-\epsilon } \epsilon }{2
   x}+O\left(\frac{1}{x^2}\right)$$
$$\text{lhs}-\text{rhs}=\frac{c e^{-\epsilon } (1-\epsilon )}{x}+O\left(\frac{1}{x^2}\right)$$
