Show that $\frac{a}{a-\epsilon} \geq \exp(b\epsilon)$ I need to show that
\begin{align}
\frac{a}{a-\epsilon} \geq \exp(b\epsilon)
\end{align}
is true, for $0<\epsilon < a<b$. Is this doable in a general way?
Edit: 
Sorry, I forgot to metion $ab=1$, which rather gives
\begin{align}
\frac{a}{a-\epsilon} \geq \exp\left(\frac{\epsilon}{a}\right)
\end{align}

It should be fairly easy for you to see that $e^x \le \dfrac 1{1-x}$ if $0 < x < 1$.  Then
$$ \frac{a}{a-\epsilon} = \frac{1}{1 - \epsilon/a} \ge \exp(\epsilon/a) \ge \exp(b\epsilon)$$ provided that $0 < \epsilon < a$ and $ab \le 1$.
It is false in general, though.
