Question: so, my problem is the following, I really tried everything I could...

Let $r\_1,r\_2 \in \mathbb{R\_+}$ be two constants and let $f: x \rightarrow \mathbb{R\_+}$, with $x\in \mathbb{R\_+}$

Define:

$$ f(x) = \frac{1- e^{-(r\_1 + r\_2)x}}{1-e^{-r\_1 x}} $$

I need to show that:

$$ 1 + r\_2/r\_1 \geq f(x) $$

For any $x\geq0$. I have no idea how to approach this... I tried differentiating, without success... Is it possible to prove this, analytically?

Answer: 
*Alternative proof*:

It suffices to prove that
$$\frac{r\_2}{r\_1} \ge \frac{1 - \mathrm{e}^{-(r\_1 + r\_2)x}}{1 - \mathrm{e}^{-r\_1 x}} - 1
= \frac{1 - \mathrm{e}^{-r\_2 x}}{\mathrm{e}^{r\_1 x} - 1}$$
or
$$\frac{r\_2}{r\_1}(\mathrm{e}^{r\_1 x} - 1) \ge 1 - \mathrm{e}^{-r\_2 x}.$$

Using $\mathrm{e}^u \ge 1 + u$ for all $u\in \mathbb{R}$ (well-known, also easy to prove),
we have $\mathrm{e}^{r\_1 x} - 1 \ge r\_1 x$ and $1 - \mathrm{e}^{-r\_2 x} \le r\_2 x$. The desired result follows.

We are done.
**Hint 1**: $$1+\frac{r\_2}{r\_1} \ge \frac{1-e^{-(r\_1+r\_2)x}}{1-e^{-r\_1x}} \\\iff \\ (1+\frac{r\_2}{r\_1})(1-e^{-r\_1x}) \ge 1-e^{-(r\_1+r\_2)x} \\ \iff \\ 
(1+\frac{r\_2}{r\_1})(1-e^{-r\_1x}) - (1-e^{-(r\_1+r\_2)x}) \ge 0$$

**Hint 2:** Apply differentiation.