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: 
The inequality is equivalent to
$$
 \frac{e^{-r\_1 x}-1}{r\_1 x} \le \frac{e^{-(r\_1+r\_2) x}-1}{(r\_1 +r\_2)x}
$$
and that is true because the function
$$
 g(t) = \frac{e^{-t}-1}{t}
$$
is *increasing* on $\Bbb R\_+$. You can verify this either by computing the derivative of $g$, or by interpreting $g(t)$ as the slope of a secant and using that $t \mapsto e^{-t}$ is *convex.*
**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.