Proving that $1 + \frac{r_2}{r_1} \ge \frac{1-e^{-(r_1+r_2)x}}{1-e^{-r_1x}} $ 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?
 A: 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.
A: 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.
A: 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.
