If it's not trivial, how can I show this inequality holds given some information I asked a question before but I think it was not clear. Here, I try to explain better and include my previous calculations.
We have constants $\alpha_1,\alpha_2,r_1,r_2,c>0$ and we know from our data that $r_2 - r_1\ge c$.
We are trying to show that
$$\alpha_1(r_2-r_1-c)\ge(\alpha_1-\alpha_2)r_2e^{\alpha_2(t-T)},\tag{1}$$
for $0\le t \le T$.
It's trivial that since $r_2 - r_1\ge c$, if $\alpha_1\le \alpha_2$, this inequality holds. But what confuses me is when $\alpha_1>\alpha_2$. What would happen then?
We have one last piece of information that I tried to use to show the inequality above holds when $\alpha_1>\alpha_2$, which is
$$r_2-r_1-c>r_2e^{\alpha_2(t-T)}-r_1e^{\alpha_1(t-T)}. \tag{2}$$
So what I did so far was to start from (2) and multiply it by $\alpha_1$
\begin{align}
\alpha_1(r_2-r_1-c)&>\alpha_1r_2e^{\alpha_2(t-T)}-\alpha_1r_1e^{\alpha_1(t-T)} \\
&>\alpha_1r_2e^{\alpha_2(t-T)}-\alpha_1r_1e^{\alpha_2(t-T)} \\
&=\alpha_1(r_2 - r_1)e^{\alpha_2(t-T)} \tag{3}
\end{align}
But as you can see, the right side of (1) and (3) are quite different so this is not conclusive.
Could someone give me a hint please?
 A: Let us prove a stronger inequality i.e. Fact 1 (2) implies (1) in the OP, given the conditions.
Fact 1: Let $c, \alpha_1, \alpha_2, r_1, r_2, t, T$ be real numbers such that
\begin{align*}
 &c > 0,\\
 &\alpha_1 > \alpha_2 > 0, \\
 &r_2 - c \ge r_1 > 0, \\
 &0 \le t < T, \\
 &r_1\left(1-e^{\alpha_1(t-T)}\right) < r_2\left(1-e^{\alpha_2(t-T)}\right)-c. \tag{1}
\end{align*}
Then
$$\alpha_1r_1e^{\alpha_1(t-T)}\le\alpha_2r_2e^{\alpha_2(t-T)}. \tag{2}$$
(The proof is given at the end.)
$\phantom{2}$

Proof of Fact 1:
From (1), we have
$$\alpha_2 >  - \frac{1}{T-t}\ln\left(1 - \frac{r_1\left(1-e^{\alpha_1(t-T)}\right) + c}{r_2}\right) := \beta.$$
Thus,
$$\beta < \alpha_2 < \alpha_1.$$
Taking logarithm, it suffices to prove that
$$\ln \left(\alpha_1r_1e^{\alpha_1(t-T)}\right)
\le \ln \alpha_2 + \ln r_2 + \alpha_2 (t - T).$$
Let
$$f(\alpha_2) := \ln \alpha_2 + \ln r_2 + \alpha_2 (t - T) - \ln \left(\alpha_1r_1e^{\alpha_1(t-T)}\right).$$
Clearly, $f(\alpha_2)$ is concave.
We have $f(\alpha_1) = \ln (r_2/r_1) > 0$.
Let us prove that $f(\beta) \ge 0$. It suffices to prove that
$$(T - t)\alpha_1r_1e^{\alpha_1(t-T)}\le (T - t)\beta r_2e^{\beta(t-T)}$$
or
\begin{align*}
 &(T - t)\alpha_1r_1e^{\alpha_1(t-T)}\\
 &\le - r_2 \left(1 - \frac{r_1\left(1-e^{\alpha_1(t-T)}\right) + c}{r_2}\right)\ln\left(1 - \frac{r_1\left(1-e^{\alpha_1(t-T)}\right) + c}{r_2}\right). 
\end{align*}
Let $g(c) := \mathrm{RHS} - \mathrm{LHS}$.
It is easy to prove that $g(c)$ is concave on $(0, r_2 - r_1]$.
We have
$$g(r_2 - r_1) = r_1 e^{\alpha_1(t-T)}\ln \frac{r_2}{r_1} > 0.$$
Let us prove that $g(0) \ge 0$.
Letting $x = r_1/r_2$, it suffices to prove that
\begin{align*}
 &x (T - t)\alpha_1e^{\alpha_1(t-T)}\\
 &\le - \left[ 1 - x\left(1-e^{\alpha_1(t-T)}\right)\right]\ln\left[ 1 - x\left(1-e^{\alpha_1(t-T)}\right)\right].
\end{align*}
Let $h(x) := \mathrm{RHS} - \mathrm{LHS}$.
It is easy to prove that $h(x)$ is concave on $(0, 1)$.
We have $h(0) = h(1) = 0$.
Thus, $h(x) \ge 0$ on $(0, 1)$.
Thus, $g(c) \ge 0$ on $(0, r_2 - r_1)$.
Thus, $f(\beta) \ge 0$.
Thus, $f(\alpha_2) \ge 0$ on $\beta < \alpha_2 < \alpha_1$.
We are done.
