I am looking for a "reversed" form of Grönwall's inequality. Let's recall the usual requirements from Grönwall's inequality. First, denote by $I\subset\mathbb{R}$ an interval of the form $[a,b]$ with $a<b$. Second, let $\alpha$, $\beta$, and $u$ be real-valued functions defined on $I$. Finally, assume that $u$ and $\beta$ are continuous and that $\alpha$ is non-decreasing.
Suppose that $\beta$ is non-negative and that $u$ satisfies the integral inequality $$ u(t) \,\le\, \alpha(t) \,+\!\int_t^b\!\!\!\beta(s)\,u(s)\,\text{d}s\quad\text{for all}\quad t\in [a,b].$$
I am looking for an upper bound on $u$ as in Grönwall's inequality. For example, is the following true: $$u(t)\,\le\, \alpha(t)\exp\Big(\int_t^b\!\!\!\beta(s)\,\text{d}s\Big)\quad\text{for all}\quad t\in [a,b].$$
My guess is that such a result can be derived using arguments as in the proof of Gronwall's inequality. But I was not able to figure it out. Could you please point me in the direction of a proof or a suitable reference?