The below is a simplification of part of a proof I'm working on, in numerical analysis. It is similar to a paper that I studied some months ago, for which I got some advice here on MathOverflow.

I have non-negative sequences $A_k, B_k, C_k$ for $k=0\dots n$, a time-step $\Delta{t}$ and a positive number $p$ such that
$$(1+p\Delta{t})A_n + B_n \le A_{n-1} + B_{n-1} + C_n\Delta{t} \tag{$\star$}$$ and **I would like to claim** that the final solution with the unknowns $A_n$ and $B_n$ is bounded by their initial solution plus a finite quantity containing the maximum value of the known quantity $C_n$ i.e. there exists a bound of the form
$$A_n + B_n \le f(\Delta t) A_0 + g(\Delta t)B_0 + h(\Delta t)\max_{0 \le k \le n}C_k \tag{$\star\star$}$$
where I need to find the forms $f(\cdot)$, $g(\cdot)$ and $h(\cdot)$.

**What I have tried:**

- I
*removed*the non-negative $B_n$ on the left side of ($\star$) and got $$(1+p\Delta{t})A_n \le A_{n-1} + B_{n-1} + C_n\Delta{t} \tag{1}$$ - Proceeding in a similar way to the linked answer, I multiplied through by $(1+p\Delta t)^{n-1}$ and set $E_n = (1+p\Delta t)^nA_n$ to get $$E_n \le E_{n-1} + B_{n-1}(1+p\Delta t)^{n-1} + C_n(1+p\Delta t)^{n-1}\Delta{t} \tag{2}$$
- Applying recursion, $$E_n \le E_0 + \max_{0\le k \le n-1} B_k \sum_{k=0}^{n-1}(1+p\Delta t)^k + \max_{0\le k \le n} C_k \Delta{t}\sum_{k=0}^{n-1}(1+p\Delta t)^k \tag{3}$$
- Now I divided by $(1+p\Delta t)^n$, and applied $A_0 = E_0$ $$A_n \le \frac{A_0}{(1+p\Delta t)^n} + \left(\max_{0\le k \le n-1}B_k + \max_{0\le k \le n}C_k\Delta{t}\right)\frac{1}{(1+p\Delta t)^n}\sum_{k=0}^{n-1}(1+p\Delta t)^k \tag{4}$$
- Simplifying and bounding the below by a geometric sum to infinity, we have $$\frac{1}{(1+p\Delta t)^n}\sum_{k=0}^{n-1}(1+p\Delta t)^k = \sum_{k=1}^{n}\frac{1}{(1+p\Delta t)^k} < \frac{1}{p\Delta t} \tag{5}$$
- Using (5) in (4) gives $$A_n \le \frac{A_0}{(1+p\Delta t)^n} + \frac{1}{p\Delta t}\max_{0\le k \le n-1}B_k + \frac{1}{p}\max_{0\le k \le n}C_k \tag{6}$$ and the problems with this approach are

(a) There is no $B_n$ term on the left since I discarded it, and (b) the $B_{k-1}$ term on the right is an unknown. Only $B_0$ can provide a useful bound here.

I think if I don't discard $B_n$ in step 1, I might find a way to use recursion and end up with $B_n$ on the left and $B_0$ on the right. I have thought about summing from $0$ to $N$ but it's unclear how to proceed. Any suggestions on how to proceed will be very much appreciated.

I eventually figured it out (see answer below), but I am not certain because @losif's answer also seems irrefutable.