How to estimate a recursive inequality with an upper bound 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.
 A: $\newcommand{\de}{\delta}$
The bound ($\star\star$) that you want is impossible in general. E.g., take $A_n=0$, $C_n=1$, $B_n=n\Delta t$ for all $n$, with $\Delta t>0$. Then ($\star$) will hold, whereas ($\star\star$) will not hold for large enough $n$, for any choice of $f,g,h$. 
Added: The bound ($\star\star$) will hold with $f=g=1$ and $h=b:=\frac{K+1}p$ if we additionally assume that $A_n$ dominates $B_n$ in the sense that $B_n\le KA_n$ for some real $K>0$ and all $n$. Indeed, let $\de:=\Delta t$, $S_n:=A_n+B_n$, and $M_n:=\max_{0\le k \le n}C_k$. Then, with such $f,g,h$, ($\star\star$) can be rewritten as 
\begin{equation}
 S_n\le S_0+bM_n. \tag{!!}
\end{equation}
On the other hand, the condition $B_n\le KA_n$ can be rewritten as $(1+p\de)A_n + B_n\ge(1+\de/b)S_n$; so, ($\star$) yields 
\begin{equation}
 (1+\de/b)S_n\le S_{n-1}+C_n\de. 
\end{equation}
Now it is easy to to prove (!!) by induction. Indeed, for $n=0$ (!!) is trivial. Assuming (!!) holds with $n-1$ in place of $n$, we have 
\begin{multline}
 (1+\de/b)S_n\le S_{n-1}+C_n\de\le S_0+bM_{n-1}+C_n\de
 \le(1+\de/b)S_0+(b+\de)M_n \\ 
 =(1+\de/b)(S_0+bM_n),
\end{multline}
so that (!!) indeed follows. 
A: Here is what I eventually did. In light of @losif's answer which I already accepted, can you please tell me what is wrong with my reasoning below?
Given ($\star$), we have
$$(1+pΔt)A_1+B_1 ≤ A_0 + B_0 + C_1\Delta{t} \tag{1}$$
and
$$(1+pΔt)A_2+B_2 ≤ A_1 + B_1 + C_2\Delta{t} \tag{2}$$
Adding a non-negative $p\Delta{t}A_1$ to the right side of (2), we get
$$(1+pΔt)A_2+B_2 ≤ (1+p\Delta{t})A_1 + B_1 + C_2\Delta{t} \tag{3}$$
and substituting (1) gives
$$(1+pΔt)A_2+B_2 ≤ A_0 + B_0 + C_1\Delta{t} + C_2\Delta{t} \tag{4}$$
Continuing in this way, we will obtain
$$(1+pΔt)A_n+B_n ≤ A_0 + B_0 + \sum_{k=1}^{n} C_k\Delta{t} \tag{5}$$
i.e.
$$A_n + \frac{1}{1+p\Delta{t}}B_n \le \frac{1}{1+p\Delta{t}}\left(A_0 + B_0 + n\Delta{t}\max_{1\le k \le n}C_k \right)$$
which means that
$$\min\left\{1,\frac{1}{1+p\Delta{t}}\right\}(A_n + B_n) \le \frac{1}{1+p\Delta{t}}\left(A_0 + B_0 + n\Delta{t}\max_{1\le k \le n}C_k \right)$$
and noting that $1 > \frac{1}{1+p\Delta{t}}$ for all $p > 0$ and $\Delta{t} >0$ leaves
$$\frac{1}{1+p\Delta{t}}(A_n + B_n) \le \frac{1}{1+p\Delta{t}}\left(A_0 + B_0 + n\Delta{t}\max_{1\le k \le n}C_k \right)$$
i.e.
$$ A_n + B_n \le A_0 + B_0 + n\Delta{t}\max_{1\le k \le n}C_k $$ which I set out to prove (except that the index of the max on the right, runs from 1, not 0), and this is okay in my case.
