<s>[INST] If the sum of the digits of $a+b, a+c, a+b$ are bounded by $k$, what is the maximum sum of digits of $a+b+c$? 
Question: For $n \in \mathbb{N}$, let $S(n)$ denote the sum of the base 10 digits of $n$. Assuming that $a,b,c \ge 0$ and
  $$
S(a+b), S(b+c),S(a+c) \le k,
$$
  What is the maximum possible value of $S(a+b+c)$ (as a function of $k$)?

This is motivated by this question.
We know that the answer is between $15k-9$ and $15k$, inclusive. Both of these bounds are found by adapting the answers to the motivating question. Specifically, we can achieve $15k-9$ by taking
\begin{align*}
a &= \underbrace{444\ldots4}_{k-1} \;\underbrace{555\ldots 5}_{k-1}\;\underbrace{555\ldots 5}_{k-1} \;5 \\\
b &= \underbrace{555\ldots 5}_{k-1}\;\underbrace{444\ldots 4}_{k-1}\;\underbrace{555\ldots 5}_{k-1}\;5\\
c &= \underbrace{555\ldots 5}_{k-1}\;\underbrace{555\ldots 5}_{k-1}\;\underbrace{444\ldots 4}_{k-1}\;5.
\end{align*}
And we can also prove that $S(x+y) \le S(x) +S(y)$ and $S(x) \le 5S(2x)$ (see my answer to the motivating question), so that
$$
S(a+b+c) \le 5S(2a+2b+2c) \le 5[S(a+b) + S(a+c) + S(b+c)] \le 15k.
$$
In the comments, it was also determined that we may assume either $S(a+b) = S(a+c) = S(b+c) = k$ and $S(a+b+c) = 15k$, or $S(a+b) = S(a+c) = k$, $S(b+c) = k-1$, and $S(a+b+c) = 15k - 5$.

I expect we can get an exact answer, and that that answer is $15k - 9$ (for $k \ge 1$).
 [/INST] We will show that the maximum is indeed $15k - 9$.
First, let's investigate the inequalities which you used, and in particular, what the effect is if either is not an equality.
In your previous answer, it was shown that $S(x) \leq 5S(2x)$ essentially by noting that this is true for each digit, and then noting that $S(2x)$ is equal to the sum over the digits $d$ of $x$ of $S(2d)$. Now we note that for individual digits, equality occurs in $S(d) \leq 5S(2d)$ precisely when $d$ is either $0$ or $5$. It follows that if any digit of $x$ (say the digit is $d$) is not $5$ or $0$, then we in fact have that $S(x) \leq 5S(2x) - 5S(2d) + S(d)$. But if $d$ is not $5$ or $0$, then we can check that $5S(2d) - S(d)$ is at least $9$, and so $S(x) \leq 5S(2x) - 9$.
The consequence of this is that if the digits of $a + b + c$ are not each equal to either $5$ or $0$, then we get that $S(a + b + c) \leq 5S(2a + 2b + 2c) - 9 \leq 15k - 9$.
We see that if $S(a + b + c) > 15k - 9$ then the digits of $a + b + c$ is each either $0$ or $5$.
If we look at your proof that $S(x + y) \leq S(x) + S(y)$, we note that equality occurs if and only if there are no carries when adding $x$ and $y$, and that in fact if there are any carries, then we can see that $S(x + y) \leq S(x) + S(y) - 9$. (A carry decreases the sum of the digits by at least $9$.)
We see that if there are any carries involved in adding $a + b$, $b + c$ and $c + a$ to get $2(a + b + c)$, then in fact $$S(a + b + c) \leq 5S(2a + 2b  + 2c) \leq 5(S(a + b) + S(b + c) + S(c + a) - 9) \leq 15k - 45.$$
This then shows that if $S(a + b + c) > 15k - 9$, then there are no carries involved in adding together $a + b$, $b + c$ and $c + a$.
Now suppose that $a$, $b$, and $c$ are as in the conditions of the problem, and that $S(a + b + c) > 15k - 9$. We can assume that $a$, $b$ and $c$ do not all end in a zero, as otherwise we could remove the trailing zero from each one. Let $M = a + b + c$, and let
$$ a = \sum_{i = 0}^\infty a_i 10^i $$
be the decimal expansion of $a$, and analogously for $b$ and $c$.
Since the last digit of $M$ is a $0$ or a $5$, the last digit of $2M$ must be a $0$. Since there are no carries involved in the addition, each of $a + b$, $b + c$, and $c + a$ must end in a $0$. We thus have that each of $a_0 + b_0$, $b_0 + c_0$, and $c_0 + a_0$ must be $0$ or $10$, and they are not all $0$. If, say, $a_0 + b_0$ were $0$, then this would imply $a_0 = b_0 = 0$. But then since at least one of $b_0 + c_0$ and $c_0 + a_0$ is not $0$, we would have that $c_0 = 10$, a contradiction. It follows that $a_0 + b_0 = b_0 + c_0 = c_0 + a_0 = 10$, and so $a_0 = b_0 = c_0 = 5$.
For will now show by induction that for each $n$, we have that $9 \leq a_n + b_n + c_n \leq 19$, and that when adding $a + b + c$ to get $M$, there is a carry of $1$ for each digit. This is true for $n = 0$ as shown above. Now suppose that for some $n$ that the claim is true for $n-1$. We will show that it is also true for $n$.
Since the digits of $M$ are each either $0$ or $5$, we have that the $n^\text{th}$ digit of $2M$ is either a $0$ or a $1$. Since there are no carries involved in the addition of $a + b$, $b + c$ and $c + a$, we have that the $n^\text{th}$ digit of each of $a + b$, $b + c$ and $c + a$ is either a $0$ or a $1$.
Now the $n^\text{th}$ digit of $a + b$ is either the units digit of $a_{n} + b_{n}$, or the units digit of $a_n + b_n + 1$. (The carry for each digit when adding $a$ and $b$ is at most $1$)
We thus have that $a_n + b_n$ must be one of $0, 1, 9, 10$ or $11$. Similarly, each of $b_n + c_n$ and $c_n + a_n$ is one of $0, 1, 9, 10$ or $11$.
Now by the inductive hypothesis, the carry involved when calculating the $n^\text{th}$ digit of $M$ is exactly $1$, and so the $n^\text{th}$ digit of $M$ is equal to the units digit of $a_n + b_n + c_n + 1$. The units digit of this must be either $0$ or $5$, and so we get that $a_n + b_n + c_n$ is one of $4, 9, 14, 19$ or $24$. Thus $2(a_n + b_n + c_n)$ is one of $8, 18, 28, 38$ or $48$. Since $a_n + b_n$, $b_n + c_n$ and $c_n + a_n$ are each at most $11$, we can eliminate $38$ and $48$ as possibilities because they are too large. It is also easy to see that $8$ is not possible, since this would require each of $a_n + b_n$, $b_n + c_n$ and $c_n + a_n$ to be less than $9$, and hence each would be at most $1$.
The remaining possibilities for $2(a_n + b_n + c_n)$ are $18$ and $28$, and so $a_n + b_n + c_n$ is one of $9$ or $14$. Each lies in the range $9$ to $19$, and together with the carry of $1$ from the previous column, we see that there will also be a carry of $1$ into the next column, so our claim is proven.
But the claim above implies that for each $n$, at least one of $a_n$, $b_n$ or $c_n$ is not zero, which is impossible since $a$, $b$, and $c$ are finite. We thus have a contradiction, and hence we must have that $S(a + b + c) \leq 15k - 9$.
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