### Step 1: Consider certain sums Define \(S_i = a_1 + a_2 + \cdots + a_i\), the sum of the first \(i\) elements of the sequence. Now consider each element \(a_i\), and study the following \(i-1\) distinct sums: \[ a_i + S_1, a_i + S_2, \dots, a_i + S_{i-1} \] Since these sums all involve distinct terms from \(a_1, a_2, \ldots, a_i\) and no element in the sequence can be expressed as the sum of distinct earlier terms, all these sums represent numbers **not in the sequence**. For the same reason, all these sums are distinct for all distinct \(i\). ### Step 2: Estimate the number Next, we estimate how many such sums there are. 1. For \(a_i \leq a_k\), we consider all the sums defined above, so for elements no greater than \(a_k\), there are \(0 + 1 + 2 + \dots + (k-1)\) sums. In other words, for elements no greater than \(a_k\), we have \(\frac{k(k-1)}{2}\) such sums. 2. For \(a_k < a_i \leq m\), we consider for each \(a_i\) only adding to the first \(k\) sums of \(S_k\) to produce the size constraint. Thus, for each element, exactly \(k\) sums are produced, and there are \(f(m) - k\) such elements \(a_i\). Overall, this produces \[ \frac{k(k-1)}{2} + k(f(m) - k) \] distinct sums. ### Step 3: Comparison with total possible sums These distinct sums are all less than or equal to \(m + S_k\). Therefore, the number of distinct sums must be bounded by the number of positive integers less than or equal to \(m + S_k\) that **are not in the sequence**. Since there are exactly \(f(m + S_k)\) elements of the sequence less than or equal to \(m + S_k\), the number of sums less than or equal to \(m + S_k\) that **are not in the sequence** is bounded by: \[ (m + S_k) - f(m + S_k) \] Thus, we have the following inequality: \[ \frac{k(k-1)}{2} + k(f(m) - k) \leq m + S_k - f(m + S_k) \] ### Step 4: Simplify and use bounds Since the function \(f(m)\) is non-decreasing, we know \(f(m + S_k) \geq f(m)\). This gives a lower bound for \(f(m + S_k)\). Furthermore, since the sequence is strictly increasing and \(a_k\) is greater than any previous element, we deduce that the sum \(S_k\) satisfies: \[ S_k + \frac{k(k-1)}{2} \leq k a_k \] This gives an upper bound for \(S_k\). Now we simplify the previous inequality: \[ k f(m) + f(m + S_k) \leq m + S_k + \frac{k(k+1)}{2} \] Using the lower bound for \(f(m + S_k)\) and the upper bound for \(S_k\), we obtain: \[ (k+1) f(m) \leq m + k a_k + k \] Finally, using \(k \leq a_k\) and dividing by \(k+1\) gives: \[ f(m) \leq a_k + \frac{m}{k+1} \] This is the desired inequality.

Unfortunately, we haven't had a complete solution between participants. It was surprising. No one could cam up with the enumerative part of the proof.