Suppose we have a fixed constant $k$, and a number $x$, where we need to partition $x$ into the sum of $k$ integers: $$x = x_1+\cdots+x_k$$ such that the geometric mean $\prod_{i=1}^k x_i$ is maximized.
If we want to partition $x+1$ into the sum of $k$ integers as well, such that the geometric mean is maximized, is it true that such a partition is simply $x_1, \ldots, (x_i+1), x_k$ for some $i \in [k]$?
Yes.
Proof:
Assume $\prod x_h$ is maximised, and there exists $i,j$ such $x_i+1\leq x_j-1$ (that is, $x_i$ and $x_j$ differ by at least 2).
Without loss of generality, let $i=1$ and $j=2$.
Let $y_h$ be such that $y_1=x_1+1$, $y_2=x_2-1$, and $y_h=x_h$ for all $h>2$.
Note that $\sum x_h = \sum y_h$, and let $\prod x_h = P$.
It's easy to show that $$\prod y_h = P\left(1+\frac{x_2-x_1-1}{x_1x_2}\right).$$
However, if $x_1+1\leq x_2-1$, then $x_2-x_1-1>0$, so $\prod y_h > P$, contradicting the assumption that $P$ is maximal.
This characterises the solutions to the puzzle: there is some $q$ for which all $x_h$ are either $q$ or $q+1$, specifically, let $x=kq+r$, with $0\leq r<k$, then $r$ of the $x_h$ are $q+1$, and $k-r$ of them are $q$.
If $x=kq+r$, and $x+1=kq^\prime+r^\prime$, then either $r^\prime=r+1$, or $r=k-1$, $r^\prime=0$ and $q^\prime=q+1$. In either of these cases, there is an $x_i$ such that $x_i=q$ and $x^\prime_i=q+1$.
