I previously asked a question Change in Partition for Geometric Mean if Sum Increases, to show that if the sum increases by 1, then all we need to do is increase one of the partitions by 1 to increase the geometric mean. I was wondering if this result extends to the weighted geometric mean case.
Running some simulations, it seems that this is true, but can't seem to derive a formal proof for it. I tried the approach by the answer to the unweighted question, but it doesn't simply extend to this case.
Expressing the problem formally: Suppose we have a fixed constant $k$, and a number $x$, and integer weights $w_1, ..., w_k \geq 1$ where we need to partition $x$ into the sum of $k$ integers: $$x = x_1+\cdots+x_k$$ such that the weighted geometric mean $\prod_{i=1}^k x_i^{w_i}$ is maximized.
If we want to partition $x+1$ into the sum of $k$ integers as well, such that the weighted geometric mean is maximized, is it true that such a partition is simply $x_1, \ldots, (x_i+1), x_k$ for some $i \in [k]$?
