In trying to find the maximum geometric mean given an integer $n$ and $k$ target partitions, where $n = \sum_{i=1}^k n_i$ and $\Pi_{i=1}^k n_i$ is maximal, we can find such $n_1, ..., n_k$ by splitting $n$ into $k$ integers that are as equal as possible.
For instance, to maximize the geometric product of 10 with 4 partitions, we get 10 = 3 + 3 + 2 + 2. To maximize the geometric product of 20 with 3 partitions, we get 20 = 7 + 7 + 6.
Is there anything that can be said about trying to find the maximum weighted geometric mean given an integer $n$, $k$ target partitions, and weight vector w = $(w_1, ..., w_k)$? That is, we want to maximize the product $$\Pi_{i=1}^k \left( n_i^{w_i} \right)$$ such that $n = \sum_{i=1}^k n_i$.
Some thoughts:
- Would splitting $n$ into $k$ integers that are as equal (weight-adjusted by division, with ties in favour of largest weights first) as possible work?
- Say for instance we have $n = 10$, $k = 2$, w = (1,2), then 10 = 3 + 7, weighted geometric product = $3^1\times 7^2 = 147$ (which is the higher than if we were to split it 4+6)
- Another example is $n = 20$, $k = 3$, w = (1,2,3), then 20 = 3 + 7 + 10, weighted geometric product = $3^1\times 7^2\times 10^3 = 147,000$ (which would be maximum)
- It seems that each of the factors (divided by their respective weights) should differ by no more than one.
I'm not sure how to begin proving this though.
Yes, your intuition is correct for integer $w$. You can mimic a proof for the $w \equiv 1$ (unweighted) case, which is to show how to improve any solution that does not have that property. That is, assume $$\frac{n_i}{w_i} - \frac{n_j}{w_j} > 1$$ for some $i$ and $j$, and modify $n_i$ and $n_j$ in a way that preserves $n_i+n_j$ but increases $n_i^{w_i} n_j^{w_j}$.