Maximum product over partition

Question

I stumbled upon the following statement:

Given a natural number $N=kq+r$ where $0 \le r<k$ which can be partitioned into $\sum_{i=0}^k N_i$, where $N_i$ are natural numbers as well, it holds that

$\prod_{i=1}^kN_i \le q^{k-r}(q+1)^r$

For $r=0$ this is obviously identical to the real case. Intuitively this result makes a lot of sense to me, you do not want to "stray too far" from partitioning $N$ into equal parts so you divide $r$ onto $r$ of the $N_i$.

I would like some pointers how to formalize this.

Answer 1

Suppose there exist $N_i, N_j$ where $N_i + 2 \le N_j$.

$(N_i+1)(N_j-1) = N_iN_j +N_j -N_i-1>N_iN_j$, because $N_j>N_i+1$.

Thus the partition can only contain numbers that have a pairwise difference of at most one if it is to be maximized. The only way to achieve this is the right hand side of the original equation.