Known upper bound for product of numbers? Is there a known method for computing the upper bound (in terms of $m$) for the product of some given numbers, $r, s, t, m... \in \mathcal{N} \cup \left\{0\right\}$ when $r + s + t \cdots \leq m$?
So what I would like to know is the tightest upper bound for $rst\cdots$.
Edit: I know there exists a trivial bound upper bound of $m^n$ where $n$ is the number of elements in the sum term $r + s + t \cdots \leq m$. But I'm wondering if there's a better bound than that.
 A: If we are allowed to choose as many factors as we desire, then it is clear that we should only use $3$'s and as few $2$'s in the product as possible. Notice first that we only need to deal with $3$ and $2$ since any $n\geq 4$ can be replaced by the two terms $2\cdot (n-2)\geq n$. However, moreover, notice that $3$ instances of $2$ gives the product $2^{3}$ which is less than $3^{2}$, meaning that any three twos should be replaced by two threes. Thus, if $f(m)$ is the maximum product obtainable, then we get:
$$f(3k)=3^k$$
$$f(3k+1)=\frac{4}3\cdot 3^{k}$$
$$f(3k+2)=2\cdot 3^k$$
If we allow only products with exactly $n$ factors, then we can make a similar argument to establish that no two terms should differ by more than $2$ (as otherwise, moving them closer together would increase their product); therefore, every term in a maximal sum must be either $\lfloor \frac{m}n\rfloor$ or $\lceil \frac{m}n\rceil$, and the exact upper bound can be obtained by knowing this. A slightly weaker, but easy to state upper bound is that a product with $n$ factors cannot exceed
$$\left\lceil\frac{m}n\right\rceil^n$$
