General problem
Let $\mathrm{DIV}$ be the integer division and $\mathrm{MOD}$ the standard remainder calculation.
Given $k$ natural numbers $n_1, ... , n_k$, what natural number $m$ minimizes $$ m+\sum_{i=1}^{k}\left(n_i\;\mathrm{DIV} \;m\right)+\left(n_i\;\mathrm{MOD} \;m\right) \;?$$ It is likely that there is no easy formula for $m$ in terms of the numbers $n_i$, but are there some lower and upper bounds for $m$ or at least an order of the size of $m$ in terms of the numbers $n_i$?
Base case ($k=1$)
For example consider the base case $k=1$, then there is a given number $n$ and I want to minimize
$$m+n\;\mathrm{DIV}\; m+n\;\mathrm{MOD} \;m.$$ By a comment of Hurkyl, the identity $n=m(n\;\mathrm{DIV}\; m)+n\;\mathrm{MOD} \;m$ leads to
$$m+n\;\mathrm{DIV}\; m+n\;\mathrm{MOD} \;m\;=\;m+n+(1-m)(n\;\mathrm{DIV}\; m).$$ I think here the choice of $m$ would be close to $\sqrt{n}$. But are there precise boundaries like $$\sqrt{n}-c_n\le m\le \sqrt{n}+c_n$$ for some small term $c_n\in o(\sqrt{n})$ ?
And for the general case, is there something similar like $$\sqrt{\sum_i n_i}-c_n\le m\le \sqrt{\sum_i n_i}+c_n\; ?$$
I have basically no idea how to even begin to prove (tight) bounds, so any help is welcome here!