# Product of Sums Minimization with Constraints

Assume $S_1+S_2+\dots+S_n=S$ is a constant. Also, assume $S_i$'s are upper bounded by $U$ where $S/n < U < 2S/n$ and lower bounded by $L$ where $0 < L < S/n$. This is obvious that the maximum of $S_1 S_2 \dots S_n$ is $(S/n)^n$. But, what is the minimum of $S_1 S_2 \dots S_n$ considering the upper bound $U$ for $S_i$'s?

Note: For $n = 2$, it is obvious that the minimum is $U(S-U)$. How about $n = 3$ or more?

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Answer to original question: (no mention of $L$)

If $(n-1)U \ge S$ and the $S_i$ must be non-negative then you can achieve $S_1 S_2 \dots S_n=0$ by setting $S_1=0$ and the other $S_i=S/(n-1)$.

Otherwise you should be looking at $S_1 S_2 \dots S_n \ge SU^{n-1}-(n-1)U^n$ for example letting $S_1=S-(n-1)U$ and the other $S_i=U$.

The minimum product comes from letting as many values as possible take the extreme values, since $(k+d)(k-d) = k^2 -d^2 \lt k^2 - e^2 = (k+e)(k-e)$ if $|e| \lt |d|$.
So let $m=\lfloor \frac{S-nL}{U-L} \rfloor$: this is the largest possible number of values which can be $U$ while respecting the bounds and the sum. Then you want the $S_i$ to be $m$ copies of $U$, $n-m-1$ copies of $L$ and one of $S-mU-(n-m-1)L$ to give a product of $U^m L^{n-m-1} (S-mU-(n-m-1)L)$.
Thanks Henry. I just edited the question and added that there is a lower bound on $S_i$'s. So, you cannot set $S_1=0$. In addition, $S_1=S-(n-1)U$ may result in a negative $S_1$. What do you think now? Set $S_1=L$ and distribute $S - L$ equally between the rest of the $S_i$'s? How can I prove such a thing? –  Helium Feb 2 '12 at 1:37