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What is the minimum of $a_1\times a_2 \times \dots \times a_n$ such that $a_1+a_2+\dots+a_n=S$ and $0 < x \le a_i \le (1+\alpha)\frac{S}{n}$?

My conjecture is that we need to set as many $a_i$'s as possible to $(1+\alpha)\frac{S}{n}$ and set the rest of $a_i$'s equally. Is that correct? How to prove it?

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Set $a_1=0$. $ {} $ $ {} $ $ {} $ $ {} $

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    @draks: Let $n=2$, $S=2$, and $\alpha = 1/3$. For sure, setting $a_1=0$ is not possible and the min occurs at $\langle 4/3, 2/3 \rangle$.2012-09-28