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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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1 Answer 1

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

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It shouldn't be that simple :)) maybe I missed an additional condition in the problem definition. I'll update the question once I figured it out. Thanks for your brief answer, anyways. –  Mohsen Sep 27 '12 at 8:30
    
I see nothing wrong with the answer. Perhaps you should add to your question that you are NOT looking for simple answers @Mohsen! –  Gigili Sep 27 '12 at 8:39
    
@Gigili: For sure the answer is correct. Actually, I made a mistake in my question. The question is related to the analysis of an algorithm and this is not an expected answer of course. The condition should be something like $x \le a_i \le (1+\alpha)\frac{S}{n}$ with $x>0$ but I need to find the value of $x$ correctly. –  Mohsen Sep 27 '12 at 8:54
    
@draks: Let $n=2$, $S=2$, and $\alpha = 1/3$. For sure, setting $a_1=0$ is not possible and the min occurs at $\langle 4/3, 2/3 \rangle$. –  Mohsen Sep 28 '12 at 4:11

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