## When is the product of a set of numbers greater than the sum of them?

This could well be too general a question, but I'd be interested in solutions to special cases too. Say you have some finite set of positive real numbers $x_i$, when is it the case that $\sum_i x_i > \prod_i x_i$? And when are they equal?

The special case that prompted this was an argument about whether any number is equal to the sum of its prime factors.

Any references or quick proofs welcome.

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I don't think it's a very deep question, but it seems well-posed enough not to deserve negative votes. – David Eppstein Feb 28 2010 at 16:45
Where you say "set" you probably mean "sequence." – Qiaochu Yuan Mar 1 2010 at 2:25
It seems that the question is about real numbers but all answers are about integers. For real numbers, the inequality defines an open, unbounded subset of $\mathbb{R}_{+}^n$. I am not sure what else can be said. – Felipe Voloch Mar 1 2010 at 3:30
I inferred from his having tagged the question "nt.number-theory" that by positive reals he meant positive integers. Of course it is just as likely that he did mean positive reals and the tag was a mistake... – Ben Linowitz Mar 1 2010 at 3:35
I was in fact interested in both questions. I tagged number theory because I didn't know what the appropriate arXiv tag would be for the general reals question. I don't think anything hangs on my saying set rather than sequence? Surely "set" makes the question more general, but perhaps certain properties of sequences allow better answers in certain cases... – Seamus Mar 1 2010 at 14:32

If you have a set of positive integers (that is, no duplicates are allowed) then the sum is greater than the product if and only if the set is of the form {1,x}. The sum is equal to the product only for singleton sets {x} and the set {1,2,3}.

For, examining the remaining cases:

• If the set is empty the sum is 0 and the product is 1, so sum < product
• If the set has two elements {x,y}, neither of which is 1, then $xy\ge 2\max(x,y)>x+y$.
• If the set has three elements {1,2,x}, with $x>3$, the sum is $x+3$ and the product is the larger number $2x$.
• If the set has any other three elements then its product is at least three times its max and its sum is less than that.
• If the set has {1,2,3,x} then the product is 6x and the sum is x+6, smaller for all $x\ge 4$.
• If the set has any other form with $k>3$ elements then by induction the sum of the smallest $k-1$ items is less than their product. Multiplying or adding the largest item doesn't change the inequality.
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 Ah, I see the importance of set vs sequence now. Maybe I do mean sequence to allow for duplicates... – Seamus Mar 1 2010 at 14:34

The "special case" is not a special case, since only squarefree numbers equal to the product of their prime factors (I guess you forgot that primes can occur with multiplicities), and the product of a finite multiset of integers > 1 is always greater or equal to their sum, with equality only if the multiset is [2, 2] (proof by induction). So it is not really clear to me what you actually want.

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 I'm not sure why this answer didn't receive more votes; the last part answers the OP's last question quite easily. All you need to do is observe that for any primes p, q we have (p-1)(q-1) \ge 1, with equality only when p = q = 2. This implies pq \ge p+q, again with equality when p = q = 2. – Qiaochu Yuan Mar 1 2010 at 2:29

Hi,

Was just looking online to see if I could find a result I did for my college course many years ago.

i found that the product of a collection of poitive reals P is related to the sum of the same collection S via this inequality.

P >= S multiplied by e to the power of 1 divided by e.

Can't remember how I proved it, but it did include one of Abel's theorems.

Happy times!

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 This is not true: $x\cdot x< e^{1/e}(x+x)$ for any $0< x< 2e^{1/e}$. – Emil Jeřábek Mar 6 2012 at 17:19