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I want to find an inequality between the to terms... Assuming all the $a_i $ to be positive, the one where $1$ is added to all the $a_i$ is bigger, but is there a real relationship? It doesn't have to be an inequality because I need something stronger than that...

  • Ohh no... So yea I am researching an relationship... –  Sep 02 '16 at 19:30
  • Do you want a lower bound for the latter term in terms of the former or something like that? – Milo Brandt Sep 02 '16 at 19:39
  • I would realy like a formula for the RHS ... That would help me so much –  Sep 02 '16 at 19:40
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    Is there a part of this problem you're not telling us? Earlier today these exact products came up with the additional information that the $a_i$ are positive, $a_1a_2\cdots a_n=1$ and we were asked to show that $(a_1+1)\cdots(a_n+1)\geq 2^n$. Is that what your problem is as well? – Arthur Sep 02 '16 at 19:45
  • Yes it is in depth, but I didn't understand the solution and wanted something more specific... –  Sep 02 '16 at 19:48
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    The solution is simple: $(a_i+1)\geq 2\sqrt{a_i}$, so $(a_1+1)\cdots(a_n+1)\geq 2^n\sqrt{a_1\cdots a_n}=2^n$, as was written in the answer here. If there is something specific you didn't understand, please ask about that instead. Write a comment to that answer asking for clarification, whatever. Asking a new question without context, and without asking about what you're really wondering about is bad. – Arthur Sep 02 '16 at 19:53
  • @TheodoreAlberter check my answer on your original question, may clear some of your doubts. –  Sep 02 '16 at 20:23

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A simple example should indicate a general pattern:

$${(1+a)(1+b)(1+c)\over2^3}={1+a+b+c+ab+bc+ca+abc\over8}\ge\sqrt[8]{1\cdot a\cdot b\cdot c\cdot ab\cdot bc\cdot ca\cdot abc}=\sqrt[8]{a^4b^4c^4}=\sqrt{abc}$$

The key step is the Arithmetic-Geometric Mean inequality.

Barry Cipra
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  • Thanks! But is there a way to prove that $(1+a_1)\cdot\dots\cdot(1+a_n)$ has, when decomposed, $2_n$ terms? Please... –  Sep 02 '16 at 19:55
  • @TheodoreAlberter, you can prove it by induction on $n$: If $(1+a_1)\cdots(1+a_{n-1})$ expands out to $2^{n-1}$ terms, then, when multiplied by $(1+a_n)$, there will be $2^{n-1}$ terms with an $a_n$ and $2^{n-1}$ terms without, for a total of $2\cdot2^{n-1}=2^n$ terms in all. – Barry Cipra Sep 02 '16 at 20:02