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Let $n\in\mathbb{N}$ and $x_1,x_2,x_3.....,x_n \in\mathbb{Q}$ with $x_i > 0$ and $\prod_{i=1}^{n} x_i = 1$

prove that $\sum_{i=1}^{n}x_i \geq n$ (hint:use induction).

Been stuck on that for hours. It seems as if the terminology of the question is not well defiened because they use "Let $n\in\mathbb{N}$" and not "For each $n\in\mathbb{N}$"

Could I get some insight and help please?

trizz
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  • Why must you use induction? – Dr. Sonnhard Graubner Mar 23 '19 at 16:44
  • This is the inequality between the arithmetic mean and geometric mean. – Angina Seng Mar 23 '19 at 16:45
  • @Dr.SonnhardGraubner they hinted it. could you show how to proove it without induction? – trizz Mar 23 '19 at 16:49
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    Possible duplicate of [Prove that $a_1+\cdots+a_n \geq n$ if $a_1$, $a_2$, ... $a_n$ are positive real numbers and their product is $1$](https://math.stackexchange.com/questions/1553204/prove-that-a-1-cdotsa-n-geq-n-if-a-1-a-2-a-n-are-positive-real) – the inductive proof is here: https://math.stackexchange.com/a/1553811/42969. – Martin R Mar 23 '19 at 17:14

1 Answers1

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Assume it works with $k$ terms; it certainly does when $k=1$. Consider $k+1$ terms, and permute them so $x_1$ is the greatest and $x_2$ the least. Since by hypothesis the $x_i$ have geometric mean $1$, $x_1\ge 1\ge x_2$. Now consider a sum of $k$ terms, $x_1x_2+\sum_{i=3}^{k+1}x_i\ge k$ by the inductive hypothesis. On the other hand, since $(x_1-1)(1-x_2)$ is a product of non-negative terms we have $$(x_1-1)(1-x_2)=x_1+x_2-x_1x_2-1\ge 0\implies x_1+x_2-x_1x_2\ge 1.$$Finally, $$\sum_{i=1}^{k+1}x_i=(x_1+x_2-x_1x_2)+(x_1x_2+\sum_{i=3}^{k+1}x_i)\ge 1+k$$as required.

J.G.
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  • First of all, thanks alot for the help! But could you elaborate some more about the steps that you've taken? – trizz Mar 23 '19 at 17:11
  • @trizz See edit. – J.G. Mar 23 '19 at 17:15
  • Any other elaborations? – trizz Mar 23 '19 at 18:24
  • @trizz Maybe you should tell me the first place it's not clear to you. – J.G. Mar 23 '19 at 18:37
  • I just feel bad for asking you over and over again. When I tried to solved it, I tried using trichotomy. When $x_n_+_1 = 1 $ the case is obvious. when $x_n_+_1 > 1 $ its also easy. but when $x_n_+_1 < 1 $ its not clear. – trizz Mar 23 '19 at 20:02
  • @trizz All you need is that $(x_\max<1\implies\prod_i x_i<1)\implies (\prod_i x_i=1\implies x_\max\ge 1)$, then similarly with $x_\min$. – J.G. Mar 23 '19 at 20:35
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/91455/discussion-between-trizz-and-j-g). – trizz Mar 23 '19 at 20:38