Let $a_1, a_2, \dots, a_n > 0$. I'm trying to prove that if $a_1a_2\cdots a_n = 1$, then $a_1 + a_2 + \cdots + a_n \geq n$ by mathematical induction without using the AM-GM inequality. So far I've got for the base case, it is clear that if $a_1 = 1$, then $a_1 \geq 1$. For the induction hypothesis, assume that $a_1a_2\cdots a_n$ implies that $a_1 + a_2 + \cdots + a_n \geq n$. Now I just need to show that $a_1a_2\cdots a_na_{n+1} = 1$ implies that $a_1 + a_2 + \cdots + a_n + a_{n+1} \geq n+1$. Could anyone show me show me how to do this? Thanks in advance.

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    $\begingroup$ Aside: Technically, this amounts to proving AM-GM, since AM-GM follows from this lemma fairly trivially. $\endgroup$ – Thomas Andrews Apr 21 '14 at 12:45
  • $\begingroup$ @ThomasAndrews : Technically, this amounts to proving AM-GM, because this IS AM-GM after a re-phrasing (i.e. if $a_1 a_2 \cdots a_n = c$, replace $a_i$ by $a_i/\sqrt[n]c$ and this is exactly AM-GM. $\endgroup$ – Patrick Da Silva Apr 21 '14 at 12:59
  • $\begingroup$ @PatrickDaSilva Umm,that was my point. $\endgroup$ – Thomas Andrews Apr 21 '14 at 13:03
  • $\begingroup$ You said it followed... I'm saying it's equivalent. Whatever, let's not argue. $\endgroup$ – Patrick Da Silva Apr 21 '14 at 13:21

We may suppose that $a_{n+1}$ is the largest element in $(a_1,\ldots,a_{n+1})$ and $a_n$ is the smallest one. Thus $a_{n+1}\geq 1\geq a_n$. with this information and the induction hypothesis we have $$ a_1+\cdots+a_{n-1}+a_na_{n+1}\geq n $$ and since $(a_{n+1}-1)(1-a_n)\geq 0$ we have $a_n+a_{n+1}-1-a_na_{n+1}\geq 0$. Just add this inequality to the preceding one and we are done.

  • $\begingroup$ +1. $n-1$ should be $n$ though. $\endgroup$ – WimC Apr 21 '14 at 13:08
  • $\begingroup$ @WimC I don't think so. $\endgroup$ – MCT Apr 21 '14 at 13:19
  • $\begingroup$ @WimC Yes, thanks, it is corrected. $\endgroup$ – Omran Kouba Apr 21 '14 at 13:27
  • $\begingroup$ -1. You can prove by induction but the way it is used here is incorrect. When you call the induction hypothesis, we have $a_1 a_2 \ldots a_n = 1$, whereas now you have $a_1 a_2 \ldots a_n a_{n+1} = 1$. $\endgroup$ – user141421 Apr 21 '14 at 13:32
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    $\begingroup$ @user141421 You apply the induction to $a_1,a_2,\ldots,a_{n-1},(a_na_{n+1})$, these are $n$ terms whose product is $1$. $\endgroup$ – Omran Kouba Apr 21 '14 at 13:34

Wlog. $a_{n+1}=\max\{a_1,\ldots,a_{n+1}\}$ and hence $a_{n+1}\ge 1$. Let $c=\sqrt[n]{a_{n+1}}$, and $b_i=ca_i$ for $1\le i\le n$. Then $b_1b_2\cdots b_n=a_1a_2\cdots a_nc^n=1$, hence by induction hypothesis $b_1+\cdots + b_n\ge n$. Then $$a_1+\cdots+a_n+a_{n+1}=c(b_1+\cdots+b_n)+a_{n+1}\ge cn+c^n.$$ Clearly, $f(1)=n+1$ and the derivative $f'(c)=n+nc^{n-1}$ is strictly positive (for $c\ge 1$), hence $f(c)\ge n+1$ for all $c\ge 1$.

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    $\begingroup$ Can you state what's $f$ before computing its derivative? It's clear from the context, but… $\endgroup$ – egreg Apr 21 '14 at 12:58
  • $\begingroup$ Why introduce a derivative? Since $c \ge 1$, $cn + c^n \ge n + 1^n = n + 1$. $\endgroup$ – 6005 Apr 21 '14 at 13:00
  • $\begingroup$ @Goos Also, $c\le d$ implies $d^n-c^n\ge0\ge cn-dn$, so the derivative is really unnecessary. $\endgroup$ – egreg Apr 21 '14 at 13:01

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