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I think it's familiar, but if not, here it is.

Given $n$ positive real numbers $a_1, a_2, ... a_n$:
Prove that, if: $$a_1\cdot a_2\cdot\cdot\cdot a_n=1$$ Then: $$a_1+a_2+\ldots+a_n\geq n$$ I need proof by induction, not by using the AM-GM inequality (the latter is easy).

Thanks in advance.

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    $\begingroup$ Use en.wikipedia.org/wiki/… $\endgroup$ – lab bhattacharjee Oct 24 '16 at 8:18
  • $\begingroup$ what do you tried? What is exactly your problem using induction here? $\endgroup$ – Masacroso Oct 24 '16 at 8:23
  • $\begingroup$ @Masacroso I tried proving that it works for n=k+1 when we know that it works for n=k. I took the case where at least one number is 1. k numbers are left, whose sum is >= k. Add 1 to both sides and the sum of the k+1 numbers is >= k+1; thus, proven for this case. In the case of all numbers different from 1, I tried with saying that at least one number is smaller than 1, but my inequalities led to nowhere... $\endgroup$ – AAN4EVA Oct 24 '16 at 8:30
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    $\begingroup$ In fact in the big thread with AM-GM proofs you can find several proofs which go by induction. In fact, this answer starts by stating and proving the claim from your question as an auxiliary lemma. $\endgroup$ – Martin Sleziak Oct 24 '16 at 9:55
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Assume that the inequality is true for $n-1$.

Without loss of generality, assume that $a_1$ and $a_n$ are respectively the maximum and minimum among $a_1, a_2, \dots, a_n$.

Note that $a_1 \ge 1$ and $a_n \le 1$

It thus follows that $$(1-a_1)(a_n-1) \ge 0 \Leftrightarrow a_1+a_n \ge a_1a_n+1$$

Note that since $a_2 \times a_3 \times \dots \times a_{n-1} \times a_1a_{n}=1$, our indutive hypotheis implies $$n \le a_2+a_3\dots+a_{n-1}+a_1a_n+1 \le a_1+a_2 +\dots+a_n$$

We are done.

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  • $\begingroup$ Thank you, sir! It did indeed have something in common with the proof by induction of AM-GM's inequality, according to Wikipedia. $\endgroup$ – AAN4EVA Oct 24 '16 at 8:50
  • $\begingroup$ Hey, also, I am asked to prove that the inequality turns to equality if and only if all the numbers are one. If I say $(1-a_1)(a_n-1)=0$, I understand that either the max is 1, or the min is 1. But if the max is 1, then [how to prove this formally?] the min is also 1, and vice versa. Thus proved. (Help with square brackets question?) $\endgroup$ – AAN4EVA Oct 24 '16 at 9:12
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    $\begingroup$ @AAN4EVA To put it formally: In order for there to be a inequality, note that $(1-a_1)(a_n-1) = 0$, and thus we have that either $a_1=1$ or $a_n=1$. If $a_1=1$, and $a_n < 1$ this would imply $a_1a_2 \dots a_n \le a_n <1$. we have a contradiction. Thus $a_n=1$. $\endgroup$ – S.C.B. Oct 24 '16 at 9:17

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