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.

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.