TITLE: $a_1a_2\cdots a_n = 1 \implies a_1 + a_2 + \cdots + a_n \geq n$ if $a_1, a_2, \dots, a_n > 0$
QUESTION [5 upvotes]: 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.

REPLY [7 votes]: 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.