$a_1a_2\cdots a_n = 1 \implies a_1 + a_2 + \cdots + a_n \geq n$ if $a_1, a_2, \dots, a_n > 0$

Question

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.

Answer 1

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 $c\ge 1$.

Answer 2

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.