Prove that if n real numbers have product 1, then they have sum greater than n

Given $a_1, \ldots, a_n \in \mathbb{R}_{>0}$ . Then if

$\prod_{k=1}^n a_k = 1,$

prove that

$\sum_{k=1}^n a_k \geq n$

with equality if and only if $a_1 = \cdots = a_n = 1$ .

Proof. We consider two cases: the first one is the case that $a_1 = \cdots = a_n = 1$ , and the second is the case that $a_k \neq 1$ for at least one $k$ in $1, \ldots, n$ .
Case #1: If $a_1 = \cdots = a_n = 1$ , then

$\prod_{k=1}^n a_n = \prod_{k=1}^n 1 = 1 \qquad \text{and} \qquad \sum_{k=1}^n a_k = \sum_{k=1}^n 1 = n.$

Thus, the inequality holds (in fact, equality holds).
Case #2: Now for the case that at least one of the $a_k \neq 1$ . The proof is by induction on the number of terms $n$ . For the case $n=2$ , we have $a_1 \cdot a_2 = 1$ , and by assumption at least one of these is not equal to 1. But, if one of them is not 1, then neither is the other (since the product must equal 1). So,

$\begin{align*} a_1 \cdot a_2 = 1 &&\implies \ a_2 &= \frac{1}{a_1} \\ &&\implies \ a_1 + a_2 &= a_1 + \frac{1}{a_1} \\ &&\implies &= \frac{a_1^2 + 1}{a_1} \\ &&\implies &= \frac{a_1^2 - 2a_1 + 1 + 2a_1}{a_1} \\ &&\implies &= \frac{(a_1 - 1)^2}{a_1} + 2 \\ &&\implies &> 2. \end{align*}$

Where the final inequality follows since $\frac{(a_1 - 1)^2}{a_1} > 0$ since $a_1 > 0$ by assumption. Therefore, the inequality indeed holds for the case $n=2$ .
Assume then that the inequality holds for some $n = k \in \mathbb{Z}_{>0}$ . Then, let $a_1, \ldots, a_{k+1} \in \mathbb{R}_{>0}$ with $a_i \neq 1$ for at least one $i = 1, \ldots, k+1$ . If $a_i < 1$ , then there must be some $j \neq i$ such that $a_j > 1$ (otherwise, if $a_j \leq 1$ for all $j \neq i$ , then $a_1 \cdot \cdots \cdot a_{k+1} < 1$ ). Similarly, if $a_i > 1$ , then there is some $j \neq i$ such that $a_j < 1$ . Hence, we have a pair, $a_i, a_j$ with one member of the pair greater than 1, and the other less than 1. Without loss of generality (since multiplication and addition are commutative), let this pair be $a_1$ and $a_{k+1}$ . Then, define $b = a_1 \cdot a_{k+1}$ . So,

$\underbrace{b \cdot a_2 \cdot \cdots \cdot a_k}_{k\text{-elements}} = 1 \quad \implies \quad b_1 + a_2 + \cdots + a_k \geq k. \qquad \qquad (\text{Ind. hyp.})$

Further, since $(1-a_1)(1-a_{k+1}) < 0$ (since one of $a_1, a_{k+1}$ is greater than 1 and the other is less than 1, so one of $(1-a_1), (1-a_{k+1})$ is positive and the other is negative), we have