Let $n\ge 2$ and let $a_1\le a_2\le\cdots\le a_n$ be positive integers. If $$\sum_{i=1}^n a_i=\prod_{i=1}^n a_i$$ then prove that $$\sum_{i=1}^n a_i\le 2n$$ and determine when equality holds.
Putting $$S=\sum_{i=1}^n a_i=\prod_{i=1}^n a_i$$ and using AM-GM gives $$S\ge nS^{\frac 1n} \implies S\ge n^{\frac{n}{n-1}}$$ but this is a lower bound on $S$ and we need an upper bound. I can't see a way to achieve that.
Write $a_i = 1 + b_i$ and define
$$ \begin{align*} e_1 &= \sum_{i=1}^{n} b_i, & e_2 &= \sum_{1 \leq i < j \leq n} b_i b_j. \end{align*} $$
What we need to prove is $e_1 \leq n$. To this end, note that the equality yields
$$\begin{align*} n + e_1 = \sum_{i=1}^{n} (1 + b_i) = \prod_{i=1}^{n} (1 + b_i) \geq 1 + e_1 + e_2. \end{align*}$$
This implies $e_2 \leq n - 1$. We also observe:
Claim. $b_{n-1} \geq 1$.
Proof. Assume otherwise. Then $b_i = 0$ for all $i < n$, hence $$n + b_n = \sum_{i=1}^n (1 + b_i) = \prod_{i=1}^{n} (1 + b_i) = 1 + b_n. $$ This contradicts the condition $n \geq 2$.
Using this observation, we have
$$ \begin{align*} n \geq e_2 + 1 &\geq \sum_{i=1}^{n-2} b_{n-1} b_i + (b_{n-1}b_n + 1) \\ &\geq \sum_{i=1}^{n-2} b_i + (b_{n-1}b_n + 1) \\ &= \sum_{i=1}^n b_i + (b_{n-1} - 1)(b_n - 1) \geq e_1. \end{align*} $$
as desired. Furthermore, the equality holds if all the intermediate inequalities hold true. A moment of thought reveals that they hold precisely when
$$ b_1 = \ldots = b_{n-2} = 0, \qquad b_{n-1} = 1, \qquad b_{n-1} = n-1, $$
or equivalently, $(a_1, \ldots, a_n) = (1, \ldots, 1, 2, n)$.
