Set with Sum Equal to Product

Solution 1: Without loss of generality, we may assume that $x_i$ are arranged in decreasing order. We have

$\begin{aligned} 1 & = \sum_{i=1}^{60} \frac {x_i} {\prod_{j=1}^{60} x_j } \\ & = \sum_{i=1}^{60} \frac {1}{ \prod_{j\neq i} x_j }\\ & \leq \frac {1}{x_2} + \frac {1}{x_1} + \frac {58}{x_1x_2} & & \mbox{since} \ x_i \geq 1 \\ \end{aligned}$

Thus $(x_1 - 1)(x_2 - 1) \leq 59$ . If $x_2 =1$ , then by assumption $x_i = 1$ for $i \neq 1$ , and the equation $x_1 + 59 = x_1$ has no solution.

Thus $x_2 \geq 2$ , and so $x_1 - 1 \leq 59 \Rightarrow x_1 \leq 60$ . It is clear that $x_1 = 60, x_2 = 2$ and $x_i = 1$ for $i \geq 3$ satisfies the equation.

Therefore the maximum value of $x_1$ is $60$ .

Solution 2: (By William Gan) We will first use induction to show that for positive integers $n \geq 2$ , $\sum \limits_{i=1}^n x_i - \prod \limits_{i=1}^n x_i \leq n-1$ .

Base case: For $n=2$ , $x_1 + x_2 - x_1 x_2 = 1 - (x_1 - 1)(x_2 - 1) \leq 1$ as $x_1, x_2$ are positive integers.

Induction step: Assume it is true for $n=k$ for some positive integer $k>1$ . We aim to show that it is true for $n=k+1$ . $\begin{aligned} \sum \limits_{i=1}^{k+1} x_i - \prod \limits_{i=1}^{k+1} x_i & = x_{k+1} + \sum \limits_{i=1}^k x_i - x_{k+1} \prod \limits_{i=1}^k x_i \\ & \leq x_{k+1} + \prod \limits_{i=1}^{k} x_i + (k-1) - x_{k+1} \prod \limits_{i=1}^k x_i\\ &= (k+1) - 1 - (x_{k+1} - 1)(\prod \limits_{i=1}^{k} x_i - 1) & \leq (k+1) - 1 \end{aligned}$ where the last inequality is done in a similar manner as $n=2$ . Thus, this proves the induction step.

Therefore, by mathematical induction the statement is true for all positive integers $n \geq 2$ .

Now, let $\prod \limits_{i=2}^{75} x_i = p$ and $\sum \limits_{i=2}^{75} x_i = s$ . If $p=1$ , then all the integers $x_i, i\geq 2$ must be equal to 1. We can check that this has no integer solution for $x_1$ , hence $p \geq 2$ .

This substitution gives $x_1 p = x_1 + s$ $\Rightarrow x_1 = \frac{s}{p-1} \leq \frac{p+(74-1)}{p-1}$ by the above lemma. We continue to simplify this to obtain $x_1 \leq \frac{p+73}{p-1} = 1 + \frac{74}{p-1} \leq 1 + \frac{74}{2-1} = 75$ .

One possible solution is $x_1=75, x_2=2, x_3=x_4=...=x_{75}=1$ . Hence, the maximum value of $x_1$ is indeed 60.

This is by no means a proper proof, a lot of it are based on intuition. It is widely know that $a\times b>a+b$ for positive a and b most of the time, with the exception when $a=b=2$ (equality) or at least one of a or b equals to 1 (reverse inequality). So clearly $x_2,x_3,\dots,x_{75}$ cannot be all equal to 1. Intuitively, to make $x_1$ biggest, $x_2,x_3,\dots,x_{75}$ should be smallest. And yet they can't be 1. So we let $x_3,x_4,\dots,x_{75}$ be 1 and $x_2$ be something bigger. Similarly, for $x_1$ to be big $x_2$ should be small. We let $x_2=2$ and find that $x_1=75$ works.