While I was working on the lonely runner conjecture, I stumbled upon this neat little problem:
Question: Given $n$ numbers $a_1, a_2, ... a_n$ ∈ $[0, \frac{1}{2}]$, and their product $P= a_1 a_2 ... a_n = \prod{a_i}$ what is the minimal number $P(n)$ for which $P⩾ P(n)$ implies that all $a_i ⩾ \frac{1}{n}$?
For example, knowing $P = \frac{1}{2^n}$ would absolutely guarantee that $a_i ⩾ \frac{1}{n}$, since all $a_i$ would need to be $\frac{1}{2}$. But $P = \frac{1}{n^n}$ does not guarantee it, since some $a_i$ might be lower or higher than $\frac{1}{n}$. So, $P(n)$ lies somewhere between those values, likely close to one half.
The smallest possible bound is $$P(n)={1\over n2^{n-1}}.$$ Suppose $B<{1\over n2^{n-1}}$. Then we can set $a_k=\frac12,\ k=1,\dots,n-1$ and $$a_n=2^{n-1}B<\frac1n,$$ and we have $$\prod_{k=1}^na_k=B.$$ On the other hand, when we take $P(n)={1\over n2^{n-1}},$ even if $n-1$ of the factors are as large as possible, the $n$th one must be at least $\frac1n$.