TITLE: Lower bound on product of distances from points on a circle
QUESTION [8 upvotes]: Let $C$ be a circle of radius $r$ with $n$ points. Prove that there is a point on the circle such that the product of the distances from this point to the other $n$ points is greater than $r^n$. So we seem to be looking at chords that are of length $\leq 2r$. To show the existence of such a point, would you use a constructive argument? Or more of an indirect argument? Likewise, to show that the inequality holds, would we invoke certain "famous" inequalities such as AM-GM or Cauchy-Schwarz?

REPLY [2 votes]: Using complex numbers, but a more elementary (and constructive) approach than the Maximum Modulus theorem.
Consider the unit circle. It is enough if we show that the product of distances is greater than $1$. 
In fact, we will show that the product of distances of some point is at least $2$.
Let the points be $\displaystyle z_1, z_2, \dots z_n$. Rotate the circle so that $\displaystyle (-1)^n \prod_{j=1}^{n} z_j = 1$
Let $\displaystyle P(z) = \prod_{j=1}^{n}(z-z_j)$.
The product of distances from $\displaystyle  z$ to $\displaystyle z_j$ is given by $\displaystyle |P(z)|$.
Now let $\displaystyle \omega_j$ be the $\displaystyle n$ $n^{th}$ roots of unity.
Since for $\displaystyle 1 \le k < n$, we have that $\displaystyle \sum_{j=1}^{n} (\omega_j)^k = 0$
we have that
$\displaystyle \sum_{j=1}^{n} P(w_j) = 2n$
Thus $\displaystyle \sum_{j=1}^{n} |P(w_j)| \ge |\sum_{j=1}^{n} P(w_j)| = 2n$
Hence there is some $j$ for which $|P(w_j)| \ge 2$.
So we have shown that there is a point whose product of distances from the $n$ points is at least $2r^n$.
In fact, I believe we can also show that if for all points, the product of distances is  $\leq 2r^n$, then the $n$ points must be equally spaced! (I will leave that for you :-)).