Show that if $\omega$ is a primitive $2m^\text{th}$ root of unity, where $m\geq 1$ is an integer, then for all $i\geq 0$ and $x\in [0,1]$, we have the inequality $$\prod_{j=0}^{2m-1}|1-x^{(2i+1)m+j}\omega^j|\leq 1.$$
I am convinced of this result, and numerical experimentation seems to confirm it, although I have yet to find a simple proof of it. I came across this result trying to show that a certain complex analytic function has a natural boundary of the unit circle, but the statement itself seems to be of independent interest (or at least independent musing-ability).
I believe the following to be true: the quantity on the left is a decreasing function of $x$ for fixed $i$ and an increasing function of $i$ for fixed $x$; if we can prove either of these assertions, then we will be done, but neither seems to be very tractable. Further, no such result is true (even with minor modifications to account for indexing problems) for primitive odd roots of unity.
Here's one proof attempt: one could try to compare with the product $$\prod_{j=0}^{2m-1}|1-x^{(2i+1)m}\omega^j|=|1-x^{2(2i+1)m^2}|\leq 1,$$ but this doesn't work because in general it is not true that $$\prod_{j=0}^{2m-1}|1-x^{(2i+1)m+j}\omega^j|\leq \prod_{j=0}^{2m-1}|1-x^{(2i+1)m}\omega^j|.$$ However, this proof attempt works for $m\in \{1,2\}$ (for which this last inequality does hold), and gives some insight to why such a result could be true in general. Note that the limit as $x\to 1^{-}$ of the quantity on the left hand size is $0$, so the question is really only challenging when $x$ (or more specifically $x^i$) is small.