For $x,y\in\mathbb R^n$ define $P(z) = \sum_{i=1}^n z^i x_i y_i$. I want to minimize the function $$ \frac{1-P(z)^2}{P'(z)^2} =\frac{1-(\sum_{i=1}^n z^{i} x_i y_i)^2}{(\sum_{i=1}^n i z^{i-1} x_i y_i)^2} $$ for a given $z\in[-1,1]$ over all $x$ and $y$ restricted by $\|x\|_2=\|y\|_2=1$.
Conjecture: For any $z\in[-1,1]$ there is some $i\in[1,n]$, such that the optimal solution is to put all the weight of $x$ and $y$ on some $i$. That is, let $x_i=y_i=\pm 1$, and put $0$ on the remaining coordinates. I believe this is known as a ''bang-bang'' solution.
I can prove the conjecture for $z$ close to $0$: Using a Taylor expansion $$ \frac{1-P(z)^2}{P'(z)^2} =\frac{1-P(0)^2}{P'(0)^2} + O(z) =\frac{1}{(x_1y_1)^2} + O(z), $$ it's clear that in the case $z=0$ we should put all the weight on the first two coordinates.
Numerically I have verified the conjecture for many other choices of $z$ and $n$, but I barely know where to start for the proof.
Update: By the suggestion of Michael Grant we can define $$ f_z(w) = \sum_{i=1}^n z^i w[i] \quad\text{and}\quad \bar{f_z}(w) = \sum_{i=1}^n i z^{i-1} w[i], $$ and the problem then reduces to showing that the function $$\frac{\bar{f_z}(w)^2}{1-f_z(w)^2}$$ is convex in $w\in\mathbb R^d$, such that $\|w\|_1\le 1$. Numerically I have verified this is true for $n$ up to 4.
Update 2: For any $\alpha\ge 0$, the set of $w$ such that $\frac{\bar{f_z}(w)^2}{1-f_z(w)^2}\le \alpha$ is the same as the set such that $\bar{f_z}(w)^2+\alpha f_z(w)^2\le \alpha,$ which is clearly convex, as the function $\bar{f_z}(w)^2+\alpha f_z(w)^2$ is a positive sum of convex functions.
That shows $\frac{\bar{f_z}(w)^2}{1-f_z(w)^2}$ is quasi-convex, which should be enough to prove the original conjecture!