Ratio of differences of any three numbers is high Let $n$ be a positive integer. Prove that we can find $n$ real numbers such that for any three distinct $x,y,z$ among them, we have $\max\left(\left|\dfrac{x-y}{x-z}\right|,\left|\dfrac{x-z}{x-y}\right|\right)>1+\dfrac{1}{n^{1.6}}$.
One idea is to choose $a_1<a_2<\ldots<a_n$ such that the gap  $a_{i+1}-a_i$ is much larger than $a_i-a_{i-1}$. But the problem there is that while it is true that the ratio between $a_k-a_x$ and $a_y-a_z$ is high for any $k>x,y,z$, the ratio between $a_k-a_x$ and $a_k-a_y$ is low, for $k>x,y$.
 A: Instead of fixing $n$ and seeing how high we can make that minimax, we will start by fixing the minimax target and seeing how many points we can generate.
Fix $\alpha > 1$. Start with the set $S=\{0,1\}$. Add to $S$ all $\alpha^{-r}$ satisfying $\alpha^{-r} \ge \frac{\alpha}{\alpha+1}$. The specific choice $\frac{\alpha}{\alpha+1}$ will be needed below; for now, note that $\frac{\alpha}{\alpha+1} > \frac12$.
Given any $y,z \in S$ with $y = \alpha^{-r}, z = \alpha^{-s}$, the smallest value of $\max\left(\left|\dfrac{x-y}{x-z}\right|,\left|\dfrac{x-z}{x-y}\right|\right)$ will be attained for $x = 0$ (because $0$ is the point in $S$ unambiguously furthest away from both $y$ and $z$). But if $x=0$, this value is simply $\alpha^{|r-s|}$, which is at least $\alpha$.
Now we can add the mirror images $1-\alpha^{-r}$ of all these points, and we still have that $\max\left(\left|\dfrac{x-y}{x-z}\right|,\left|\dfrac{x-z}{x-y}\right|\right) \ge \alpha$ for all $x,y,z \in S$. The only case that doesn't follow directly by symmetry is when $y$ and $z$ are the smallest $\alpha^{-r}$ and the largest $1-\alpha^{-r}$, with $x = 0$ or $1$. But then
$$\frac{\alpha^{-r}}{1-\alpha^{-r}} \ge \frac{\alpha/{(\alpha+1)}}{1-\alpha/{(\alpha+1)}} = \alpha$$
So for given $\alpha>1$, if $\alpha^{-r} \ge \frac{\alpha}{\alpha+1}$ we can construct a set $S$ containing the $2r+2$ points $\{0,1-\alpha,...,1-\alpha^{-r},\alpha^{-r},...,\alpha,1\}$.
Returning to the original problem: given $n$, how large can we make $\alpha$? Suppose for the moment that $n$ is even. Then we can choose any $\alpha$ such that $\alpha^{-r} \ge \frac{\alpha}{\alpha+1}$, where $r = \frac12(n-1)$. $\frac{\alpha}{\alpha+1} < \frac12\alpha$, so this is the case if $\alpha^{-(r+1)} \ge \frac12$, i.e. $\alpha^{-\frac12n} \ge \frac12$, or $\alpha^n \le 4$.
Thus (if $n$ is even) we can choose $n$ real numbers such that for any three distinct $x,y,z$ among them, we have
$$\max\left(\left|\dfrac{x-y}{x-z}\right|,\left|\dfrac{x-z}{x-y}\right|\right) \ge 4^{1/n}$$
This is asymptotically much better than the requested bound of $1 + n^{-1.6}$; indeed, $4^{1/n}$ behaves like $1 + \dfrac{\ln 4}{n}$ for large $n$.
If $n$ is odd, we can use the above construction for $n+1$, and simply remove a point.
