Sides are smaller and one of diagonals is bigger. What happens to other diagonal?

Question

For any two point $x,y\in \mathbb{R}^n$, denote the euclidean distance of $x$ and $y$ by $xy$. Let $a,b,c,d,a',b',c',d'$ be eight points in $\mathbb{R}^n$ for some positive $n$. If $ab \le a'b'$, $bc \le b'c'$, $cd \le c'd'$, $da \le d'a'$, and $ac > a'c'$, then can we deduce that $bd \le b'd'$?

Answer 1

You cannot. The inequality $bd \le b'd'$ can fail.

To construct a counterexample in $\mathbb{R}^3$, consider following family of $4$-tuples of points: $$\begin{cases} A(t) = (t,0,0),\\ B(t) = (0,t,\sqrt{1-3t^2}),\\ C(t) = (-t,0,0),\\ D(t) = (0,-t,\sqrt{1-3t^2}) \end{cases}\quad\text{ for } t \in (0,\frac{1}{\sqrt{3}})$$ It is easy to see $$AB = BC = CD = DA = \sqrt{1-t^2} \quad\text{ and }\quad AC = BD = 2t$$ As functions of $t$, $AB, BC, CD, DA$ is strictly decreasing while $AC$, $BD$ is strictly increasing.

Take $(a,b,c,d) = (A(t),B(t),C(t),D(t))$ and $(a',b',c',d') = (A(t'),B(t'),C(t'),D(t'))$ for any $0 < t' < t < \frac{1}{\sqrt{3}}$, one find $$ab < a'b', bc < b'c', cd < c'd', da < d'a';\; ac > a'c'\quad\text{ but }\quad bd > b'd'$$