I want to prove/disprove that for a set $S = R \cup B$ of $n \geq 3$ colored (red and blue) points in general position (no 3 are collinear, no 4 are cocircular, and the distances between the $n \choose 2$ pairs of points are all distinct), if (r, b) is a pair, with $r \in R$ and $b \in B$, whose distance between them is the second closest distance between a red point and a blue point, then the edge (r, b) must be Delaunay.
An initial idea I had was perhaps converting the second-closest red-blue problem to the red-blue problem itself by changing the red-blue labels of the points. However, it leads nowhere as the new labels may cause changes to the closest pair information.
Another idea was to use contradiction, where I assume that (r, b) is not delaunay and I try to reach some contradiction. Even with this, I am unable to reach anywhere.
Any hints to approach this problem will be appreciated!
Turns out, the statement is inaccurate and can be disproved using an example. Take the following example: let 4 points be $S = \{A, B, C, D\} = \{(0, 0), (\frac{3\sqrt{3}}{2}, \frac{3}{2}), (3+\epsilon_3, 0), (\frac{3\sqrt{3}}{2}+\epsilon_1, -\frac{3}{2}+\epsilon_2)\}$. For the sake of this example, lets assume $\epsilon_1, \epsilon_2 = 0.05, \epsilon_3=0.001$. Let D be the sole red point, and the rest blue. We therefore get $\{|AD|, |BD|, |CD|\} = \{3.0190, 2.9504, 1.4923\}$. Based on the location of these points, the delaunay triangulation won't include $\bar{BD}$. But B is the second-closest point to D. Therefore, a point can be second-closest, yet not make a delaunay edge. Hence, disproved.