Example 3 Let $A, B, C, D$ be four points in space, and connect $AB, AC, AD, BC, BD, CD$ where at most one of these has a length greater than 1. Try to find the maximum value of the sum of the lengths of the six segments.

Connect $AB, AC, AD, BC, BD, CD$ where at most one of these has a length greater than 1, try to find the maximum value of the sum of the lengths of the six segments.
Let $AD$ be the longest of the six line segments.
(1) Fixing the lengths of the other five line segments, it is evident that $AD$ attains its maximum value when $A$ and $D$ are opposite vertices of the parallelogram $ABDC$.
(2) Fixing the positions of $B$ and $C$, then $A$ and $D$ must lie within the intersection of two circles with radii of 1, centered at $B$ and $C$ respectively. The pair of points with the maximum possible distance are precisely the two intersection points of the circles. When $A$ and $D$ are these two points, $AB$, $BD$, $AC$, and $CD$ all reach their respective maximum values of 1.
(3) The remaining issue is that when $BC$ decreases, $AD$ increases; and when $BC$ increases, $AD$ decreases. Therefore, we should explore the maximum value of $BC + AD$ under the condition that the other four sides are fixed at 1.

Let $\angle ABO = \theta$. Given $AB = 1$ and $0 < BC \leq 1$, we have $\theta \in \left[\frac{\pi}{3}, \frac{\pi}{2}\right)$.
Thus, $AD + BC = 2(\sin \theta + \cos \theta) = 2\sqrt{2} \sin \left(\theta + \frac{\pi}{4}\right)$ attains its maximum value when $\theta = \frac{\pi}{3}$. At this point, the sum of the six line segments is $4 + 2\left(\sin \frac{\pi}{3} + \cos \frac{\pi}{3}\right) = 5 + \sqrt{3}$.