Example 24 (1985 US Olympiad Problem) Let $A, B, C, D$ be four points in space, and the lines $AB, AC, AD, BC, BD, CD$ have at most one length greater than 1. Find the maximum value of the sum of the lengths of these six segments.

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Let $AD$ be the longest of the six segments.
(i) When the lengths of the other five segments are fixed, it is easy to see that $AD$ reaches its maximum value when $A$ and $D$ are opposite vertices of the planar quadrilateral $ABCD$.
(ii) Fixing the positions of $B$ and $C$, since the lengths of $AB$, $AC$, $BD$, and $CD$ do not exceed 1, $A$ and $D$ must lie within the intersection of two circles with centers at $B$ and $C$ and radii of 1. At this time, the only pair of points with the maximum possible distance are the two intersection points of the circles. When $A$ and $D$ are these two points, $AB$, $BD$, $AC$, and $CD$ are all 1, achieving their respective maximum values.
(iii) The remaining issue is that when $BC$ decreases, $AD$ increases; when $BC$ increases, $AD$ decreases. Therefore, under the condition that the other four sides are fixed at 1, we need to consider the maximum value of $BC + AD$.

Let $\angle ABO = \theta$ (where $O$ is the intersection point of $AD$ and $BC$). Given $AB = 1$ and $0 < BC \leq 1$, we know $\frac{\pi}{3} \leq \theta < \frac{\pi}{2}$. Thus, $AD + BC = 2(\sin \theta + \cos \theta) = 2\sqrt{2} \sin \left(\theta + \frac{\pi}{4}\right)$. Noting that $\sin \left(\theta + \frac{\pi}{4}\right)$ is a decreasing function in the interval $\frac{\pi}{3} \leq \theta < \frac{\pi}{2}$, the maximum value is achieved when $\theta = \frac{\pi}{3}$. At this time, the sum of the six segments is $4 + 2\left(\sin \frac{\pi}{3} + \cos \frac{\pi}{3}\right) = 5 + \sqrt{3}$.

Note: For problems involving multiple variable objects, first adjust a few of these objects while keeping the others temporarily unchanged to achieve a partial solution. After several such local adjustments, gradually approach the goal and ultimately solve the entire problem. This is undoubtedly a strategy of breaking down the whole into parts and simplifying the difficulty.