[Example 3.2.2] Let $A, B, C, D$ be four points in space, and suppose that among the line segments $A B, A C, A D$, $B C, B D, C D$, at most one has a length greater than 1. Find the maximum possible value of the sum of the lengths of these six line segments.
Among the six line segments given in the problem, at most one has a length greater than 1. We can start with this longest line segment.
Let $AD$ be the longest of the six line segments.
Step 1: We freeze the other 5 line segments except $AD$.
In this way, we can consider two triangles, $\triangle ABC$ and $\triangle DBC$, sharing the common side $BC$, folded along $BC$. At this point, we look for the longest $AD$ in the folding of the two triangles.

Obviously, when $\triangle ABC$ and $\triangle DBC$ are in the same plane, and $A, D$ are the opposite vertices of the quadrilateral $ABDC$, $AD$ is the longest.

Step 2: Fix the positions of points $B$ and $C$. At this point, since the lengths of $AB, AC, BD, CD$ are all less than or equal to 1, and only $AD$ may be greater than 1, we have
$$
AB = AC = BD = BC = 1.
$$

Step 3: The problem now is to find the maximum value of $BC + AD$, which depends on the length of $BC$.
$A$ and $D$ must lie within the circles centered at $B$ and $C$ with a radius of 1. To maximize $AD$, $A$ and $D$ should be the intersection points of $\odot B$ and $\odot C$.
Since $BC \leq 1, AB = 1, AC = 1$.
If we set $\angle ABC = \theta$, then
$$
\angle BAC \leq \frac{\pi}{3}, \frac{\pi}{3} \leq \theta < \frac{\pi}{2}.
$$

Because
$$
\begin{array}{l}
BC = 2AB \cos \theta = 2 \cos \theta, \\
AD = 2AB \sin \theta = 2 \sin \theta,
\end{array}
$$

then $BC + AD = 2 \sin \theta + 2 \cos \theta = 2 \sqrt{2} \sin \left(\theta + \frac{\pi}{4}\right)$,
and $\quad \frac{7 \pi}{12} \leq \theta + \frac{3 \pi}{4} < \frac{3 \pi}{4}$,
while $\sin \left(\theta + \frac{\pi}{4}\right)$ is a decreasing function in the interval $\left[\frac{7 \pi}{12}, \frac{3 \pi}{4}\right) \subset \left(\frac{\pi}{2}, \pi\right)$, so $\sin \left(\theta + \frac{\pi}{4}\right)$ reaches its maximum value when $\theta + \frac{\pi}{4} = \frac{7 \pi}{12}$, i.e., $\theta = \frac{\pi}{3}$, which is
$$
\sin \frac{7 \pi}{12} = \frac{\sqrt{6} + \sqrt{2}}{4}.
$$

Thus, $BC + AD$ reaches its maximum value
$$
2 \sqrt{2} \cdot \frac{\sqrt{6} + \sqrt{2}}{4} = \sqrt{3} + 1.
$$

At this point, the six line segments form a rhombus with acute angles of $60^\circ$ and obtuse angles of $120^\circ$, and their sum is maximized. The maximum sum of the six line segments is $\sqrt{3} + 5$ (at this time, $AB = AC = BC = BD = CD = 1, AD = \sqrt{3}$).

In solving this problem, we first fix the lengths of $AB, AC, BC, BD, CD$ to find the maximum value of $AD$, then fix the positions of $B$ and $C$ to find the maximum value of $AB, AC, BD, DC$, and finally let the positions of $B$ and $C$ change to find the maximum value of $AD + BC$.