For the tetrahedron $A B C D$, we know that

$$
A B>C D>A C>B C>A D>B D .
$$

Among the segments connecting the midpoints of opposite edges of the tetrahedron, which one is the shortest?
Let the midpoints of the tetrahedron's edges be marked as $J, K, E, O, Z, X$ as shown in the diagram. The segments connecting the midpoints of adjacent edges on each triangular face of the tetrahedron are the midlines of these triangles, hence they are parallel to the corresponding edges of the tetrahedron, and their lengths are half of the corresponding edges. Therefore,

$$
\begin{array}{ll}
Z X=E O=\frac{A B}{2} ; & Z E=O X=\frac{C D}{2} \\
K Z=O J=\frac{A C}{2} ; & K X=E J=\frac{B C}{2} \\
E K=X J=\frac{A D}{2} \quad \text { and } & Z J=K O=\frac{B D}{2}
\end{array}
$$

From this, it also follows that the quadrilaterals $E O X Z, K Z J O$, and $K E J X$ are parallelograms, because their opposite sides are parallel and of equal length.

It is known that the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. Therefore,

$E X^{2}+Z O^{2}=E O^{2}+O X^{2}+Z X^{2}+Z E^{2}=\frac{1}{2}\left(A B^{2}+C D^{2}\right),(1) Z O^{2}+K J^{2}=Z J^{2}+O J^{2}+K O^{2}+K Z^{2}=\frac{1}{2}\left(A C^{2}+B D^{2}\right)$,

Subtracting equation (2) and (3) from equation (1) and using the inequalities between the edge lengths of the tetrahedron, we get

$E X^{2}-K J^{2}=\frac{1}{2}\left[\left(A B^{2}+C D^{2}\right)-\left(A C^{2}+B D^{2}\right)\right]=\frac{1}{2}\left[\left(A B^{2}-A C^{2}\right)+\left(C D^{2}-B D^{2}\right)\right]>0, Z O^{2}-K J^{2}=\frac{1}{2}\left[\left(A B^{2}+C D^{2}\right)-(B C\right.$

Thus, $E X>K J$ and $Z O>K J$.

Therefore, among the segments connecting the midpoints of opposite edges, the segment connecting the midpoints of the longest edges $A B$ and $C D$ is the shortest.

Based on the work of Zám Katalin (Eger, Szilágyi E. Gymnasium, 1st year)

Note. From the system of equations (1)-(3), it is easy to express the lengths of the segments connecting the midpoints of opposite edges in terms of the tetrahedron's edges. For example:

$$
K J^{2}=\frac{1}{4}\left(A C^{2}+B C^{2}+A D^{2}+B D^{2}-A B^{2}-C D^{2}\right)
$$