11.13. a) Consider for each internal point of a regular tetrahedron the sum of distances from it to its vertices. Prove that this sum will be the smallest for the center of the tetrahedron.

b) Two opposite edges of the tetrahedron are equal to $b$ and $c$, while the other edges are equal to $a$. What is the minimum value of the sum of distances from an arbitrary point in space to the vertices of this tetrahedron?
11.13. a) Let's draw planes through the vertices of a regular tetrahedron \(ABCD\) parallel to the opposite faces. These planes also form a regular tetrahedron. It follows that the sum of the distances from any internal point \(X\) of the tetrahedron \(ABCD\) to these planes is constant (Problem 8.1, a). The distance from point \(X\) to such a plane does not exceed the distance from point \(X\) to the corresponding vertex of the tetrahedron, and the sum of the distances from point \(X\) to the vertices of the tetrahedron equals the sum of the distances from point \(X\) to these planes only if \(X\) is the center of the tetrahedron.

b) Let in the tetrahedron \(ABCD\) the edges \(AB\) and \(CD\) be equal to \(b\) and \(c\), and the other edges be equal to \(a\). If \(M\) and \(N\) are the midpoints of the edges \(AB\) and \(CD\), then the line \(MN\) is the axis of symmetry of the tetrahedron \(ABCD\). Let \(X\) be an arbitrary point in space; point \(Y\) is symmetric to it with respect to the line \(MN\); \(K\) is the midpoint of the segment \(XY\) (it lies on the line \(MN\)). Then \(XA + XB = XA + YA \geq 2KA = KA + KB\). Similarly, \(XC + XD \geq KC + KD\). Therefore, it is sufficient to determine the minimum value of the sum of distances to the vertices of the tetrahedron for points on the line \(MN\). For points on this line, the sum of distances to the vertices of the tetrahedron \(ABCD\) will not change if the segment \(AB\) is rotated around it so that it becomes parallel to \(CD\). In this case, we obtain an isosceles trapezoid \(ABCD\) with bases \(b\) and \(c\) and height \(MN = \sqrt{a^2 - (b^2 + c^2) / 4}\). For any convex quadrilateral, the sum of distances to the vertices reaches its minimum value at the point of intersection of the diagonals; in this case, it is equal to the sum of the lengths of the diagonals. It is easy to verify that the sum of the lengths of the diagonals of the resulting trapezoid \(ABCD\) is \(\sqrt{4a^2 + 2bc}\).