Problem 3
Prove that the sum of the distances of the vertices of a regular tetrahedron from the center of its circumscribed sphere is less than the sum of the distances of these vertices from any other point in space.
Step 4 of 6: Add the four distance inequalities
In plain words
One perpendicular-bisector plane already rules out every point outside it as a possible minimizer.
Detailed analysis
Summing the inequalities from steps 2 and 3 shows that replacing by its projection onto never increases the total distance to the four vertices, with equality iff .