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 5 of 6: Repeat for all three pairs of opposite edges

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In plain words

Three independent constraints, one per pair of opposite edges, pin the minimizer down to a single point instead of a whole plane.

P\in\Pi_{AB,CD}\cap\Pi_{AC,BD}\cap\Pi_{AD,BC}

Detailed analysis

A regular tetrahedron has three pairs of opposite edges ( $AB\,\&\,CD$ , $AC\,\&\,BD$ , $AD\,\&\,BC$ ), each giving a perpendicular-bisector plane that step 4 shows any minimizing point must lie in. By the symmetry of the regular tetrahedron, these three planes meet in exactly one point, the center $O$ .