Can you somehow conclude that the surface area of the tetrahedron's circumsphere is greater than or equal to the area of its faces?
I tried to go from the 2D version (i.e. the diameter of the circumcircle of a triangle is always greater than or equal to the triangle's edges), but is kinda stuck of how to make the proper connection. In the 2D version, the proof is rather straightforward by drawing a diameter that intersects with one of the vertices and using the observation that the hypotenuses are always the largest edge of a square triangle.
Can we make use of that argument in 3D?