Problem Description:
The radii $r$ of 4 spheres are $r_1,r_2,r_3,r_4>0$ .
The center points of the spheres are not on a plane.
The spheres can overlap.
The distance to the surface of the sphere is positive in the case a sphere encloses the point. The cost-function is therefore always positive or 0 (in the case all spheres intersect at exactly one point).
Intuition:
There has to be a unique point where the sum of the distances from the point to the surfaces of the 4 spheres is minimal.
Q1: Is there a proof that a unique minimum exists?
Q2: Is the cost-function monotonically decreasing when approaching this point?
Edit:
- In the comments it has been concluded, that the function can not be smooth.
- In the comments it has been concluded, that in case of equal radii the point is the Fermats point.
I expect there is the following counterexample. Let the centers of the spheres are the vertices of a regular tetrahedron inscribed in the sphere of radius $1$ centered at the origin $O$ and each sphere has radius $3$. If the cost function has a unique minimum at a point $P$ then by the symmetry $P$ should be the origin. The value of the cost function at the origin is $8$. Pick any vertex $V$ of the tetrahedron. Let $Q$ be the intersection point of the ray from $O$ towards $V$ with the sphere of radius $3$ centered at $V$. Then the value of the cost function at $Q$ should be $3\left(\frac{\sqrt{46}}2-3\right)\approx 1.173$.
Here is the graph of a two-dimensional counterpart.