This is an excellent problem. And surprising too.

Note that \(S\) is the area of the triangle whose sides are the products of the opposite pairs of edges of \(MABC\). According to the tetrahedron circumradius formula \(6VR=S\). Thus we need to prove only \(R\ge \frac 1 {\sqrt 3}\) which is obvious since the circumradius of \(ABC\) is \(\frac 1 {\sqrt 3}\). Equality holds iff the circumcenter of \(MABC\) lies on \(ABC\).