Let T be a possibly irregular tetrahedron, let V be its volume, let S be
its surface area, and let R be the radius of the largest sphere inscribed
in T. Express the ratio V/S in terms of R.

Let the four faces of T have areas A_{1}, A_{2},
A_{3} and A_{4}. Then:

S = A_{1} + A_{2} + A_{3} + A_{4}

Now, decompose T into four smaller tetrahedra, each one having a face of T
as base, and the center of the inscribed sphere as the opposite point.
The height of each of these tetrahedra is R, so:

V | = A_{1}R/3 + A_{2}R/3 + A_{3}R/3 + A_{4}R/3 |

| = SR/3 |

So...

V/S = R/3