In fact, this topic is to find the center of a regular tetrahedron. If the radii of the four spheres are all r and tangent, the connecting lines of the centers of the four spheres form a regular tetrahedron with a side length of 2R.

As shown in the figure, ABCD is the vertex of the regular tetrahedron, O is the center, AO = DO = X, OP = Y.

According to the properties of regular tetrahedron, P is the center of △BCD, then PD=√3/3*CD=2√3/3R=√3/3.

In RT△OPD, OD2-OP2=PD2.

That is X2-Y2= 1/3.

In RT△APD, AP2+PD2=AD2, that is, (X+Y)2+ 1/3= 1.

X=√6/3 can be obtained by the above two equations.

So the maximum radius that can be placed in the middle of four balls =X-R=(2√6-3)/6.

I hope this helps.