 Solution

Compute twice the four opposite face areas and their weighted vertex average:These positive weights make $I$ an interior convex combination. To verify the incenter of a tetrahedron, put $\Delta =|(B-A)\cdot ((C-A)\times (D-A))|$. The altitude from $A$ to its opposite face is $\Delta /w_A$. Signed distance to that face is affine and vanishes at the other three vertices, so $I$ has distance $(w_A/w)(\Delta /w_A)=\Delta /w$. The same argument works for every face. HenceThe sphere of radius $r_{\mathrm{in}}$ centered at $I$ is tangent to all four faces. Equality of the four distances conversely forces these face-area weights, proving uniqueness.