Let $V$ be the volume of the tetrahedron, and let $A$, $B$, $C$, and $D$ be the areas of the faces of the tetradhedron. Let $h_A$, $h_B$, $h_C$, and $h_D$ be the altitudes to the corresponding faces of the tetrahedron. Let $r$ be the inradius. Since \[V=\frac{1}{3}Ah_A=\frac 13Bh_B=\frac 13Ch_C=\frac 13Dh_D\] we have that \[V=\frac{Ah_A+Bh_B+Ch_C+Dh_D}{12}\ge \frac{A+B+C+D}{12}\] We also have that \[V=\frac{1}{3}(A+B+C+D)r\] so \[r=\frac{3V}{A+B+C+D}\ge \frac{1}{4}\] Thus, a sphere of radius $\frac 14$ can be placed in any such tetrahedron. It only needs to be verified that a sphere of radius $\frac 14$ is the largest possible sphere that can be placed in a regular tetrahedron with altitudes of length $1$, which is true, as desired.