Tetrahedron Inscribed spheres

Take the altitude from vertex $A$ onto the plane $BCD$ , call this altitude $h_1$ , and let the altitude of the small tetrahedron with vertex at $A$ and base parallel to $BCD$ be $h'_1$ , then

$h_1 = 2 r + h'_1$

where $r$ is the radius of the original tetrahedron's in-sphere, hence

$r = \dfrac{3V}{\Sigma A_i}$

Here, $\{ A_i \}$ is the area of the face opposite to $i$ -th vertex, for example $A_1$ is the area of $\triangle BCD$ .

Since, we also have $V = \dfrac{1}{3} A_1 h_1$ , then the above relation becomes,

$h'_1 =\displaystyle h_1 - 2 r = h_1 - 2 \left( \dfrac{ 3V} {\Sigma A_i} \right) = h_1 - 2 \left( \dfrac{A_1 h_1}{\Sigma A_i} \right )$

So that,

$h'_1 =\displaystyle h_1 \left(1 - 2 \left( \dfrac{A_1}{\Sigma A_i} \right) \right)$

Using exactly the same reasoning as above, we deduce that

$h'_2 =\displaystyle h_2 \left(1 - 2 \left( \dfrac{A_2}{\Sigma A_i} \right) \right)$ , $h'_3 =\displaystyle h_3 \left(1 - 2 \left( \dfrac{A_3}{\Sigma A_i} \right) \right)$ , $h'_4 =\displaystyle h_4 \left(1 - 2 \left( \dfrac{A_4}{\Sigma A_i} \right) \right)$

and from similarity it follows that $\dfrac{ h'_1}{h_1} =\dfrac{ r'_1}{r} , \hspace{9pt}\dfrac{ h'_2}{h_2} =\dfrac{ r'_2}{r}, \hspace{9pt}\dfrac{ h'_3}{h_3} =\dfrac{ r'_3}{r}, \hspace{9pt} \dfrac{ h'_4}{h_4} =\dfrac{ r'_4}{r}$

Hence,

$r'_1 + r'_2 + r'_3 + r'_4 = r \left(4 - 2 \left( \dfrac{A_1 +A_2+A_3+A_4}{\Sigma A_i} \right) \right) = r (4 - 2) = \boxed{2 r}$