<s>[INST] Sum of radii of exspheres I am interested in finding some results for tetrahedron that would be analoguous to known results for triangle. In triangle with circumradius R and inradius r, if we consider the excircles $r_i$, then for their  sum the formula
$$\sum r_i  = 4R + r$$ holds.
I wonder if there is similar formula for the sum of the
exradii of the exspheres of the tetrahedron in terms
of the radii of the circumradius and inradius of the circumscribed and inscribed spheres?
$$\sum r_i = ? \quad i = 1,..,4.$$
Maybe you have seen similar problems or you can give me
references where to search further. Thank you.
 [/INST] Such a relation does not and cannot exist.
Let us first write down something we know about in-sphere and ex-spheres for higher dimensions.
For any $d \ge 2$, consider $d+1$ points $\vec{v}_1, \vec{v}_2,\ldots,\vec{v}_{d+1} \in \mathbb{R}^{d}$ in general position.
Let $\Delta$ be the $d$-simplex formed by these $d+1$ points. Let $A_k$ be the area of the face on $\partial\Delta$ opposite to $\vec{v}_k$.
Let $A = \sum\limits_{k=1}^{d+1} A_k$ be the total "area" and $V$ be the "volume" of the simplex $\Delta$.
Let $r$ and $\vec{c}$ be the radius and center of the in-sphere. 
Let $r_k$ and $\vec{c}_k$ be the radius of center of the ex-sphere on the face opposite to vertex $\vec{r}_k$. 
In terms of $A_k$ and $V$, they are given by
$$\begin{cases}
r       =& \frac{dV}{A}\\
\\
\vec{c} =& \frac{1}{A}\sum\limits_{k=1}^{d+1} A_k \vec{v}_k\\
\end{cases}
\quad\text{ and }\quad
\begin{cases}
r_k     =& r\frac{A}{A-2A_k} = \frac{dV}{A - 2A_k}\\
\\
\vec{c}_k =& \frac{r_k}{r} \vec{c} + \left(1 - \frac{r_k}{r}\right) \vec{v}_k\\
\end{cases}$$
In particular, this means in any dimension, we have a relation between $r_k$ and $r$.
$$\sum_{k=1}^{d+1} \frac{1}{r_k} = \frac{d-1}{r}$$
Apply $AM \ge HM$ to the list of exradii $r_k$, we obtain a lower bound for their sum:
$$\frac{1}{d+1}\sum_{k=1}^{d+1}r_k \ge \frac{1}{\frac{1}{d+1}\sum_{k=1}^{d+1}\frac{1}{r_k}}
\quad\implies\quad
\sum_{k=1}^{d+1}r_k \ge \frac{(d+1)^2}{d-1} r$$
Let us back to our original problem at $d = 3$. We will like to show 

For any $R > 3$ and $K \gtrsim \frac{(d+1)^2}{d-1} = 8$, there is a
  tetrahedron whose circumradius is $R$, inradius $1$ and
  $\sum\limits_{k=1}^{4} r_k = K$ whenever $K$ is sufficiently close to $8$.

For any $0 < s < p < R$, let $q = \sqrt{R^2-p^2}$ and $t =  \sqrt{R^2 - s^2}$. Consider
the tetrahedron formed by following 4 points on a sphere of radius $R$:
$$\vec{v}_1 = (p, -q, 0),\; \vec{v}_2 = (p, q, 0),\;\vec{v}_3 = (-s,0,t)\;\text{ and }\;
\vec{v}_4 = (-s,0,-t)$$
It is easy to check
$$V = \frac23 (p+s)qt \quad\text{ and }\quad 
\begin{cases}
A_1 = A_2 = t\sqrt{(p+s)^2 + q^2}\\
A_3 = A_4 = q\sqrt{(p+s)^2 + t^2}
\end{cases}
$$
This implies
$$r = \frac{3V}{A} = \frac{(p+s)qt}{t\sqrt{(p+s)^2 + q^2} + q\sqrt{(p+s)^2 + t^2}}\\
\sum_{k=1}^4 r_k = \sum_{k=1}^4\frac{3V}{A-2A_k} = 2(p+s)qt \left(\frac{1}{t\sqrt{(p+s)^2 + q^2}} + \frac{1}{q\sqrt{(p+s)^2+t^2}}\right)$$
Introduce a small parameter $\sigma$ such that $K = \frac{8}{1-\sigma^2}$. The condition that $r = 1$ and $\sum_{k=1}^4 r_k = K$ can be rewritten as
$$
\begin{cases}
t\sqrt{(p+s)^2 + q^2} = \frac{1+\sigma}{2} (p+s)qt\\
q\sqrt{(p+s)^2 + t^2} = \frac{1-\sigma}{2} (p+s)qt
\end{cases}
\quad\iff\quad
\begin{cases}
\frac{1}{q^2} + \frac{1}{(p+s)^2} = \left(\frac{1+\sigma}{2}\right)^2\\
\frac{1}{t^2} + \frac{1}{(p+s)^2} = \left(\frac{1-\sigma}{2}\right)^2\\
\end{cases}
$$
Let $p+s = 2\mu$, we can rewrite the condition on RHS as
$$
\begin{cases}
R^2 - p^2 = q^2 = 4\mu^2\Lambda_{+}(\mu)\\
R^2 - s^2 = t^2 = 4\mu^2\Lambda_{-}(\mu)
\end{cases}
\quad\text{ where }\quad
\begin{cases}
\Lambda_{+}(\mu) = \frac{1}{(1+\sigma)^2\mu^2 - 1}\\
\Lambda_{-}(\mu) = \frac{1}{(1-\sigma)^2\mu^2 - 1}
\end{cases}
$$
This will lead to a solution for $(p,s)$  of the from
$$\begin{cases}
p = \mu \left( 1 + \Lambda_{-}(\mu) - \Lambda_{+}(\mu) \right)\\
s = \mu \left( 1 - \Lambda_{-}(\mu) + \Lambda_{+}(\mu) \right)
\end{cases}
$$
provided one can find a root $\mu$ for following consistency condition:
$$R^2 = p^2 + 4\mu^2\Lambda_{+}(\mu) = \mu^2 \left( 1 + 2( \Lambda_{+}(\mu) +  \Lambda_{-}(\mu)) + (\Lambda_{+}(\mu) - \Lambda_{-}(\mu))^2\right)$$
Let $\Omega(\mu)$ be the horrible expression at RHS. Notice in the limit of small $\sigma$,
we have
$$\Omega(\sqrt{3}) \sim 3\left( 1 + 2( \frac{1}{3-1} + \frac{1}{3-1} )\right) = 3^2 < R^2$$
Since $\Omega(R) > R^2$, there is a $\mu \in (\sqrt{3}, R)$ which satisfy $R^2 = \Omega(\mu)$. The corresponding $(p,s)$ will then give us
a tetrahedron with fixed circumradius and inradius and yet the sum of exradii is sort of arbitrary.
</s>