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Let us consider $n$-dimensional simplex $K$ in $n$-dimensional Euclidean space. Let $r_0$ be the radius of the inscribed sphere of $K$, and let be $r_1, r_2, \cdots, r_{n+1}$ be each radius of the exsphere of $K$. Then, here is my question.

Question : How can we represent $r_0$ by $r_1, r_2, \cdots, r_{n+1}$?

Motivation : I've known the followings :

In the $n=2$ case, $$r_0=\frac{1}{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}}.$$ In the $n=3$ case, $$r_0=\frac{2}{\frac{1}{r_1}+\frac{1}{r_2}+\frac{1}{r_3}+\frac{1}{r_4}}.$$

Then, I reached the following conjecture (of course, this is just a conjecture) : $$r_0=\frac{n-1}{\sum_{i=1}^{n+1}\frac{1}{r_i}}.$$

I don't have any good idea for $n$ in general. Can anyone help?

Update : I crossposted to MO.

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I'll prove the formula for a triangle, and it'll be "obvious" that the argument extends to any-dimensional space.

Consider $\triangle ABC$ with incenter $O$ and inradius $r$. Recall the fundamental observation that point $O$ helps dissect the triangle into sub-triangles of equal height ($r$), so that $$\begin{align} |\triangle ABC| &= |\triangle OBC| + |\triangle OCA|+ |\triangle OAB| \\[6pt] &= \frac{1}{2}r\;|BC| + \frac{1}{2}r\;|CA| + \frac{1}{2}r\;|AB|\\[6pt] &=\frac{r}{2}\left( a + b + c \right) \end{align}$$

Similarly, if $O_A$ is the excenter on the opposite side of $\overline{BC}$ from $A$, and if $r_A$ is the corresponding exradius, then we have this formula: $$\begin{align} |\triangle ABC| &= -|\triangle O_A BC| + |\triangle O_A CA| + |\triangle O_A AB| \\[6pt] &= \frac{r_A}{2}\left(-a + b + c \right) \end{align}$$ Likewise, for exradii $r_B$ and $r_C$: $$|\triangle ABC| = \frac{r_B}{2}\left( a - b + c \right) = \frac{r_C}{2}\left( a + b - c \right)$$ Thus, $$\frac{1}{r_A}+\frac{1}{r_B}+\frac{1}{r_C} = \frac{(-a+b+c)+(a-b+c)+(a+b-c)}{2|\triangle ABC|} = \frac{a+b+c}{r \left(a+b+c\right)} = \frac{1}{r}$$ as expected.

The higher-dimensional cases proceed according to the same idea: the incenter and each excenter provides a dissection of the simplex into equal-height sub-simplices. The resulting "content" formulas are identical, except for a change in sign (and the radius involved). The extra coefficient appears as a natural consequence of the increasing failure of negative components to cancel positive ones.

Here's a run-through of the case of tetrahedron $ABCD$, where I'll write $w$, $x$, $y$, $z$ for the areas of faces opposite $A$, $B$, $C$, $D$. For incenter $r$, we have

$$\begin{align}|\operatorname{tet}ABCD| &= |\operatorname{tet}OBCD|+|\operatorname{tet}OCDA|+|\operatorname{tet}ODAB|+|\operatorname{tet}OABC| \\ &= \frac{1}{3} r \;\left( w+x+y+z \right) \end{align}$$

And for excenter $r_A$ opposing $A$, we have: $$\begin{align}|\operatorname{tet}ABCD| &= -|\operatorname{tet}O_A BCD|+|\operatorname{tet}O_A CDA|+|\operatorname{tet}O_A DAB|+|\operatorname{tet}O_A ABC| \\[6pt] &= \frac{r_A}{3} \;\left(-w+x+y+z \right) \end{align}$$ Also, $$|\operatorname{tet}ABCD| = \frac{r_B}{3} \;\left(w-x+y+z \right) = \frac{r_C}{3} \;\left(w+x-y+z \right) = \frac{r_D}{3} \;\left(w+x+y-z \right)$$ whence $$\begin{align} \frac{1}{r_A} + \frac{1}{r_B} + \frac{1}{r_C} + \frac{1}{r_D} &= \frac{(-w+x+y+z)+\cdots+(w+x+y-z)}{3|\operatorname{tet}ABCD|} = \frac{2(w+x+y+z)}{r(w+x+y+z)} \\[6pt] &= \frac{2}{r} \end{align}$$

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I'm posting an answer just to inform that the question has received an answer by Ilya Bogdanov on MO.

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