Maximum volume ball in tetrahedron Consider the problem :
    "Find the maximum volume of the three dimensionnal ball inscribed to a tetraedron." 
The tetraedron have the following summit (just to have a concrete case):
$(0,0,0), (0,\pi,0),(\pi,0,0),(\pi,\pi,\pi)$.
Geometrically, from the optimization point of view, the problem can be formulated as
$
\begin{align}
&\max_{a,b,c,r\in \mathbb{R}}\qquad r^3\\
&r-|c|<0\\
&r-|a-b|/\sqrt{2}<0\\
&r-|b-\pi|<0\\
&r-|a-c|/\sqrt{2}\\
&a-\pi<0\\
&b-\pi<0\\
&c-\pi<0\\
&a>0\\
&b>0\\
&c>0\\
&r>0\\
\end{align}
$
($r$ stands for the radius of the ball of center $(a,b,c)$).
The optimal volume is obtained for $r\simeq 0.65$. 

Now, I have difficulty in formulating the problem with a ball of dimension $4$ in a tetraedron of dimension $4$ because I don't have the geometric point of view and I don't know how to write the constraints. 
 A: Alright, still not sure what the problem is; however...
To find the circle inscribed in a triangle in the plane, draw the angle bisectors through each vertex. These meet in the center of the inscribed circle.
In $\mathbb R^3,$ for each pair of planes bounding the tetrahedron, draw the plane that bisects the angle between the two planes. The six new planes meet in the center of the inscribed sphere.
In higher dimension, for each pair of hyperplanes bounding the simplex, construct the hyperplane bisecting that angle; these new hyperplanes meet in the center of the inscribed sphere.
There is a little linear algebra/geometry involved in describing these bisecting hyperplanes... In $\mathbb R^n,$ a plane containing a face of the simplex is an $(n-1)$-plane. Two of them meet in an $(n-2)$-plane, call it $P.$ The orthogonal complement of $P$ (through a favorite point) is a $2$-plane, call it $O.$ The intersection of $O$ with the two bounding hyperplanes is just two intersecting lines; we construct the angle bisector between these two lines, and take the sum of that with $P,$ getting us back up to a bisecting  $(n-1)$-plane. 
