I'm looking for a proof for the extremum problem on regular simplex.



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> **Question.** Let $\mathcal{A}=A\_0A\_1...A\_n$ be a regular simplex in $\Bbb E^n$. $P$ is a point inside and on boundary of $\mathcal{A}$. Prove that the product $$\prod\_{i=0}^n PA\_i$$ attains maximum iff $P$ is one of the midpoints of edges of this simplex.
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Extremum problem on regular simplex
It looks like for small $n$, your assertion is true. For large $n$, there are symmetrically placed pairs on the side achieving the maximum. Please see [this paper](https://link.springer.com/article/10.1007/s00022-006-1813-7) for details. (I do not have access to this paper yet.)


[Previuos comments]
The result is clear for $n=1$ from the AM-GM inequality. The case $n=2$ can be shown by complex analysis: Represent the regular triangle by the roots of unity $1,\zeta,\zeta^2$, where $\zeta =-\frac 12+\frac {\sqrt{3}}2i$. Then the required function is represented by $$p(z)=|(z-1)(z-\zeta)(z-\zeta)|=|z^3-1|.$$ By the maximum modulus principle and by symmetry we may assume that $z$ lies on the side joining $\zeta$ to $\zeta^2$, i.e. $z$ is of the form $z=-\frac 12+ti,$ where $-\frac{\sqrt 3}2\leq t\leq\frac{\sqrt 3}2$. By calculus, the maximum happens when $t=0$ and $z=-1/2$, which is the midpoint of a side of the regular triangle.


EDIT. Here is a suggestion for the general case. Taking logarithm, the problem is equivalent to finding the maximum of the function $$f=\frac 12\sum\_{i=0}^n \log [(x\_1-a\_1^i)^2+\cdots+(x\_n-a\_n^i)^2],$$ where $(x\_1,\cdots,x\_n)$ is the coordinates for $P$, and $(a\_1^i,\cdots,a\_n^i)$ represents the vertices with $i=0,\cdots,n$. If $n>2$, then $\Delta f>0$, so $f$ is a subharmonic function and the maximum principle applies. This does not solve the problem yet, but it follows that the maximum occurs on the boundary faces. 


