For an integer $n \geq 2$, define $f\_n(\alpha\_0, \alpha\_1, \ldots, \alpha\_{n-1}) = \prod\limits\_{0 \leq i < j < n}\sin^2\left(\alpha\_i - \alpha\_j\right)$ and $$M\_n = \max\limits\_{(\alpha\_0, \alpha\_1, \ldots, \alpha\_{n-1}) \in \mathbb R^n}\{f\_n(\alpha\_0, \alpha\_1, \ldots, \alpha\_{n-1})\}.$$


I *think* that the maximum of this function is achieved at $(\alpha\_0, \alpha\_1, \ldots, \alpha\_{n-1}) = \left(0 + c, \frac{\pi}{n} + c, \ldots, \frac{(n-1)\pi}{n} + c\right)$ for any real number $c$, but I don't know how to prove this.


That being said, I do know how to prove that $f\_n\left(0, \frac{\pi}{n}, \ldots, \frac{(n - 1)\pi}{n}\right) = \frac{n^n}{2^{n(n - 1)}}$.



Maximizing $\prod_{i < j} \sin^2(\alpha_i - \alpha_j)$
Denote $z\_k=e^{2i\alpha\_k}$, then you want to maximize $\prod\_{j<k}|z\_j-z\_k|$, i. e., the product of all sides and diagonals of an inscribed to the unit circle $n$—gon $A\_1\ldots A\_n$. For any given $j=1,2,\ldots, n-1$ the product of $A\_kA\_{k+j}$ over all $k=1,2,\ldots,n$ (indices are cyclic mod $n$) is maximized for a regular polygon, as follows from Jensen inequality for $\log \sin$ (the sum of corresponding arcs taken counterclockwise equals $2\pi j$, so, it is constant). Thus the result.


