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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 $f_n\left(0, \frac{\pi}{n}, \ldots, \frac{(n - 1)\pi}{n}\right) = \frac{n^n}{2^{n(n - 1)}}$.

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    $\begingroup$ Denote $x_i=a_i-a_{i+1}$. Then you are studying $$ \prod_{0\le i< j< n}\sin^2(x_i+x_{i+1}+\cdots+x_{j-1}), $$ and you conjecture that the maximum is attained when $x_0=\cdots=x_{n-1}=\frac{\pi}n$. This is a bit more symmetric conjecture, so perhaps something can be done. In any case, using this formulation, I checked the result for $n=3$ with almost no calculations, but you perhaps did that yourself already. $\endgroup$
    – Vladimir Dotsenko
    Commented Oct 21, 2022 at 18:46
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    $\begingroup$ You should be able to prove this by writing this as $\exp(\sum_{0 \leq i < j \leq n} \log \sin^2 (\alpha_i - \alpha_j))$ and using Jensen's inequality. A similar sort of result is given in Proposition 5.1 of a paper of mine: doi.org/10.1142/S1793042113500978 $\endgroup$
    – Peter Humphries
    Commented Oct 21, 2022 at 20:06

1 Answer 1

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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.

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