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This is probably too hard for math.stackexchange, so I migrated it here.

For $\lambda_i \in S^1 \subset \mathbb{C}$, consider the functional $H(\{\lambda_1, \ldots, \lambda_n\}):= \sum_{j < k} | \lambda_j - \lambda_k | $. I want to show that $H$ is globally maximized by some picket fence configuration, that is, $\lambda_j = e^{2 \pi i (\alpha + j/n)}$, for some $\alpha \in \mathbb{R}$.

Let $\lambda_j = e^{i \theta_j}$, with $0 \le \theta_1 \le \ldots \le \theta_n < 2 \pi$. Using the identity $| e^{i \theta_1} - e^{i \theta_2}| = | 2 \sin \frac{\theta_1 - \theta_2}{2}|$ and the fact that $0 \le \theta_k - \theta_j < 2 \pi$, I reduce the problem to maximizing over $\theta_j$ satisfying the above constraints, the following function: $$ \sum_{j < k} \sin \frac{\theta_j - \theta_k}{2}.$$ Assuming $\theta_j$'s are distinct for now, differentiating gives me $$ \sum_{\ell < k} \cos \frac{\theta_\ell - \theta_k}{2} - \sum_{j < \ell} \cos \frac{\theta_j - \theta_\ell}{2} = 0, \qquad \forall \ell=1,\ldots, n.$$

But how do I finish?

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Use Hadamard's inequality for the determinant $$ | \det (\mathbf{a}_1, \mathbf{a}_2, \ldots , \mathbf{a}_n ) | \leq \| \mathbf{a}_1 \| \| \mathbf{a}_2 \| \cdots \| \mathbf{a}_n \| , $$ with equality if and only if $\mathbf{a}_1, \mathbf{a}_2, \ldots , \mathbf{a}_n $ forms an orthogonal system. Then $$ |H(\lambda_1, \ldots ,\lambda_n )| = abs \det \left( \begin{array}{cccc} 1 & \lambda_1 & \cdots & \lambda_1^{n-1} \\ 1 & \lambda_2 & \cdots & \lambda_2^{n-1} \\ \vdots & \vdots & \ddots & \vdots \\ 1 & \lambda_n & \cdots & \lambda_n^{n-1} \end{array} \right) \leq n^{\frac{n}{2}} $$ with equality if and only if \begin{align*} & \lambda_1+ \lambda_2 + \cdots + \lambda_n = 0 \\ & \lambda_1^2+ \lambda_2^2 + \cdots + \lambda_n^2 = 0 \\ & \vdots \\ & \lambda_1^{n-1} + \lambda_2^{n-1} + \cdots + \lambda_n^{n-1} = 0 . \end{align*} Equivalently elementary symmetric polynomials up to order $n-1$ vanish, i.e., \begin{align*} \sum_{k=1}^n \lambda_k = 0, \quad \sum_{1\leq k_1< k_2 \leq n} \lambda_{k_1} \lambda_{k_2} = 0, \ldots , \\ \lambda_2 \lambda_3 \cdots \lambda_n + \lambda_1 \lambda_3 \cdots \lambda_n + \cdots + \lambda_1 \lambda_2 \cdots \lambda_{n-1} = 0 . \end{align*} This means $\lambda_1, \lambda_2 , \ldots , \lambda_n $ forms a picket fence in your notation.

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    $\begingroup$ Nice try. But note I have sum, not product. So Vandermonde doesn't apply here? $\endgroup$
    – John Jiang
    Feb 22, 2019 at 7:02

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