Find the smallest real $C$ (as a function of $n$) such that $$\sum\limits_{1\leq i<j\leq n}|z_i-z_j|\leq C\sum\limits^n_{i=1}|z_i|$$ for all complex numbers $z_k$'s with $\sum_{i=1}^n z_i = 0$.
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I repost it since it is meaningful to related inequalities.
Using the triangular inequality, we have $\sum_{1\leq i<j\leq n}|z_i-z_j| \le \sum_{1\leq i<j\leq n} (|z_i| + |z_j|) = (n-1)\sum_{i=1}^n |z_i|$.
Letting $z_k = \mathrm{e}^{\mathrm{i}\frac{2\pi (k-1)}{n}}, ~ k=1, 2, \cdots, n$, we have $\sum_{k=1}^n z_k = 0$ and $\sum_{1\leq i<j\leq n}|z_i-z_j| = n \cot \frac{\pi}{2n}$ which gives $C \ge \cot \frac{\pi}{2n}$.
Also, I found the following result (however the bound is weak for the case here):
Theorem 3.4 in [1]: Let $a_1, a_2, \cdots, a_n$ be complex numbers. Let $r \in \mathbb{R}$, $r\ne 0$ and $s \in \mathbb{R}_{>0}$. Then $$\left(\left|\sum_{i=1}^n a_i\right|^r + \sum_{1\le i < j\le n} |a_i - a_j|^r\right)^{1/r} \le C \left(\sum_{i=1}^n |a_i|^s\right)^{1/s}$$ where $$C = \left\{\begin{array}{ll} m^{1/r - 1/2}n^{1/2 + 1/p - 1/s} & \mathrm{if} ~ r \le 2,~ p = \min(2, s),~ m = n(n-1)/2 \\[9pt] n^{1/r + 1/q - 1/s} & \mathrm{if} ~ r > 2, ~ q = \min(s, r'), ~ \frac{1}{r} + \frac{1}{r'} = 1. \end{array} \right.$$
Question: What is the minimum of $C$, as a function of $n$ ?
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Reference
[1] Lech Maligranda and Lars-Erik Persson, “On Clarkson's Inequalities and Interpolation,” Mathematische Nachrichten, vol. 155, no. 1, pp. 187-197, Jan. 1992.
It's much simpler than everyone thought. Take $z_1 = z = - z_2$ and everything else $0.$ Then: $$\sum_{i<j}|z_i-z_j| = 2(n-1)|z|\leq C\sum_{i}|z_i| = 2C|z|\implies C\geq n-1.$$ But the trivial triangle inequality bound prove this one is optimal.