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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.

River Li
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    If all $z_j$ have the same modulus then $\cot(\pi/(2n))$ is an upper bound, compare https://math.stackexchange.com/q/2048082/42969. This may or may not help to solve the general problem. – Martin R Aug 20 '22 at 14:58
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    Side note: I would avoid using $i$ as an index when dealing with complex numbers :) – Martin R Aug 20 '22 at 14:59
  • @MartinR By the way, I use $\mathrm{i}$ such as $\mathrm{e}^{\mathrm{i} x}$, and never use $i$ for complex numbers. (Also, I use $\mathrm{e}$, and don't use $e$ for the constant.) – River Li Aug 20 '22 at 15:08
  • @MartinR The question in your link is interesting. It reminds me the conjecture below. Is there some relation between them? Let $0\le a_1 \le a_2 \le \cdots \le a_n \le \pi$. Prove or disprove that $$\sum_{1\le i < j \le n} \sin (a_j - a_i) \le \frac{n}{2}\cot \frac{\pi}{2n}.$$ – River Li Aug 20 '22 at 15:17
  • Using triangular inequality directly, you will have $n-1$? Or am I making mistake? – Kroki Aug 20 '22 at 15:33
  • @Youem You are right. The bound there is too weak for that case. – River Li Aug 20 '22 at 23:18
  • nice problem. Looks like you can square and bash to show that the cotangent bound is optimal for $n = 3.$ – dezdichado Aug 21 '22 at 00:16
  • @dezdichado If so, it is helpful. – River Li Aug 21 '22 at 00:18
  • never mind - it's much simpler than we thought. – dezdichado Aug 21 '22 at 00:36
  • @MartinR Thanks for the link. I can prove this inequality https://math.stackexchange.com/questions/2458827/inequality-with-condition-sum-limits-1-leq-i-j-leq-n1-x-ix-j-sum-limits – River Li Aug 23 '22 at 02:15

1 Answers1

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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.

dezdichado
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