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If $z_1,..z_n$ are complex number, then there is a subsequence $n_i$ such that $$\pi \left|\sum_i z_{n_i}\right| \geq \sum_k^n |z_{k}|.$$ How to prove this inequality?

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  • $\begingroup$ There is no further information about the sequence? $\endgroup$ – gbox Oct 18 '17 at 7:39
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Let $z_k=|z_k|e^{ia_k}$ and $S(\theta)$ be the set of all $k$ such that $\cos(a_k-\theta)>0$. $$\left|\sum\limits_{S(\theta)}z_k\right|=\left|\sum\limits_{S(\theta)}e^{-i\theta}z_k\right| \geq Re\sum\limits_{S(\theta)}e^{-i\theta}z_k=\sum\limits_{k=1}^{n}|z_k|\min(0,\cos(a_k-\theta))$$ We choose some $\theta$ to maximize the term $\sum\limits_{k=1}^{n}|z_k|\min(0,\cos(a_k-\theta))$ By integrating it, $$\int_{-\pi}^{\pi}\sum\limits_{k=1}^{n}|z_k|\min(0,\cos(a_k-\theta))d\theta=\sum\limits_{k=1}^{n}|z_k|\int^{a_k+\pi/2}_{a_k-\pi/2}\cos(a_k-\theta)d\theta=2\sum\limits_{k=1}^{n}|z_k|$$ So the integral average value of $\sum\limits_{k=1}^{n}|z_k|\min(0,\cos(a_k-\theta))=\frac{1}{\pi}\sum\limits_{k=1}^{n}|z_k|$, there is a $\theta_0$ such that $\sum\limits_{k=1}^{n}|z_k|\min(0,\cos(a_k-\theta_0)) \geq \frac{1}{\pi}\sum\limits_{k=1}^{n}|z_k|$. Hence we choose the subset $S(\theta_0)$.

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  • $\begingroup$ But can someone explains the intuitive meaning of the proof? I just copy this from the book and I find it so amazing. $\endgroup$ – mnmn1993 May 1 '18 at 8:19

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