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Assmue that give the positive integer number $n$,Find the largerst the constant $\lambda(n)$,such for any complex $z_{1},z_{2},\cdots,z_{n}(z_{i}\neq 0,i=1,2,\cdots n)$,have $$\sum_{k=1}^{n}|z_{k}|^2\ge\lambda(n)\cdot\min_{1\le k\le n}{|z_{k+1}-z_{k}|^2}$$ where define $z_{n+1}=z_{1}$

This problem is from china TST 2014,

before I have post this simarler problem: Finding the biggest $k$ such $\sqrt{x_1^2+x_2^2+\dots+x_{n-1}^2+x_n^2} \geq k\min(|x_1-x_2|,|x_2-x_3|,\dots,|x_{n-1}-x_n|,|x_n-x_1|)$

and

How find this inequality $\max{\left(\min{\left(|a-b|,|b-c|,|c-d|,|d-e|,|e-a|\right)}\right)}$

In fact, this two different problem.

maybe this idea can help solve this problem? Thank you for you help

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math110
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1 Answers1

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Let $\mu=\min_{1\le k\le n}{|z_{k+1}-z_{k}|^2}$. When $n$ is even, achille hui’s proof can be used again without changes : we have

$$ |z_{2k}|^2+|z_{2k+1}|^2 =\frac{|z_{2k}-z_{2k+1}|^2+ |z_{2k}+z_{2k+1}|^2}{2}\geq \frac{|z_{2k}-z_{2k+1}|^2}{2} \geq \frac{\mu}{2} $$

and summing on $k$ we have that $\sum_{k=1}^{n} |z_k|^2 \geq \frac{n\mu}{2}$. This becomes an equality when $z_k=1,z_{2k+1}=-1$ for every $k$, so $\lambda(n)=\frac{n}{2}$ when $n$ is even.

When $n$ is odd, however, achille hui’s proof uses the ordering of the real numbers which does not exist in $\mathbb C$.

Ewan Delanoy
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