It also was in our winter camp in this year.

Where can I see the problem? Thank arqady.

Oh, I am sorry. My mistake. Our problem is the following. Find a maximal value of $k$, for which an inequality \[\sqrt{x_1^2+x_2^2+...+x_n^2}\geq k\cdot\min\{|x_1-x_2|, |x_2-x_3|, ..., |x_{n-1}-x_n|, |x_n-x_1|\}\] holds for all given real numbers $x_1$, $x_2$,...,$x_n$. They are different problems of course.

Nice problem, sqing. Here is my proof: First case: $n$ is even. Obviously $|x_i-x_{i+1}|\le |x_i|+|x_{i+1}|\le 2$. If we define $a_k=(-1)^k$ it follows that $a_1+...+a_n=0$ and $|a_i-a_{i+1}|=2$ for $i=1,2..n-1$. Therefore, if n is even, $\max \{ \min\{ |x_i-x_{i+1}| \} \}=2$. Second case: $n$ is odd, with $n=2m+1$ We will prove that $\max \{ \min\{ |x_i-x_{i+1}| \} \}= \frac{n}{\left\lceil{\frac{n}{2}\right\rceil}}$. Assume, on the contrary, that there exist numbers $x_1,..x_n$ satisfying condition and $\min|x_i-x_{i-1}|>\frac{n}{n-1}$. WLOG, we can assume that $x_1>0$. Then, from assumption we know that $x_{2k+1}> 0$ and $x_{2k}< 0$. Now, assumption implies: $x_1>\frac{n}{m+1}+x_2$ and $x_{2k+1}>\frac{n}{m+1}+x_{2k}$ for $k=1,..m$. Finally, from the above implication, we have: $\sum_{k=1}^{n} x_k> \left(\frac{n}{m+1}+x_2 \right) + x_2+ ...+x_{2m}+ \left(\frac{n}{m+1}+x_{2m} \right) = n+x_2+2\sum_{k=1}^{m}x_k \ge 0$ Contradiction! Therefore, $\max \{ \min\{ |x_i-x_{i+1}| \} \}\le \frac{n}{\left\lceil{\frac{n}{2}\right\rceil}}$ However, numbers: $x_{2k+1}=\frac{m}{m+1}$ for $k=0,..,m$ and $x_{2k}=-1$ for $k=1,...m$ satisfy condition and $|x_i-x_{i+1}|=\frac{n}{\left\lceil{\frac{n}{2}\right\rceil}}$, so: $\max \{ \min\{ |x_i-x_{i+1}| \} \}= \frac{n}{\left\lceil{\frac{n}{2}\right\rceil}}$. $\blacksquare$

MathUniverse wrote: Second case: $n$ is odd, with $n=2m+1$ We will prove that $\max \{ \min\{ |x_i-x_{i+1}| \} \}= \frac{n}{\left\lceil{\frac{n}{2}\right\rceil}}$. Assume, on the contrary, that there exist numbers $x_1,..x_n$ satisfying condition and $\min|x_i-x_{i-1}|>\frac{n}{n-1}$. Are you mean $\max\min|x_i-x_{i-1}|>\frac{2n}{n+1}$?

I suppose he does @arqady the answer should be:2(n is even),2n/(n+1)(n is odd)

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