TITLE: Finding the biggest $k$ such $\sqrt{x_1^2+x_2^2+\dots+x_{n-1}^2+x_n^2} \geq k\min(|x_1-x_2|,|x_2-x_3|,\dots,|x_{n-1}-x_n|,|x_n-x_1|)$
QUESTION [6 upvotes]: Given $n$ natural. Find the biggest real $k$ so for all $n$ real numbers $x_1,x_2,\dots,x_n$:
$$\sqrt{x_1^2+x_2^2+\dots+x_{n-1}^2+x_n^2} \geq k\min(|x_1-x_2|,|x_2-x_3|,\dots,|x_{n-1}-x_n|,|x_n-x_1|).$$ 
Tried by AMGM. But failed.

REPLY [3 votes]: For any $n \ge 2$, let $k_n$ be the biggest $k$ that makes the inequality
$$\sqrt{\sum_{i=1}^n x_i^2} \ge k \min\bigg\{ |x_1-x_2|, \ldots, |x_{n-1}-x_n|,  |x_{n}-x_1| \bigg\}$$
valid for all $(x_1,\ldots,x_n) \in \mathbb{R}^n$. We are going to show $k_n$ is given by
the following formula:

$$k_n = \begin{cases}  \sqrt{\frac{n}{4}},& n \text{ even}\\
\sqrt{\frac{n+5}{4}}, & n \text{ odd} \end{cases}$$

Let $\mathcal{C}_n, \mathcal{C}'_n \subset \mathbb{R}^n$ be the set of $n$-tuples defined by
$$\begin{align}
\mathcal{C}_n &= \bigg\{ (x_1, \ldots, x_n )\in\mathbb{R}^n : | x_i - x_{i+1} | \ge 1, \forall 1 \le i < n \bigg\}\\
\mathcal{C}'_n &= \bigg\{ (x_1, \ldots, x_n )\in \mathcal{C}_n : |x_1 - x_n | \ge 1 \bigg\}
\end{align}$$
For any $x = (x_1, \ldots, x_n) \in \mathbb{R}^n$, let $\mathcal{V}(x)$ be the "unscaled " variances of $x_i$. More precisely,
$$\mathcal{V}(x) = \sum_{i=1}^n (x_i - \bar{x})^2 = \sum_{i=1}^n x_i^2 - n \bar{x}^2
\quad\text{ where }\quad \bar{x} = \frac{1}{n}\sum_{i=1}^n x_i$$
If $x_i \ne x_{i+1}$ for $1 \le i < n$ and $x_1 \ne x_n$, then we can scale $x$ by 
$$\mu = \min\bigg\{\;|x_2 -x_1|, \ldots, |x_n - x_{n-1}|, |x_1 - x_n|\;\bigg\}$$
and make $\quad\displaystyle\frac{x}{\mu} = \left(\frac{x_1}{\mu},\ldots,\frac{x_n}{\mu}\right)\;\;$ and 
$\;\;\displaystyle\frac{x-\bar{x}}{\mu} = \left(\frac{x_1-\bar{x}}{\mu},\ldots,\frac{x_n-\bar{x}}{\mu}\right) \in \mathcal{C}'_n$.
Notice 
$$\mu^{-2}\sum_{i=1}^n x_i^2 \ge \mu^{-2}\sum_{i=1}^n (x_i - \bar{x})^2 =
\mu^{-2}\mathcal{V}(x-\bar{x}) = \mathcal{V}\left(\frac{x-\bar{x}}{\mu}\right) =
\mathcal{V}\left(\frac{x}{\mu}\right)$$
and this inequality becomes an equality when $\bar{x} = 0$. If one 
compares this with the definition of $k_n$, we find:
$$k_n = \sqrt{ \inf \bigg\{\;\mathcal{V}(x) : x \in \mathcal{C}'_n\; \bigg\} }$$
There is another property of $\mathcal{V}(x)$ we need. 
For any $x = (x_1,\ldots,x_m) \in \mathbb{R}^m$, $y = (y_1,\ldots,y_n)$$ \in \mathbb{R}^n$, we have
$$\mathcal{V}(x \oplus y) \ge \mathcal{V}(x) + \mathcal{V}(y)\quad\text{ where }\quad
x\oplus y = (x_1,\ldots,x_m, y_1,\ldots,y_n) \in \mathbb{R}^{m+n}$$
and the equality holds if and only if $\bar{x} = \bar{y}$.
One consequence of this is for any $x \in \mathcal{C}_{2k}$, $\mathcal{V}(x) \ge \frac{2k}{4}$.
We can prove this by induction on $k$. 
The start case $k = 1$ is obvious. Assume this is true for a particular $k$. Then for any $x \in \mathcal{C}_{2k+2}$, we will rewrite it as $y \oplus z$ for some $y \in \mathcal{C}_2$, $z \in \mathcal{C}_{2k}$. This leads to
$$\mathcal{V}(x) = \mathcal{V}(y \oplus z) \ge \mathcal{V}(y) + \mathcal{V}(z) = \frac{2k}{4} + \frac{2}{4} = \frac{2k+2}{4}$$
and complete the induction step. Since $\mathcal{C}'_{n} \subset \mathcal{C}_{n}$, 
we find for any even $n$,
$$\mathcal{V}(x) \ge \frac{n}{4} \quad\text{ for } x \in \mathcal{C}'_{n}.$$
In above inequality, the case of equality can be achieved by the configuration $$\left(\frac12,-\frac12,\ldots,\frac12,-\frac12\right) \in \mathcal{C}'_n,$$ 
From this, we can conclude $k_n = \sqrt{\frac{n}{4}}$ for even $n$.
Let us switch to the case for odd $n \ge 5$. For any $x = (x_1, \ldots, x_n) \in \mathcal{C}'_n$, consider following cyclic list of $n$ numbers:
$$x_2 - x_1,\;x_3 - x_2,\;\ldots,\;x_n - x_{n-1},\;x_1 - x_n$$
Since $n$ is odd, there is at least one successive pair whose sign is the same.
Since $\mathcal{V}(x)$ is invariant under cyclic permutation of components and under the
map $x \mapsto -x$. Without loss of generality, we only need to consider the case $x_1 < x_2 < x_3$. Once again,
rewrite our $x$ as $y \oplus z$ where
$$y = (x_1, x_2, x_3) \in \mathcal{C}_3 \quad\text{ and }\quad
  z = (x_4, \ldots, x_n ) \in \mathcal{C}_{n-3}$$
It is clear $\mathcal{V}(y) \ge 2$ and by the above discussion, $\mathcal{V}(z) \ge \frac{n-3}{4}$. Combining these, we get
$$\mathcal{V}(x) \ge \frac{n+5}{4}\quad\text{ for } x \in \mathcal{C}'_n$$
Once again, the case of equality is achievable. Consider the following configuration by
combining the equal-spacing and zero-hugging strategies in John Bentin's answer:
$$\left(-1,0,1,-\frac12,\frac12,\ldots,-\frac12,\frac12\right) \in \mathcal{C}'_{n}$$
From this, we can conclude $k_n = \sqrt{\frac{n+5}{4}}$ for odd $n \ge 5$. It is easy to check this formula also works for the case $n = 3$.