Finding the biggest $k$ such $\sqrt{x_1^2+x_2^2+\dots+x_{n-1}^2+x_n^2} \geq k\min(|x_1-x_2|,|x_2-x_3|,\dots,|x_{n-1}-x_n|,|x_n-x_1|)$ 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.
 A: 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}$$
From this, we can conclude $k_n = \sqrt{\frac{n+5}{4}}$ for odd $n \ge 5$. It is easy to check this formula also works for the case $n = 3$. 
A: You need a positive lower bound for the differences $|x_j-x_{j+1}|$; otherwise there is no biggest $k$. Once such a bound, say $2a$, has been set, the answer depends on whether $n$ is odd or even. The easy case is when $n$ is even: then the LHS is minimized by $x_j=(-1)^ja$, and $k=\sqrt n/2$.
Now consider the case when $n$ is odd. Let us write $\mathbf{x}:=(x_1,...,x_n),$   $x_{n+1}:=x_1,$ and $\Bbb R_\neq^n:=\{x\in \Bbb R:|x_j-x_{j+1}|\neq 0\;(j=1,...,n)\}.$ For $\mathbf x\in \Bbb R_\neq^n,$ define $\lVert\mathbf x\rVert:=\sqrt{x_1^2+\cdots+x_n^2},$ $ \mu(\mathbf x):=\min\{|x_j-x_{j+1}|:j=1,...n\},$ and
$$\nu(\mathbf x):=\dfrac{\lVert\bf x\rVert}{\mu(\mathbf x)}.$$
The task is to minimize $\nu(\mathbf x)$ over $\mathbf x\in\Bbb R_\neq^n.$
Note first that $\nu(\xi \bf x) =\nu(\bf x)$ for any nonzero scalar $\xi.$ Thus we may scale $\bf x$ so that $\mu(\mathbf x)=2.$ Also, shifting the $x_j$ by a constant does not affect $\mu(\mathbf x);$ so we may take the mean of the $x_j$ to be $0$, to minimize the sum of their squares and hence $\lVert \bf x\rVert.$ In general, we try to follow the even case and place each $x_j$ as close to zero as possible; but the oddness of $n$ forces an outlier. So, taking $x_1,...,x_n$ to be regular and $x_n$ to be the outlier, we may choose
$$x_j=(-1)^j-\dfrac{3}{n}\;(j=1,...,n)\quad\text{with}\quad x_n=3-\dfrac{3}{n}$$
(the $-3/n$ term brings the mean to $0$). A straightforward calculation shows that $$\nu(\mathbf x)=\sqrt{\dfrac{(n-1)(n+9)}{4n}}.$$
However, observe that the outlier term $3-3/n$ is $2.4$ when $n=5,$ and its square makes a major contribution of 5.76 to the total sum of squares $\lVert \mathbf x\rVert^2=11.2.$ In this case, we can do better by equally spacing the $x_j$ around zero, with $x_1=-2, x_2=0, x_3=2, x_4=-1,$ and $x_5=1,$ so that $\lVert \mathbf x\rVert^2=10.$ When $n$ is odd and exceeds $5$, the equal-spacing strategy fails to beat zero-hugging with an outlier. Thus, for equal spacing, with $n\;(\geqslant 5)$ odd, we may take $x_1=(1-n)/2$ and $x_{j+1}\equiv x_j+2$ (mod $n$) for $j=1,...,n$ with $x_j\in\{(1-n)/2,...,(n-1)/2,$ and then the formula $\sum_{i=1}^ri^2=r(r+1)(2r+1)/6,$ where $r=(n-1)/2,$ gives
$$\nu(\mathbf x) = \sqrt{\dfrac{(n-1)n(n+1)}{48}},$$
and it is clear that
$$\sqrt{\dfrac{(n-1)(n+9)}{4n}} < \sqrt{\dfrac{(n-1)n(n+1)}{48}}$$ 
when $n\geqslant 7$.
In summary, $\nu(\bf  x)$ minimizes to $\sqrt n/2$ when $n$ is even, to $\sqrt 5/\sqrt 2$ when $n=5$, and to $\sqrt{(n-1)(n+9)/4n}$ when $n$ is odd and not $5.$
