For Euclidean vectors $x_1,\ldots,x_n$ with $\|x_i-x_j\|\geq 2$ for $i\neq j$, show $\sup_{i\in\{1,\ldots,n\} }\|x_i\|\geq \sqrt{2(n-1)/n}$ Can you please help me to prove this inequality?

Let $E$ be an euclidean space and $x_1,...,x_n$ a vectors of $E$ such that:$$\forall i,j\in\{1,...,n\},~~i\neq j\Rightarrow\|x_{i}-x_{j}\|\geq 2$$
  Let $R=\sup_{i\in\{1,...,n\} }\|x_i\|$. Show that $R\geq \sqrt{\dfrac{2(n-1)}{n}}$.

Thank you.
 A: Let $u=\sum_{i=1}^n x_i$. We know that $\langle u,u\rangle \geq 0,$ which means
$$
\left\langle \sum_{i=1}^n x_i , \sum_{j=1}^n x_j\right\rangle 
=\sum_{i=1}^n \sum_{j=1}^n \left\langle x_i , x_j\right\rangle 
=\sum_{i=1}^n \| x_i\|^2 + 2\sum_{i<j} \langle x_i,x_j\rangle
\geq 0
$$
We have $\|x_i-x_j\| \geq 2$, which means $\|x_i\|^2-2\langle x_i,x_j\rangle +\|x_j\|^2 \geq 4.$ If we take the sum of this inequality over all $\frac{n(n-1)}{2}$ pairs $(i,j)$ with $i<j$, we get
$$
(n-1)\sum_{i=1}^n \|x_i\|^2 - 2 \sum_{i<j}\langle x_i,x_j\rangle
 \geq 4\;\frac {n(n-1)}{2}
$$
or
$$
n\sum_{i=1}^n \|x_i\|^2 - \left(\sum_{i=1}^n \|x_i\|^2 + 2 \sum_{i<j}\langle x_i,x_j\rangle\right) \geq 2n(n-1)
$$
The term in the big parentheses is $\geq 0$. Therefore
$$
n\sum_{i=1}^n \|x_i\|^2  \geq 2n(n-1)
$$
or
$$
\sum_{i=1}^n \|x_i\|^2  \geq 2(n-1)
$$
or 
$$
nR^2 \geq 2(n-1)
$$
which means
$$
R \geq \sqrt{\frac{2(n-1)}{n}}
$$
Note that the equality is attained iff the points form the vertices of a regular simplex in $\mathbb{R}^{n-1}$ with the center in the origin.
