Let   $H$   be a Hilbert space. Let   $e_1\ \ldots\ e_n$   be vectors such that   $\forall_{k=1\ldots n}\ \|e_k\|\le 1$   for a natural   $n>1$.  I am addressing $(n-1)$-simplex for the sake of avoiding eyesores. Then:

$$ 0\ \le\ (e_1+\ldots +e_n)^2\ \ \le\ \ n\ +\ 2\cdot\sum_{j\ne k}e_j\cdot e_k $$

It follows that there exist   $j\ne k$   such that

$$ e_j\cdot e_k\ \ \ge\ \ \frac{-n}{2\cdot\binom n2}\ =\ \frac1{1-n}$$

and

$$\|e_j-e_k\|\ =\ \sqrt{e_j^2 - 2\cdot e_j\cdot e_k + e_k^2}\ \le\ \sqrt{2\cdot(1+\frac 1{n-1})}\ =\ \sqrt{\frac{2\cdot n}{n-1}}$$

i.e.

$$\|e_j-e_k\|\ \le\ \sqrt{\frac{2\cdot n}{n-1}}$$

**REMARK 0**   We get equality above   $\Leftrightarrow$   $\|e_j\|=\|e_k\|=1$   and   the   $\binom n2$   values   $e_j\cdot e_k$   are all equal one to another, i.e.   all distances   $\|e_j-e_k\|$   are all equal. (Then of course all dot products are equal to   $\frac 1{1-n}$,   and the distances to   $\sqrt{\frac{2\cdot n}{n-1}}$).

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Now let's reverse our point of view. Let &nbsp; $e_1\ldots e_k\in H$ &nbsp; be such that &nbsp; $\|e_j-e_k\|\ge s$ &nbsp; for every &nbsp; $j\ \,k = 1\ldots n$ &nbsp; such that &nbsp; $j\ne k$, &nbsp; where &nbsp; $s>0$ &nbsp; is an arbitrary positive real constant. Furthermore, we may assume that the center of a sphere which has &nbsp; $e_1\ldots e_k$ &nbsp; for its points coincides with the origin of &nbsp; $H$, &nbsp; so that &nbsp; $\|e_1\|=\ldots =\|e_n|=r$ &nbsp; for certain positive real &nbsp; $r$. &nbsp; Then

$$ s\ \ \le\ \ r\cdot \sqrt{\frac{2\cdot n}{n-1}}$$

i.e.

$$ r\ \ \ge\ \ s\cdot\sqrt{\frac{n-1}{2\cdot n}}$$

The equality holds &nbsp; $\Leftrightarrow$ &nbsp; all edges have the same length &nbsp; $\|e_j-e_k\|=s$, &nbsp; and vectors &nbsp; $e_1\ldots e_n$ &nbsp; have to belong to a common $(n-1)$-dimensional linear subspace of &nbsp; $H$ &nbsp; (substitute *linear* by *affine* in the general case).