Is it true that of all $n$-simplices with edge lengths greater than or equal to some parameter $s$, the regular simplex with edge lengths $s$ has the smallest circumradius? It seems obvious, but I haven't been able to prove this.
I would greatly appreciate if someone could supply a proof or reference. Thank you.
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 $\sqrt{\frac{2\cdot n}{n-1}}$).
Now let's reverse our point of view. Let $e_1\ldots e_k\in H$ be such that $\|e_j-e_k\|\ge s$ for every $j\ \,k = 1\ldots n$ such that $j\ne k$, where $s>0$ is an arbitrary positive real constant. Furthermore, we may assume that the center of a sphere which has $e_1\ldots e_k$ for its points coincides with the origin of $H$, so that $\|e_1\|=\ldots =\|e_n|=r$ for certain positive real $r$. 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 $\Leftrightarrow$ all edges have the same length $\|e_j-e_k\|=s$, and vectors $e_1\ldots e_n$ have to belong to a common $(n-1)$-dimensional linear subspace of $H$ (substitute linear by affine in the general case).
