Simplex with edges of length at least s having smallest circumradius

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.

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Wow! Two hours and still no answer?! I better play MO's savior :-) – Włodzimierz Holsztyński May 15 '13 at 3:08

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).

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Thank you very much for your answer! This is a bit embarrassing, but can I ask you to supply a bit more detail? In particular, where does the circumradius come into play? – user21277 May 15 '13 at 4:15
You ask for the shortest radius when the edges are long. I look above for the longest minimum of edges when the radius is short. It leads to your question. It's a standard duality for such situations. (And you can always move your simplex to position the center of the sphere at the origin of the space; I'll expand the text of my "Answer" a little later; first let my eyes rest from the $\LaTeX$ struggles and battles from other MO questions. – Włodzimierz Holsztyński May 15 '13 at 5:51
Edges are intervals $(e_j\ e_k)$ for $j\ne k$. – Włodzimierz Holsztyński May 15 '13 at 5:53
Thanks a lot! It's a very clean solution. I was trying to use the Cayley-Menger determinant and was having no success. It's surprising to me that the solution avoids such complications. – user21277 May 15 '13 at 6:48
Otherwise (if it were complex) I would not be able to provide it :-). You're most welcome. BTW. I introduced this method (most likely very well known?) for some other geometric results in the past. – Włodzimierz Holsztyński May 15 '13 at 6:57