TITLE: Maximal geometric mean of distances between points on an interval
QUESTION [5 upvotes]: Suppose I had T points in the interval $[0,1]$.  Call them $e_1, \dots, e_T$.
Question 1:
What is a good nontrivial bound on the geometric mean of $$\{|e_i - e_j| : 1 \leq i < j \leq T \}, $$ as a function of $T$, independent of our choice of $e_i$?
Question 2:
Suppose we can choose $i$ that minimizes the geometric mean of $$E_i = \{ |e_i-e_j| : 1 \leq j \leq T, j \neq i \}.$$What is a good nontrivial upper bound on the geometric mean of $E_i$ as a function of $T$?
Some context:
Suppose I were given an arithmetic circuit that computes $$f(z) = \sum_{i=1}^t c_tz^{e_t},$$ with bounds $D$ on the degree and $T$ on the number of terms, and we would like to test whether $f$ is zero.  One way to test this for $f$ over an arbitrary field is to compute images $f \bmod (z^p-1)$ for a set comprised of the smallest primes $p$ whose product exceeds $$\prod_{j \neq 1} (e_1 - e_j),$$
though one could replace $e_1$ with any of the $e_i$'s.  A naive upper bound on this product is $D^T$.  An answer to either of the questions above could improve this bound.

REPLY [5 votes]: The extremal $e_i$ for Question 1 are probably well-known: 
they are $0$, $1$, and $(1+r)/2$ where $r$ ranges over the roots of the
Gegenbauer
 polynomial $C_{T-2}^{(3/2)}$.
The product of the $|e_i-e_j|$ is then a power of $2$ times
the square root of $\mathop{\rm disc}\bigl((x^2-1)C_{T-2}^{(3/2)}(x) / c_{T-2}\bigr)$,
where $c_n$ the leading coefficient $(2n+1)!/2^n n!^2$ of $C_n^{(3/2)}$.
This discriminant can in turn be computed from the values of $C_{T-2}^{(3/2)}$
at $\pm 1$ and the discriminant of $C_{T-2}^{(3/2)} / c_{T-2}$,
which is known but somewhat complicated $-$ experimentally it seems to be
$$
\prod_{m=1}^{T-2} \frac{(m+2)^{m-2}m^m}{(2m+1)^{2m-3}}.
$$
To prove that these are the optimal $e_i$: let $P(x) = \prod_{i=1}^T (x-e_i)$ and
suppose $\{e_i\}$ maximizes $\prod_{i<j} |e_i-e_j|$.
Then clearly two of them are at $0$ and $1$.  I claim that
each of the others is a root of $P''$.  Indeed $e_i$ is a local
maximum of $P(x)/(x-e_i)$, so a root of $\frac{d}{dx}(P(x)/(x-e_i))$,
so a triple root of $(x-e_i)P'(x)-P(x)$; but then it's a root of
$\frac{d^2}{dx^2}\bigl((x-e_i)P'(x)-P(x)\bigr)$,
whose value at $x=e_i$ is $2P''(e)$.  It follows that
$P$ is a multiple of $(x^2-x)P''$, and by comparing leading coefficients
we see that the multiple is $T^2-T$.  The resulting differential equation
characterizes $(T^2-T)C_{T-2}^{(3/2)}(2x-1)$ among polynomials up to scaling, so we are done.