My question is to find the minimum of the following expression:
$$A(x,y) = \sum\_{1\leq i<j\leq n} |x\_i-x\_j|\ |y\_i-y\_j|,$$
over the set of pairs of real vectors $x=(x\_1,\dots,x\_n),y=(y\_1,\dots,y\_n)$ subject to the following conditions:
$$\sum\_{i=1}^n x\_i^2 =\sum\_{i=1}^n y\_i^2 = 1, \sum\_{i=1}^n x\_i=\sum\_{i=1}^n y\_i=0.$$


It seems that $1$ is less or equal to the minimum, but don't have a proof.



What is the minimum of this quantity on $S^{n-2}\times S^{n-2}$?
The minimum must occur at vectors $x,y$ where
$x\_i$ and $y\_i$ take only two values each.
This should make it easy to check **Neil Strickland**'s
experimental result (where $x$ and $y$ are indeed of this form).
[*EDIT*: see the answer by **Adam P. Goucher** for this check.]


It is convenient to extend $A$ homogeneously to all nonzero $x,y$
in the zero-sum hyperplane:
$$
A(x,y) = \frac1{\| x \| \, \|y\|} \sum\_{1\leq i<j\leq n} |x\_i-x\_j|\ |y\_i-y\_j|
$$
where $\|x\| = (x\_1^2 + \cdots + x\_n^2)^{1/2}$ and likewise $\|y\|$.
So we seek the largest $a$ such that
$$
(1)\qquad\qquad\qquad
\sum\_{1\leq i<j\leq n} |x\_i-x\_j|\ |y\_i-y\_j| \geq a \, \|x\| \, \|y\|
\qquad\qquad\qquad\phantom{(1)}
$$
for all $x,y$ such that $\sum\_{i=1}^n x\_i = \sum\_{i=1}^n y\_i = 0$.


Fix one of the $n!^2$ possible orders of the coordinates of $x$ and $y$.
Such a choice limits each of $x$ and $y$ to a cone. For the order
$x\_1 \geq x\_2 \geq \ldots \geq x\_n$, the cone's extreme rays are
generated by $(1,\ldots,1,1-n)$, $(2,\ldots,2,2-n,2-n)$, etc., all with
$x\_i$ or $y\_j$ taking only two values each. In general we have the image of
these generators under some coordinate permutation, so still with
$x\_i$ or $y\_j$ taking only two values each.


Given that choice of order, $\sum\_{1\leq i<j\leq n} |x\_i-x\_j| |y\_i-y\_j|$
is a bilinear form in $x,y$ because the signs of $x\_i-x\_j$ and $y\_i-y\_j$
are constant. The claim then follows by a convexity argument
(so ultimately by the triangle inequality). Indeed if we fix $x$,
and (1) holds for $(x,y)$ and $(x,y')$ for some $y,y'$ in the same cone,
then it also holds for $(x, ty+(1-t)y')$ for all $t \in [0,1]$.
So it is enough to check (1) for $y$ on an extreme ray of the cone.
Likewise we can fix $y$ and reduce to the case where $x$ is on an extreme ray.
**QED**


