Asked this over in math overflow and have refined the question a bit.
I'm working on trying to show this, but can't seem to get a proof methodology worked out. No guarantees that it is true, but counter examples are helpful too so I don't spend too much time on a proof that isn't true.
For an $N$ node graph $G=(V,E)$ define $d(i,j)$ to be the geodesic distance between two graph nodes $i$ and $j$.
Show that for distance-transitive graphs ( $u,v,u',v' \in V$ with $d(u,v) = d(u',v') \implies$ there exists some $f$ in the automorphism group of $G$ s.t. $f(u) = f(u')$ and $f(v) = f(v')$ http://en.wikipedia.org/wiki/Distance-transitive_graph) the following holds for $i,a,b \in V$:
$ d(i,a) > d(i,b) \implies \sum_{j \in V} d(i,j) d(a,j) \leq \sum_{j \in V} d(i,j) d(b,j) $
A few notes:
- I've checked this for a large number of specific examples of distance-transitive graphs and it does hold, though there are distance-regular graphs for which it does not hold ( Tutte 12 cage is one example, see Mathworld )
- My intuition is that you may be able to use properties of the intersection array (http://en.wikipedia.org/wiki/Distance-regular_graph#Intersection_numbers) of distance-transitive graphs for the proof. In particular for distance transitive graphs the first row is weakly increasing and the third row can be shown to be weakly decreasing. That said, this may be entirely on the wrong track.
Thanks for any help.
It happens that this is not true. I was able to find a large number of counter-examples. I did this by searching all known distance-transitive graphs with degree <=13. About 80% had the property hold, but the remainder it did not. In particular the (5-6) Cage (http://mathworld.wolfram.com/CageGraph.html) is one counter-example.