TITLE: Average squared distance vs diameter in vertex-transitive graphs
QUESTION [6 upvotes]: Let $X=(V,E)$ be a finite, connected graph on $n$ vertices, endowed with its graph metric $d$. The average squared distance of $X$ is $avg(d^2)=\frac{1}{n(n-1)}\sum_{x,y\in V,x\neq y} d(x,y)^2$; it satisfies the obvious bound $avg(d^2)\leq diam(X)^2$, where $diam(X)$ is the diameter of $X$. 
Now assume that $X$ is vertex-transitive. My intuition is that, in this case, for ``many'' pairs of vertices, the distance is much smaller than the diameter, which should entail an inequality $avg(d^2)\leq \lambda(diam(X))^2$, where $\lambda<1$ is some constant, maybe depending only on the common degree of the vertices. Is this intuition correct? If yes, can $\lambda$ be estimated?
EDIT: Thanks to all for your input. The example of the complete graph is somewhat embarrassing, meaning that the OP was poorly formulated. As Aaron sort of guessed, I'm interested in families of $k$-regular graphs ($k$ fixed) with number of vertices increasing to infinity. So the new question would be: does a bound $avg(d^2)\leq \lambda(diam(X))^2$ hold for $|V|$ large enough? Observe that, for vertex-transitive graphs, the lower bound $avg(d^2)\geq\frac{(diam X)^2}{8}$ holds: see proposition 3.4 in
http://toctest.cs.uchicago.edu/articles/v005a006/v005a006.pdf

REPLY [2 votes]: For Random 3-regular graphs \lambda is asymptotically  1. I suspect that for Ramanujan graphs it will be 1 as well?
Regarding random graphs papers to look  are papers citing    Bollobas and de la Vega, Combinatorica 2
(1982), 125-134.
http://www.stanford.edu/class/msande337/notes/the%20diameter%20of%20random%20regular%20graphs.pdf
which does not contain it but a related result. 
A reference which contains the claim is Remco van der Hofsted's book: see Theorem
10.15 and Theorm 10.16 (in the latest version on his webpage).