# Thread: graph: average of average degree of neighbours, is at least the average degree

1. ## graph: average of average degree of neighbours, is at least the average degree

Hello,

thanks for letting me become a member of this forum.
I would like to ask my very first question here

Let G be a simple graph, with loops, multiple edges and without weighted edges.

For every vertex $v_i$, let $d_i$ denote its degree (number of neighbours), and let $w_i$ denote the average of the degrees of neighbours of $v_i$

Now one should prove that the average $D$ of the $d_i$ is at most the average $W$ of the $w_i$, with equality if and only if the graph is regular (all $d_i$ are equal).

At first, I thought that this was simply related to the friendship paradox, where one considers

$\frac{\sum d_i}{n}\leq X=\frac{ \sum d_i^2}{\sum d_i},$
which follows essentially from Cauchy-Schwarz.

However, I noticed that the above is not quite the same.
For instance, if you take a star graph on 4 vertices, then the average degree $D$ is 1.5, while the number $W$ is 2.5.
The number $X$ from the above is 2, though.

I found a discussion here, where a graph on four vertices is considered, here $D= 2, W=2.4167$ but $X=2.25$
Whenever I tried to prove that $D \leq W$, I started struggling with a $\frac{1}{d_i}$ somewhere that I can't get rid off.