The average degree of $G = (V, E)$ is $\frac{\sum_{v\in V}d(v)}{|V|}$. Let $G = (V, E)$ be a graph with average degree $d$ and without isolated vertices.
Must there be a vertex $v ∈ V$ so that the average degree of the neighbours of $v$ is at most $d$?
I know this is a classic problem in graph theory, but I just can't seem to solve it. I already showed that if we replace "at most" with "at least" then the answer is "Yes". I also know, that the answer to this question must be "No".
If such vertex doesn't exist, then $\forall v \in V$ we have that $$\sum_{w \in \Gamma(v)}d(w)>d\cdot d(v)$$
Summing up all those inequalities gives
$$|V|\sum_{v \in V}d^2(v)>(\sum_{v \in V}d(v))^2$$
Hence by Cauchy-Schwarz we need not all of the degrees to be equal but I don't think this helps
I also tried to construct counterexamples using some form of symmetry by isolating $2$ vertices and trying to join all of the others to them, but to no avail.
Any suggestion would be appreciated!
If the answer is no, then a single counterexample suffices.
Here is one such graph (source):
Here, there are three vertices of degree $1$ and three vertices of degree $3$; the average degree is $2$. For the vertices in the middle, the average degree of their neighbors is $\frac{1+3+3}{3} = \frac73 > 2$, and for the vertices on the outside, the average degree of their neighbors is $3$.