Let $G$ be a simple, connected, undirected graph with more than one vertex. Proof that a that $G$ contains a vertex $v$ such that
$\frac{1}{|\delta(v)|} \displaystyle\sum_{w\in{}N(v)} |\delta(w)| \geq \frac{2|E(G)|}{|V(G)|}$
,where $\delta(v)$ is the degree of a vertex, $N(v)$ is the neighborhood of $v$, $E(G)$ is the number of edges in $G$ and $V(G)$ the number of vertices in $G$.
So far if have shown this to bee true for trees where there are always $V(G)-1$ edges, cycle graphs where there are $V(G)$ edges and I have looked at quite a few examples.
The intuition is that the right side is the average degree for the vertices in $G$ while the left is the average degree of vertices in the neighborhood of $v$. So proof that there is a neighborhood with above average degree.
I have also tried proving it by removing edges from a fully connected graph or adding edges to a tree but it didn't really work out.
It would be really nice if someone could give me a hint.
If you sum the quantity $$ \frac{1}{|\delta(v)|} \displaystyle\sum_{w\in{}N(v)} |\delta(w)| $$ over all vertices $v$, you will get $$ \sum_{vw \in E(G)} \left(\frac{|\delta(w)|}{|\delta(v)|} + \frac{|\delta(v)|}{|\delta(w)|}\right). $$ You can show that this sum exceeds $2|E(G)|$. Therefore its average over all $v$ is at least $\frac{2|E(G)|}{|V(G)|}$, and there must be some vertex $v$ which achieves at least this average.