Hello,


Does anyone know if there is a result that relates a quantity such as an average degree to the fact the a (simple and connected graph) has no cut vertices? 


e.g. if a graph has a Hamiltoninan cycle then it has not cut vertex. 
By Ore's theorem, if deg(v) >= n/2 for each vertex v (n is the number of vertices of a graph), then the graph is Hamiltoninan. 
Hence deg(v) >= n/2 implies no cut vertices (non-separable).


I am looking for a result similar to Ore's theorem, but w.r.t. non-separability e.g. something like if the deg(v) >= f(n) for each v (or some statement about the average degree), then the number of cut vertices is zero. 


Thank you.



Non-separable simple connected graphs.
I do not think you can do anything better than $\frac{n}{2}$. Consider the following construction: Take two complete graphs of equal size $\frac{n-1}{2}$. Add one vertex $v$ and connect this vertex to all vertices of the two complete graphs. Each vertex has degree at least $\frac{n-1}{2}$ and this graph has a cut vertex.


