TITLE: If $G$ is biconnected and $\delta(G) \geq 3 \Rightarrow \exists v: G-v$ is also biconnected.
QUESTION [18 upvotes]: If $G$ is biconnected and $\delta(G) \geq 3 \Rightarrow \exists v: G-v$ is also biconnected.
Where $\delta (G) - $ minimum degree of all vertices, $G-v$ is equal to if we remove this vertex from $G$
Give some clue please!
Thanks!

REPLY [6 votes]: Unfortunately, the most I can do is provide a reference for the result. 
A much stronger statement than the one in question is proven here in the paper Critically $n$-connected graphs by Gary Chartrand, Agnis Kaugars and Don R. Lick. 
Specifically they define a critically $n$-connected graph to be an $n$-connected graph for which the removal of any vertex will reduce the connectivity by $1$. They prove that any such graphs must have a vertex of degree less than $\frac{3n-1}{2}$ (and that this bound cannot be improved).
They do provide a reference to a paper [A theorem on on the removal of vertices from blocks by A. Kaugars] which explicitly proves the case for $n=2$ which is what we're interested in. Unfortunately, it does not seem easy to get access to said paper (unless somebody here happens to attend the Kalamazoo College).