# Prove that if $G$ is a graph of order $n \geq 3$ such that $deg$ $v \geq \frac{n}{2}$ for every vertex $v$ of $G$, then $G$ is nonseparable

I know a nonseparable graph is a connected graph with simply no cut vertices. And that a graph of order at least $3$ is nonseperable if and only if every two vertices lie on a common cycle.

I'm not sure how to advance from here though

Suppose $v$ is a cut vertex. Consider the smallest component $A$ of $G-v$. Each $a \in A$ has at most $(|A|-1)+1=|A| < \frac{n}{2}$ neighbours in $G$, contradiction.
• Is the contradiction that $|A| < \frac{n}{2}$ when originally we assumed that $deg$ $v$ $\geq \frac{n}{2}$? – atherton Mar 6 '14 at 21:03
• The contradiction is that $\deg a < n/2$ when we assumed that $\deg v \geq n/2$ for every vertex. – user133281 Mar 6 '14 at 21:05