## If $G$ is 2 - self centered graph. then how to prove that $G$ has at least $2n - 5$ edges? where $n\geq 5$ [closed]

If $G$ is 2 - self centered graph. then how to prove that $G$ has at least $2n - 5$ edges? where $n\geq 5$.

I started by assuming if number of edges $\mid E\mid\leq 2n-6$ then there exist a vertex say $u$ such that $deg = 2$ otherwise if no such vertex exists then

$\mid E\mid\geq \frac{3n}{2}>2n-5$ (I am stucked here. How to prove this. Sincerely thanks for giving me time.)

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Is there a research angle here? If not, you're in the wrong place. Try math.stackexchange.com, but you had better include a definition of "2-self centered" when you do. – Gerry Myerson May 15 2012 at 5:45
Also posted to m.se, with no indication at either site that it has been posted to both. – Gerry Myerson May 15 2012 at 12:20
Has a certain homework aroma to it... – Igor Rivin May 15 2012 at 13:51
It would also help to clean up your quantifiers. It reads like you asking for a single graph $G$ with the property that for every integer $n \ge 5$, the graph $G$ has $\ge 2n-5$ edges. – Lee Mosher May 15 2012 at 20:00
(continued) in which case I would answer: ...---•---•---•---•---•---•---... – Lee Mosher May 15 2012 at 20:01