TITLE: Relationship between triangle free graphs and their minimum degree
QUESTION [5 upvotes]: Let $T$ be a triangle-free graph on $n$ vertices with minimum degree $\delta$ (which can be $0$). How does one show that $n >2\delta -1$? It seems to be true for bipartite graphs, but I cannot see how to prove it for general triangle free graphs in general. The motivation behind this question is if this is solved, I will have a much more interesting corollary to garner from it. Thank you!

REPLY [5 votes]: In addition to the literature mentioned in the other answers, one can try some
arguments based on counting.
Let $u$ and $v$ be two of $n$ vertices in a triangle-free graph, and further assume they
are distinct and connected by an edge, with degrees $c$ and $d$ and $c$ at most $d$.
Then their neighborhoods are disjoint, so $c - 1 + d - 1$ is at most $n-2$, giving
$2\delta \leq n$.
It might be fun to recreate the $2n/5$ result: here is a start.  Take an odd
cycle from a non bipartite graph with minimal cycle length.  If the cycle
length is $7$ or greater, show that three vertices will produce either a triangle,
a shorter odd length cycle, or a degree at most $2n/5$.  Try a similar analysis
with a $5$-cycle. Enjoy!