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!



Relationship between triangle free graphs and their minimum degree
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!


