This follows from familiar results once everything is untangled.

You wish to show:

In a triangle free graph $2\delta-1 \lt n.$

A stronger result is $2\delta \ \le n$ i.e. $\delta \lt \frac{n}{2}.$ It is sufficient to prove $\delta' \lt \frac{n}{2}$ where $\delta' = \frac{2|E|}{n} $ is the average degree, since clearly $\delta \le \delta'.$

Turans theorem (proof at the link) has as a special case Mantel's theorem:

A triangle free graph on $n$ vertices has at most $\big\lfloor\frac{n^2}{4}\big\rfloor$ edges.

So the cases for a triangle free graph are

• $n$ even , $|E| \le \frac{n^2}{4}$ and $\delta'=\frac{2|E|}{n} \le \frac{n}{2}$

along with

• $n$ odd , $|E| \le \frac{n^2-1}{4}$ and $\delta'=\frac{2|E|}{n} \le \frac{n}{2}-\frac{1}{2n}.$

This follows from familiar results once everything is untangled. Suppose we have a triangle free graph with $n=2m$ or $n=2m+1$ vertices. Then $\delta \le m$ so $2 \delta \le n.$ This follows from the comments below.

You wish to show:

In a triangle free graph $2\delta-1 \lt n.$

Since $\delta$ is an integer this is the same as $2\delta \ \le n$ i.e. $\delta \le \frac{n}{2}.$ It is sufficient to prove $\delta' \le \frac{n}{2}$ where $\delta' = \frac{2|E|}{n} $ is the average degree, since clearly $\delta \le \delta'.$

Turans theorem (proof at the link) has as a special case Mantel's theorem:

A triangle free graph on $n$ vertices has at most $\big\lfloor\frac{n^2}{4}\big\rfloor$ edges.

So the cases for a triangle free graph are

• $n$ even , $|E| \le \frac{n^2}{4}$ and $\delta'=\frac{2|E|}{n} \le \frac{n}{2}$

along with

• $n$ odd , $|E| \le \frac{n^2-1}{4}$ and $\delta'=\frac{2|E|}{n} \le \frac{n}{2}-\frac{1}{2n}.$