TITLE: Graph with cycles of every length
QUESTION [6 upvotes]: Prove that in a graph on $n$ vertices with min degree more than $n \over 2$, there are cycles of every length between 3 and $n$ included.
I know that from Dirac's theorem, $G$ has a Hamiltonian cycle, i.e a cycle of length $n$. I'm trying to show that the cycle has enough chords so that there are cycles of every length, but I'm having trouble doing that exactly.

REPLY [4 votes]: Let $v$ be a vertex of $G$. Let $H=G-v$.
Then $\delta(H)\geq\frac{n-1}2$ and since $H$ has only $n-1$ elements, $H$ is Hamiltonian,
which means that $G$ has an $n-1$-cycle $C$.
Now $v$ has more than $\frac n2$ neighbours, all on $C$, so at least two of them are adjacent,
which gives us a triangle and a Hamiltonian cycle.
Let $v_1,\ldots,v_{n-1}$ be the vertices of $C$ and assume there is no cycle of length $t$ for some $3<t<n-1$.
If edge $vv_i$ exists, then edge $vv_{i+t-2}$ (indices $\pmod{n-1}$) cannot exist, since that would produce
$t$-cycle $vv_i\ldots v_{i+t-2}v$. But this means that $v$ can have degree at most $\frac{n-1}2$.
This contradiction proves the claim.