Length of a cycle is $3k$ if every cycle has length of $\geq 5$ Let $G$ be a simple graph G such that every vertex has degree at least $k\geq 3$ and every cycle of $G$ has length at least $5$. Show that $G$ contains a cycle of length at least $3k$. 
Let $P$ be a path of maximum length. By contradiction, assume that the longest cycle has length at most $3k − 1$. I want to show that $G$ has a cycle of length $4$. I tried to mimic the method in  Diestel Proposition 1.3.1 to get the proof, but it didn't work.
Any hint would be highly appreciated.
 A: As saulspatz commented, you can use Diestel $1.3.1$ and immediately obtain a cycle of length $3k-1$. But here's how to get $3k$. I talked this out with someone, and we may have arrived at a proof.
Let $P$ be a maximum path in $G$, let $v$ be an endpoint of $P$, and let $u$ be the vertex adjacent to $v$ previously on the path. Since $\deg(v) \ge 3$, $v$ has at least $2$ additional vertices on the path. Since a cycle in $G$ has length at least $5$, the first neighbor of $v$ must be at least $4$ edges away from $v$ on $P$, and every subsequent neighbor must be at least $3$ edges apart from the other neighbors (Otherwise we'd get a cycle of length $4$ or $3$.) So let's pick our neighbors in a minimum fashion, that is, in a way that we are not guaranteed a cycle of length $3k$ immediately.

So far we've got the existence of a cycle of length $3k-1$, but maybe we can increase the length of this cycle. Let's consider $2$ cases in the following way.
Since $\delta(G) \ge 3$, $u$ has at least $1$ more neighbor, $u'$.
Case $1$: $u'$ is not on $P$.

We can think of $u'$ as the endpoint of $P$, and $u'$ must have a neighbor further down the path than all the neighbors of $v$ (otherwise we obtain a cycle of length less than $5$), extending our cycle of length $3k-1$ to $3k$. (This uses the fact that $\deg(u')\ge 3$ and the neighbors of $u'$ lie on $P$).
Case $2$: All of the neighbors of $u$ are on $P$.
Notice $u$ cannot have any neighbors that are adjacent to a neighbor of $v$, otherwise we get a cycle of length less than $5$. So $u$ has a neighbor $u'$ farther on $P$ than the farthest neighbor $v'$ of $v$, and so our cycle is $u \to u' \to v' \to v \to u$, which has length greater than $3k-1$.

