Show that if $k \in \mathbb{N}$ and $\delta(G) \geq 2k+1$ then $G$ contains a cycle which has even length and length at least $k+2$. Question:

Show that if $k \in \mathbb{N}$ and $\delta(G) \geq 2k+1$ then $G$ contains a cycle which has even length and length at least $k+2$


My try: 
I've managed to show that any graph $G$ with $\delta(G) \geq 3$ has an even cycle.
So, I know that part is true but showing the length of it is stumping me.
 A: I'm showing the improved (or correct?) version of the problem, with $k + 2$ reblaced by $2k + 2$.
Consider any maximal path in $G$, that is, any sequence of vertices $v_1, \: v_2, \: \dots, \: v_n$ such that:


*

*there is an edge connecting $v_i$ and $v_{i + 1}$ for $i = 1, \: \dots, \: n - 1$;

*there is no vertex $u \ne v_i$ connected either to $v_1$ or $v_n$ (i.e. the path cannot be extended).


Now we focus on the neighbourhood of $v_1$, say $N(v_1)$. According to the second above condition, every vertex in $N(v_1)$ must be one of the $v_i$'s; and, of course, $v_2 \in N(v_1)$. Thus there exists a subset $S \subseteq \{3, \: 4, \: \dots, \: n\}$ such that $N(v_1) = \{v_2\} \cup \{v_i: \: i \in S\}$.
Assume, for the sake of contradiction, that $G$ contains no even cycles of length $\ge 2k + 2$. We will henceforth try to show that this is indeed absurd. Let $4 \le a_1 < a_2 < \cdots < a_s$ be the (possibly empty) sequence of even elements in $S$, and likewise $3 \le b_1 < b_2 < \cdots < b_t$ the (possibly empty) sequence of odd elements in $S$.
Observe that, for each $i = 1, \: \dots, \: s$, the path $v_1 \to v_2 \to \dots \to v_{a_i} \to v_1$ is an even cycle of length $a_i$. Since we assumed that every even cycle has at most length $2k$, we conclude that $s$ is at most $2k$, and therefore $\displaystyle s \le \frac{2k - 4}{2} + 1 = k - 1$.
As for the odd-indexed vertices, a slightly different reasoning is required. For any $1 \le i < j \le t$, the path $v_1 \to v_{b_i} \to v_{b_i + 1} \to \dots \to v_{b_j} \to v_1$ is an even cycle of length $b_j - b_i + 2$ and, by the same argument we used before, this should not exceed $2k$. Thus $b_t - b_1 + 2 \le 2k$, which implies $\displaystyle t \le \frac{b_t - b_1}{2} + 1 \le \frac{2k - 2}{2} + 1 = k$.
Finally, we have $$\deg(v_1) = \lvert N(v_1)\rvert = 1 + s + t \le 1 + (k - 1) + k = 2k$$ and we are done because this contradicts $\delta(G) \ge 2k + 1$.
