The question is to prove that if graph G has at least one cycle and that the girth of G (Girth= length of shortest cycle) is at least 2k, then the diameter of G is at least k.
My attempt:
I tried to show both cases if we have even cycle and odd cycle. Let C be the cycle with girth at least 2k. If C is even, then maximum distance between 2 random vertices on C is (2K/2). If C is odd, then the maximum distance between two random vertices on C is at least (2k/2)+1.
Therefore, proved. Any suggestion or modification?
Let $C$ be a smallest-length cycle of length $m$ and let $x$ and $y$ be two points furthest from each other on $C$. So the distance between $x$ and $y$ in $C$ is $\lfloor \frac{m}{2} \rfloor$, which, as $m$ is at least $2k$, is at least $k$. So let $P_1$ be a path from $x$ to $y$ on $C$ of length $\lfloor \frac{m}{2} \rfloor$.
Now suppose the diameter of $C$ is less than $\lfloor \frac{m}{2} \rfloor$. Then there is a path $P_2$ of length $l$ $< \lfloor \frac{m}{2} \rfloor$ from $x$ to $y$. Then the graph $P_1 \cup P_2$ [a vertex $v$ is in $P_1 \cup P_2$ iff $v$ is on either $P_1$ or $P_2$ and an edge is in $P_1 \cup P_2$ iff $e$ is in either $P_1$ or $P_2$] has no more than $(\lfloor \frac{m}{2} \rfloor +1)+l-1$ vertices [make sure you see why], and has at least one cycle. [Indeed, as $P_1$ is strictly longer than $P_2$, there is at least one vertex $x'$ in $P_1$ but not on $P_2$. So writing $P=x_1x_2\ldots x_{\lfloor m/2\rfloor}$ let $i$ and $j$ be integers $i+1<j$ such that $x_i$ and $x_j$ is in $P_2$ but $x_{i'} $ is not for $i' \in \{i+1,\ldots , j-1\}$. Next let $P'_2$ be the subpath of $P_2$ with endpoints $x_i$ and $x_j$. Then $P'2$ intersects $x_ix_{i+1}\ldots x_j$ at precisely $x_i$ and $x_j$, and so $P'_2 \cup x_ix_{i+1}\ldots x_j$ is a cycle.]
This however, implies that there is a cycle of length no more than the number of vertices in $P_1 \cup P_2$ which is no more than $(\lfloor \frac{m}{2} \rfloor +1)+l-1$ $> m$, which is a contradiction.