I was going through a graph theory book and found this
Let $G$ be a simple undirected graph. $\kappa(G) <\frac{n}{2}$ . If $G$ is connected, then $G$ must contain a simple path of length at least $2\kappa(G)$
Where $\kappa(G)$ is defined to be $G$'s vertex-connectivity.
and I wasn't able to prove this formally .
Suppose to the contrary that all the simple paths are of length $<2K(G)<n$. Let $P$ be the longest path in the graph, then the ends of $P$ (say $a$ and $b$) doesn't have any neighbors outside $P$, and since $l(P)\le n-1$ then there is a vertex $v$ such that $v\not\in P$.
Also $d(v)\ge K(G)$ and all the neighbors of $v$ are in $P$ (but not $a$ and not $b$). So the neighbors of $v$ in $P$ are more than half the number of vertices in $P$ (i.e. $d(v)>\frac{v(P)}{2}$).
So you can find $2$ consecutive vertices in $P$ that are neighbors to $v$, and thus you can find a path larger than $P$, a contradiction.