TITLE: Graph Path Length Problem
QUESTION [5 upvotes]: Let $G$ be a graph such that $\delta(G) \geq k$.

Prove that $G$ has a path of length at least $k$.
Solution: We know that 
$\delta(G) = \min\lbrace \deg(v) \mid v \in V(G) \rbrace$ 
If $\delta(G) = k$ then there exists some $v \in V(G)$ such that $\deg(v) = k$.  This means all other vertices $u \in V(G)$ have $\deg(u) \geq k$.
Now I know this must be part of the proof. How would I prove that there exists a path of at least $k$?
If $k \geq 2$, prove that $G$ has a cycle of length at least $k+1$.

REPLY [2 votes]: a) Take a maximal path $P=v_0v_1 \ldots v_l$. As $P$ is maximal, all neighbourghs of $v_0$ are in $P$. As $v_0$ has at least $k$ neighbourghs, then $l \geq k$.