Let $\delta$ denote the minimum degree of graph G. Show for every graph G, if G is connected and |V|>2$\delta$, then G has a path of length 2$\delta$.
I started this way:
Let P be the longest path in G. If P has length k but does not use all the vertices, I tried using this Theorem (Let G be a graph of order n$\ge3$. If deg(x) + deg (y)$\ge$ n for all pairs of non-adjacent vertices x,y, then G is Hamiltonian) to find a cycle of length k+1, and use the cycle to find a longer path.
I do not know where to go from here.
Let $P=v_1\ldots v_m$ be a path of maximal length and assume $m<|V|$. Let $S$ be the vertex set of $P$.
Note that the subgraph induced by $S$ cannot contain a cycle of length $m$: there must be a vertex $w$ that is not on this cycle and since $G$ is connected there is a shortest path from $w$ to the cycle. Now cutting the cycle next to the attachment point of this shortest path would give a path that is longer than $P$. Contradiction.
Both $v_1$ and $v_m$ must have all their neighbours on $P$ (or otherwise we can make $P$ longer), so specifically $v_1$ must have at least $\delta$ neighbours in $v_2\ldots v_m$.
Now assume $v_i$ is a neighbour of $v_m$. We claim that $v_{i+1}$ cannot be a neighbour of $v_1$: if it was, we find a cycle $v_1\ldots v_iv_mv_{m-1}\ldots v_{i+1}$, which was disallowed. So for each neighbour of $v_m$ on $P$ we find a non-neighbour of $v_1$ in $v_2\ldots v_m$.
Now we can resume: $v_2\ldots v_m$ contains at least $\delta$ neighbours of $v_1$ and at least $\delta$ non-neighbours of $v_1$. So $P$ must have at least $2\delta+1$ vertices, i.e. length at least $2\delta$.