Dirac's theorem says that
IF you have $\delta(G)\geq \frac{n}{2}$ THEN the graph is hamiltonian.
$P\implies Q$ is not the same thing as $\neg P\implies \neg Q$.
Just because $\delta(G)<\frac{n}{2}$ it is not implied that the graph is not hamiltonian.
Take for trivial counterexample the graph $C_{100}$, the cycle of length 100. This graph is clearly hamiltonian since the graph itself is a hamiltonian cycle, yet the degree of every vertex is $2$ which is much less than $\frac{100}{2}=50$.
The information you have given us so far is not enough to confirm whether the graph does or does not have a hamiltonian cycle.
I tried to give as much details as possible :
Suppose that $G$ is hamilton-connected, i.e. for any pair of vertices $u$, $v$, there exists an Hamiltonian path from $u$ to $v$.
Suppose that in $G$, there exists a vertex $v$ of degree only 2, and call its neighbours $x$ and $y$.
If $G$ is hamiltonian, then it must exist an hamiltonian path from $x$ to $y$. In order to be hamiltonian, this path must include $v$ (as well as all other vertices of G, once only).
In the path $(x\ldots y$), suppose that $v$ is not in second position, i.e. the path is $(x\ldots av\ldots y)$. But then this means that $a$ is connected to $v$. $a$ can be only equal to $x$ or $y$ (the only neighbours of $v$). $a$ cannot by $y$ as an Hamiltonian path includes each vertex only once (and this one also ends by $y$). But $a$ cannot be $x$ too. This is a contradiction. Hence $v$ is in second position in the Hamiltonian path from $x$ to $y$. Therefore the path must be $(xv\ldots y)$.
But from $v$, the only vertex we can reach is $y$, therefore the path is exactly $(xvy)$
However we have $|V(G)|\geq 4$, therefore this path is not hamiltonian, (it would need to go throught every vertex in order to be hamiltonian). this is a contradiction.
Therefore there exist no hamiltonian path from $x$ to $y$, and then $G$ is not Hamilton-connected
Best Answer
Actually, there is: every graph with minimum degree at least $\frac{n+1}{2}$ is Hamilton-connected!
To see this, let $G$ be such a graph and note that $G$ is Hamilton-connected if and only if for any pair of vertices $u,v$ of $G$, the graph $G' = G + x + ux + vx$ is Hamiltonian, where $x$ is a new vertex. The key ingredient now is the Bondy-Chvátal theorem, which says that if $u',v'$ are non-adjacent vertices in a graph such that the sum of their degrees is at least the number of vertices, then adding the $u'v'$ edge to the graph doesn't change whether it has a Hamiltonian cycle. In our case, $G'$ is a graph on $n+1$ vertices and each vertex in $G'-x$ has degree at least $\frac{n+1}{2}$. This means that we can add an edge between every pair of vertices in this subgraph, so that $G'$ becomes a complete graph plus a vertex of degree two. But such a graph is clearly Hamiltonian, which is what we wanted to show.
In fact, using the same idea the Bondy-Chvátal theorem itself has the following analogue for Hamilton-connected graphs: if $G$ is a graph on $n$ vertices and $u,v$ are non-adjacent vertices with $\deg(u) + \deg(v) \geq n+1$, then $G$ is Hamilton-connected if and only if $G+uv$ is Hamilton-connected.