The proof is similar to Dirac theorem In an $n$-vertex graph in which each vertex has degree at least $\frac {n}{2}$ must have a Hamiltonian cycle.
So, we can say that If a graph contain Hamiltonian cycle, it will surely contain a Hamiltonian Path.
But the converse of this is not true.
Here, consider a graph with $4$ vertices and $6$ edges which is $K4$ and the degree of each vertex is $3$ $\left(\text{i.e} > \frac{n}{2}\right).$
So, the graph contains $a \ b \ c \ d$ one path.
$b \ c \ d \ a$ is another and even more paths exist.
And it even contains a Hamiltonian cycle.