maximum Hamilton cycle and NP-completeness
A decision problem is NP-complete iff it is in NP and it is NP-hard.
The decision version of Max-Hamiltonian cycle would be &quot;given a weighted graph $G$ and a value k, decide whether there exists a Hamiltonian cycle of weight at least $k$ in $G$&quot;.
Your problem is clearly in NP (since a yes-certificate is the sought Hamiltonian cycle). Moreover it is also NP-hard since it is a generalization of the classical Hamiltonian cycle problem on unweighted graphs (simply set all edge weights to $1$ and $k=n$ to obtain an instance of your problem).
We can then conclude that this problem is NP-complete.