Let $C$ be a Hamiltonian cycle of a graph $G$. Call an edge $e$ of $G$ a chord if $e\not\in C$. Let each edge of $C$ be weighted $1$ and each chord be weighted $2$. The weight of a path or cycle of this graph is the sum of the weights of its edges. For example, cycle $C$ has weight $n$, where $n$ is the order of $G$.
Is the following statement true on the graph $G$ under the conditions given above? Or maybe there is a counter-example?
If at least one chord leaves every vertex of the graph $G$, then there exists a cycle in $G$ (which need not be Hamiltonian) whose weight exceeds $n$.
I posted this question on MSE. But it doesn't seem like such a simple question.
What I know is that if the number of chords for each vertex is odd, then it follows from Thomason's theorem $G$ has a Hamiltonian cycle $C'$ different from $C$, hence $C'$ contains chords. It follows that the weight of $C'$ is greater than $n$.
In general, however, a graph can have a single Hamiltonian cycle (such graphs are called uniquely Hamiltonian), even if the degree of each vertex is at least three. The smallest uniquely Hamiltonian graph with minimum vertex degree at least $3$ contains $18$ vertices and is shown in the figure. This graph was constructed by Gordon Royle. The cycle $$ C'=\{1,5,6,\ldots,17,4,3,2,1\} $$ has weight $19$. This graph also has a cycle with a weight of $25$ (which seems to be the maximum possible).
It's worth noting that this very problem is formulated in Tony Huynh's answer to a question the length of cycles in a $2$-connected simple gragh. Here is the very last sentence of his answer "I think it should be easy to show that such a graph has a cycle longer than $C$". The cycle $C$ is the Hamiltonian cycle of the graph and the length is the weight of the cycle in our terminology.
