# Proving a graph with $11$ vertices and $53$ edges is Hamiltonian

I have a graph with $11$ vertices and $53$ edges and I'm trying to prove it is Hamiltonian. I know that a graph is Hamiltonian if $n \geq 3$ and $d(v) \geq \frac{n}{2}$.

I'm just having trouble proving that the smallest possible degree of a vertex in a graph with $11$ vertices and $53$ edges is greater than $5.5$.

Note that $K_{11}$ has $\binom{11}{2} = 55$ edges. So you remove two edges from a single vertex. In this case, $\delta(G) = 8 > 5.5$, where $\delta(G)$ denotes the minimum degree of $G$. It follows that $\delta(G) \geq 8$ in the general case.
Hint: how many edges does $K_{11}$ have?