Suppose $H$ and $G$ are connected of order $n$. Let $c_k(G)$ and $c_k(H)$ denote the number of closed walks of length $k$ in $G$ and $H$, respectively. Let $m(G)$ and $m(H)$ are the number of edges in $G$ and $H$, respectively. I was trying to prove the following:
If $m(H)<m(G)$, then $$c_k(H)\leq c_k(G)$$ for all $k\geq 2$. Perhaps it's not true.
It's obviously true if the edge set $H$ is the subset of the edge set of $G$, since any walk in $H$ is a walk in $G$ (assuming the same vertex set).
The simplest counterexamples involve bipartite graphs: they don't have closed walks of odd length. So a 5-cycle has a closed walk of length 5, but any bipartite graph on 5 vertices does not.
From left. A bipartite graph on $5$ vertices with $6$ edges $-$ this has no closed walks of length $5$.
From right. A $5$-cycle (with $5$ edges) $-$ this has closed walks of length $5$.