WOWII Conjecture 143:

For a simple connected graph $G$, $\mathrm{tree}(G) \ge (\mathrm{girth}(G) + 1) / \sigma(G)$, where $\mathrm{tree}(G)$ is the largest induced tree size, $\mathrm{girth}(G)$ is the length of the shortest cycle, and $\sigma(G) = G.\mathrm{secondSmallestDegree}$ is the second-smallest degree of $G$'s degree sequence (per WOWII defEntry 65). We state the inequality in denominator-free form to avoid the $\sigma = 0$ corner case ($n \le 1$).

The proof splits into the acyclic case and two cyclic cases. For an acyclic graph the girth is zero. If $\sigma \ge 2$, deleting two consecutive edges from a shortest cycle leaves an induced tree on $\mathrm{girth}(G)-1$ vertices, and the factor $\sigma \ge 2$ yields the required inequality. If $\sigma = 1$, there are two degree-one vertices. A maximal induced tree containing them must have an external vertex with two attachments; those attachments create a cycle whose length forces the tree to contain at least $\mathrm{girth}(G)+1$ vertices.