Erdős problem 621

Let $G$ be a graph on $n$ vertices, $\alpha_1(G)$ be the maximum number of edges that contain at most one edge from every triangle, and $\tau_1(G)$ be the minimum number of edges that contain at least one edge from every triangle.

Is it true that$$\alpha_1(G)+\tau_1(G) \leq \frac{n^2}{4}?$$

A problem of Erdős, Gallai, and Tuza [EGT96], who observe that this is probably quite difficult since there are different examples where equality hold: the complete graph, the complete bipartite graph, and the graph obtained from $K_{m,m}$ by adding one vertex joined to every other.

This is true, and was proved by Norin and Sun [NoSu16], who in fact proved that$$\alpha_1(G)+\tau_B(G) \leq \frac{n^2}{4},$$where $\tau_B(G)$ is the minimum number of edges that need to be removed to make the graph bipartite.

Here $\alpha_1(G)$ and $\tau_1(G)$ are taken over subsets of the edge set of $G$, and the inequality is stated multiplied through by $4$ so that it lives in the natural numbers.

The linked file states $\tau_1$ as the least number of edges whose deletion leaves $G$ triangle-free, which is the same as meeting every triangle of $G$, and quantifies over an arbitrary Fintype V rather than Fin n.