Remove the maximum-degree restriction from the total 2-coalition bound
Prove that every graph G with minimum degree \(\delta(G)\ge 2\) and maximum degree \(\Delta(G)\) satisfies \(\mathrm{TC}_2(G)\le \left\lfloor\frac{\delta(G)}{2}\right\rfloor\bigl(\Delta(G)-2\left\lfloor\frac{\delta(G)}{2}\right\rfloor+1\bigr)+\left\lceil\frac{\delta(G)}{2}\right\rceil\), without assuming the restriction \(\Delta(G)\ge 4\left\lfloor\frac{\delta(G)}{2}\right\rfloor-2\).
References
It is possible that Theorem~\ref{thm:upperbound_delta} is true even if the restriction $\Delta(G)\ge 4\lfloor \delta(G)/2\rfloor-2$ is omitted. In spite of extensive investigations, we could not find a counterexample to that statement. In addition, we base our suspicion that the restriction can be omitted because this is true in the case when $\delta(G)\le 5$, which follows from Theorem~\ref{Combination}. Based on the above discussion, we propose the following:
If $G$ is a graph with $\delta=\delta(G)\ge 2$ and $\Delta=\Delta(G)$, then $$\TC_{2}(G)\leq \left\lfloor\frac{\delta}{2}\right\rfloor(\Delta-2\left\lfloor \frac{\delta}{2}\right\rfloor+1)+\left\lceil\frac{\delta}{2}\right\rceil.$$