Let $G$ be a simple graph such that $\omega(G)\leq\lfloor\frac{\Delta(G)+1}{2}\rfloor+1$ where $\Delta(G)$ is the maximal degree of $G$. Is it true that \begin{equation} \chi(G)\leq \lfloor\frac{\Delta(G)+1}{2}\rfloor+2? \end{equation} A similar problem can be found here, where, ignoring the exceptional case of odd cycles, it is conjectured that

Conjecture (Reed).For any graph $G$ of maximum degree $\Delta(G)$, $\chi(G)$ is at most $\lceil\frac{\Delta(G)+1+\omega(G)}{2}\rceil$.

**UPDATE:** In the case of graphs that satisfy $\omega(G)\leq\lfloor\frac{\Delta(G)+1}{2}\rfloor+1$, the statement in my problem is stronger than Reed's conjecture. For example suppose $\Delta=100$ and $\omega=50$; then by Reed's conjecture we need $76$ colors, while if my conjecture is true, we need just $52$ colors!

convex combinationof its obvious lower bound $\omega(.)\colon\mathsf{Graphs}\to\omega$ and its obvious upper bound $\Delta(.)+1$, since $0\cdot x+1\cdot y$ is a convex combination, (1) (King-)Reed proved $\exists\varepsilon>0$ s.t. $\epsilon\cdot \omega+ (1-\epsilon)\cdot(\Delta+1)$ is a bound for $\chi$ too, (2) Reed's conj. is that $\epsilon:=\frac12$ works (it's open, and a bit 'hopeful'). $\endgroup$ – Peter Heinig Oct 18 '17 at 7:21