TITLE: Upper bound for chromatic number of graphs with $\omega(G)\leq\lfloor\frac{\Delta(G)+1}{2}\rfloor+1$
QUESTION [7 upvotes]: 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!

REPLY [2 votes]: i can't comment yet, but the answer of Ztas Nellets is correct, this conjecture is false, the probabilistic construction in my linked paper with Cranston is from Reed's discussion of variants and tightness of his conjecture.
In regards to the paper of Kostochka, i proved a generalization here https://doi.org/10.1002/jgt.21634
which also implies the slightly stronger result Peter Heinig mentions. The proof is essentially the same as Kostochka's original in the paper nobody can find (i had a hardcopy at some point, but cannot find it).