TITLE: Chromatic number of union of graphs
QUESTION [5 upvotes]: I have two graphs, $G$ and $G'$ on the same vertex set $V$. I have already disproven that $\chi (G\cup G')\leq \chi G + \chi G'$.
Now I have to prove or disprove that $\chi (G\cup G')\leq \chi G \cdot \chi G'$. I think that it is true, but I don't know how to start proving this. Can someone help me? Thanks!

REPLY [2 votes]: Provided that @Ben's answer is better (i.e. it gives more insight of what is going on, at least in my opinion), you can also prove your claim as a corollary of your previous result.
Suppose that $\chi(G)=1$, then $G$ necessarily has no edges. Hence, $\chi(G\cup G')=\chi(G')=\chi(G)\chi(G')$. By symmetry, we can then assume that both $\chi(G)$ and $\chi(G')$ are larger than $1$. But under these assumptions, $\chi(G)\chi(G')\geq\chi(G)+\chi(G')$, and hence the claim follows.