Prove or disprove that if $G$ and $H$ are two edge-disjoint graphs on the same vertex set $V$, then $\chi(G\cup H) \le \chi(G)+ \chi(H)$ If $G$ and $H$ are two edge-disjoint graphs on the same vertex set $V$, then $\chi(G\cup H) \le \chi(G)+ \chi(H)$. (Here $G\cup H$ is the graph with vertex set $V$ and edge set $E(G)\cup E(H)$.)

I feel like this is true as i don't see why the chromatic number of $G\cup H$ could get so large that the statement is false, but not sure how i would go about proving this. Any help is appreciated.
Counterexample. Let $G$ be a complete bipartite graph $K_{n,n}$ and let $H=\overline G,$ the complement of $G.$ Then $\chi(G)=2$ and $\chi(H)=n$ but $\chi(G\cup H)=\chi(K_{2n})=2n\gt2+n=\chi(G)+\chi(H)$ if $n\gt2.$

The correct inequality is $\chi(G\cup H)\le\chi(G)\chi(H).$
A colouring of $G$ using colours $c_1,\ldots, c_m$ and a colouring of $H$ using colours $c_1', \ldots, c_n'$ gives us a colouring of $G\cup H$ using $c_1,\ldots, c_m,c_1',\ldots, c_n'$.