Brooks's Theorem - If $G$ is a connected graph other than a complete-graph or an odd cycle, then $$\chi(G)\le\Delta(G)$$
I am not able to think for an counter example for not having this theorem of iff type. So let for a graph $G$ given that $\chi(G)\le\Delta(G)$, cant we say that this graph neither a complete graph nor an odd cycle?
You are correct that this is an "if and only if" theorem; it's just that the direction that isn't stated is fairly trivial.
Essentially, you want to prove that if $G$ is a complete graph or an odd cycle, then $\chi(G) > \Delta(G)$. Well, if $G$ is a complete graph on $n$ vertices, then $\chi(G) = n$ and $\Delta(G) = n-1$. If $G$ is an odd cycle, then $\chi(G) = 3$ and $\Delta(G) = 2$. Simple as that!