For a finite graph $G$ without odd cycles with length at least 5, we have $\chi(G)=\omega(G)$.

Question

Let $G$ be a finite graph. Assume that $G$ and its complement $\bar{G}$ do not contain the odd cycles of length at least 5. Please use the fact that "the complement of a perfect graph is perfect and show that $\chi(G)=\omega(G)$. We recall that $\chi(G)$ is the chromatic number and $\omega(G)$ is the clique number of $G$.