We have a finite undirected graph $G := (V, E)$ and its complementary graph $\overline G := (\overline V, \overline E)$. How do we show that $\chi (G) \cdot \chi (\overline G) \ge |V|$?
We know that there is a subset V' in $V$ with |V'| \ge \frac{|V|}{\chi (G)} and it also applys to $\overline G$. I started with transforming the given inequality to $\chi(G)$, respectivly $\chi(\overline G)$, so that \chi(G) \ge \frac{|V|}{|V'|} , and \chi(\overline G) \ge \frac{|V|}{|\overline V'|} . And put both together: \chi(G) \cdot \chi(\overline G) \ge \frac{|V|^2}{|V|\cdot|\overline V'|} . But how do I eleminate the |V|\cdot|\overline V'|, or show that \frac{|V|^2}{|V|\cdot |\overline V'|} \ge |V|?