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 $$ \frac{|V|^2}{|V|\cdot |\overline V'|} \ge |V|?$$