TITLE: the Nordhaus-Gaddum problems for chromatic number of graph and its complement
QUESTION [10 upvotes]: Is there any relation between the chromatic number of a graph $G$ and its complement $G'$ that are always true?
I saw these ones: $\chi(G)\chi(G')\geq n$ and $\chi(G)+\chi(G')\geq 2n$,
but I'm not pretty sure about them.

REPLY [5 votes]: $(a)$ Prove that $\chi(G)\cdot \chi(G')\geq n$.
Proof: For every graph $G$ and $G'$ we know that $\chi(G)\geq {n\over \alpha(G)}$ and $\chi(G')\geq \omega(G')=\alpha(G)$ where $\alpha(G)$ and $\omega(G)$ denote the independence number and clique number of $G$. So $\chi(G)\cdot \chi(G')\geq {n\over \alpha(G)}\cdot \alpha(G)=n$. Thus $\chi(G)\cdot \chi(G')\geq n$.
$(b)$ Prove that $\chi(G)+\chi(G')\geq 2\sqrt{n}$.
Proof: Since both $\chi(G)$ and $\chi(G')$ are at least one we know that $(\chi(G)-\chi(G'))^2\geq 0$ and so $\chi(G)^2-2\chi(G)\cdot \chi(G')+\chi(G')^2\geq 0$. Adding $4\chi(G)\cdot \chi(G')$ to both sides of the equation we obtain $\chi(G)^2+2\chi(G)\cdot\chi(G')+\chi(G')^2\geq 4\chi(G)\cdot\chi(G')$ and factoring the left-hand side we now have $(\chi(G)+\chi(G'))^2\geq 4\chi(G)\cdot\chi(G')$. Taking the square root of both sides gives us $\chi(G)+\chi(G')\geq 2\sqrt{\chi(G)\cdot \chi(G')}$ and since $\chi(G)\cdot \chi(G')\geq n$ by the previous result in $(a)$ we can conclude that $\chi(G)+\chi(G')\geq 2\sqrt{n}$.