chromatic number of a graph versus its complement What can be said about the rate of growth of $f(n)$, defined by 
$$f(n) = \min_{|V(G)|=n} \left[ \chi(G) + \chi(\bar{G}) \right],$$ where the minimum is taken over all graphs $G$ on $n$ vertices.
Two observations.
(1)  Either $G$ or $\bar{G}$ contains a clique on roughly $\log{n}$ vertices by Ramsey theory, so $f(n) \ge c_1 \log n $ for some constant $c_1 > 0$.
(2)  If $G = G(n,1/2)$ is a random graph, then $\chi(G) \approx \chi(\bar{G}) \approx n / \log{n}$ almost surely, so we also have $f(n) \le c_2 \,  n/ \log n$ for some constant $c_2 > 0$.
These bounds seem hopelessly far apart.

Can we improve on the bounds $$ c_1 \log n \le f(n) \le c_2 \, n / \log n$$ for all sufficiently large $n$?

 A: Although there is already an accepted answer, I answer to give a bit more information, for what I have to say would not fit in a comment.
This is the Nordhaus-Gaddum Theorem:

If $G$ is a graph of order $n$, then
  
  
*
  
*$2 \sqrt{n} \leq \chi(G) + \chi(\bar{G}) \leq n+1$
  
*$n \leq \chi(G) \cdot \chi(\bar{G}) \leq \left(\frac{n+1}{2}\right)^2$
  

And, there is no possible improvement of any of these bounds.  In fact, much more can be said:

Let $n$ be a positive integer.  For every two positive integers $a$
  and $b$ such that
  
  
*
  
*$2 \sqrt{n} \leq a + b \leq n+1$
  
*$n \leq a b \leq \left(\frac{n+1}{2}\right)^2$
  
  
  there is a graph $G$ of order $n$ such that $\chi(G) = a$ and
  $\chi(\bar{G}) = b$.

Source: Graphs & Digraphs (Fifth edition) by Chartrand, Lesniak, and Zhang
A: (Answering for posterity, don't select this as best.)
First, extend @ccc's proof to show that if $pq\geq n$ then $f(n)\leq p+q$.  You do this first by taking a graph on $n$ nodes consisting of (at most) $p$ cliques of at most $q$ nodes each, and noting that $\chi(G)\leq p$ and $\chi(\bar G)\leq q$.
Now, as shown by ccc in coments above,
$$\lceil 2\sqrt{n} \rceil \leq f(n) \leq 2 \lceil \sqrt{n}\rceil$$
This reduces to at most two values.
In general, if $\lceil 2x \rceil < 2\lceil x \rceil$, then the fractional part of $x$ is in $(0,\frac{1}{2})$.  So we only need to consider the cases where the fractional part of $\sqrt n$ is in this range, that is, $\sqrt n =p + \delta$ where $p$ is an integer and $0<\delta<\frac{1}{2}$.
Claim: $p(p+1)\geq n$.
We know that $p(p+1)+\frac{1}{4}=(p+\frac{1}{2})^2>(p+\delta)^2=n$.  Since $n$ and $p(p+1)$ are integers, this means $p(p+1)\geq n$.
Therefore, $f(n)\leq 2p+1 = \lceil 2\sqrt{n}\rceil$ 
So we know that $f(n)$ is always $\lceil 2\sqrt{n}\rceil$
