If $n \geq 6$, $G$ or $G_c$ contains a cycle of length $3$ That is the statement, if the order of $G$ is greater or equal to $6$, $G$ or its complementary contain a cycle of length $3$. 
I don't really know where to start, I have drawn a lot of examples but I cant understand the essence of the proof. 
Any help is welcome!

Here are a few assorted hints.
Hint 1: Is it clear that the $n=6$ case implies the $n>6$ case?
Hint 2: This problem is usually phrased in slightly different language. Let me present it to you, as it may help you think about things in a more clear fashion. 
Consider the complete graph with six vertices: $K_6$. Take a palette containing two colours, red (R) and blue (B), and colour each edge either R or B. Then the two following statements are equivalent:


*

*given a graph $G$ on six vertices, either $G$ or its complement must contain a 3-cycle

*there will always be either a blue triangle or a red triangle, no matter how I choose to colour the edges of my $K_6$.
You may wish to try a few more examples using this new language, you will very quickly find that the theorem is always true.
In order to find a proof, try choosing a fixed vertex in your graph, and thinking about what possibilities there are for colouring the edges incident to your chosen vertex, as well as the consequences of each possible choice.
General comments: The essence of this problem is surprisingly deep, and a whole theory of surprising and (in my opinion) beautiful mathematics called Ramsey Theory has been built upon this simple example. The general thrust of the theory is about studying the emergence of small pockets of order in chaotic settings - the natural generalisation of your problem is to ask whether there exists a number $R(n)$ such that every red-blue edge-colouring of graphs of order greater than $R(n)$ necessarily contains a monochromatic copy of $K_n$. Ramsey Theory tells us: yes!

Explicit proof: We proceed by contradiction, attempting to colour the edges of the complete graph without creating a red triangle or a blue triangle. We may assume without loss of generality that $n=6$ using the observation of hint 1 (if $n > 6$ just consider a subgraph with 6 vertices). 
Pick a vertex $v$. This vertex has 5 incident edges, each of which is coloured either R or B, so we must have either 3 red edges incident to $v$ or 3 blue edges incident to $v$. Without loss of generality, we may assume that we have three red edges incident to $v$. 
Call the three vertices connected to $v$ via a red edge $x$, $y$ and $z$. Then if any of the edges $xy$, $yz$, $xz$ are red, we have found a red triangle ($vxy$, $vyz$, or $vxz$ respectively). So all of these three edges must be blue. But this means we have found a blue triangle $xyz$. 
We conclude that there must always be either a red triangle or a blue triangle, proving the result. 
