Show that a triangled is formed My math prof just gave us this homework. I really tried to solve it but, you know...

Let $n$ be a positive integer such that $n\ge2$. We have $2n$ points in space that are joint by at least $n^2+1$ line segments. Show that at least one triangle is formed.

I would appreciate any help, but if you please, post hints rather than the full answer. Thnx :D
 A: Here is a way of thinking to get started.
Suppose you have a triangle-free graph. If you remove a vertex and the associated edges, it remains triangle-free.
Suppose your original graph had the greatest possible number of edges for a triangle-free graph on that number of vertices. If you can add an edge to the reduced graph (without creating a triangle), and try to rebuild the original, you may have a triangle, but only if the removed point was originally joined to both ends of the edge you added, and you can avoid the triangle by deleting one of the edges involving the removed point. The number of edges remains the same, but you are now working with a maximal triangle-free graph which has fewer vertices than the original.
Now you have to establish a starting point and work out how to add a vertex in the best way.
A: This could be done also with Turan's theorem or by induction. 
Induction step: Say we have $2n + 2$ point and  $(n+1)^2 + 1 = n^2 +2n +
2$ edges. Choose two connected points $A$ and $B$. There remain a set $S$ with $2n$ points. If among them there is $n^2 + 1$ edges then we are done, by induction hypothetis. 
Else we have $d_A + d_B \geq 2n + 1$, where $d_X$ is number of edges between $X\in\{A,B\}$ and $S$. If $A$ and $B$ don't have common neighbor in $S$
then we have $d_A + d_B \leq 2n $ which is in a contradiction with previous inequality.
A: You can note that the complete bipartite graph $K_{n,n}$ has $n^2$ edges. Also that other bipartite graphs on the same number of vertices have fewer.
