Here's a proof, from Wikipedia, that $K_6$ has at least 2 monochromatic triangles. You may find that this method of proof generalizes to give you the lower bound you want.
"An alternate proof works by double counting. It goes as follows: Count the number of ordered triples of vertices $x, y, z$ such that the edge $(xy)$ is red and the edge $(yz)$ is blue. Firstly, any given vertex will be the middle of either $0\times5=0$ (all edges from the vertex are the same color), $1\times4=4$ (four are the same color, one is the other color), or $2\times3=6$ (three are the same color, two are the other color) such triples. Therefore there are at most $6\times6=36$ such triples. Secondly, for any non-monochromatic triangle $(xyz)$, there exist precisely two such triples. Therefore there are at most 18 non-monochromatic triangles. Therefore at least 2 of the 20 triangles in the $K_6$ are monochromatic."
On the other hand, if you want all the work done for you, there is an answer in A W Goodman, On sets of acquaintances and strangers at any party, Amer. Math. Monthly 66 (1959) 778-783, also in Frank Harary, The two-triangle case of the acquaintance graph, Math. Mag. 45 (1972) 130-135.
Assume we have a complete graph on $k(r(k-1)-1)+2$ vertices and $k$ edge colours [I am European]. Choose an arbitray vertex $v$. Divide the other $k(r(k-1)-1)+1$ vertices according to colour that connects them to $v$. At least one of these $k$ sets must have $r(k-1)$ vertices. The vertices in this set are all connected to $v$ by the same colour. That colour cannot be used between the vertices in that set (or else we find a triangle). So the set is internally coloured with $k-1$ colours, and must contain a monochromatic triangle.
This proof is very similar to the classic one that shows that $R(3,3,3) \le 17$, based on $R(3,3)=6$.
Best Answer
Let $K_n$ be 2-colored. A bi-angle is a configuration of two edges with different colors meeting at a vertex. If $n=2m+1$ then the greatest possible number of bi-angles involving any particular vertex is $m^2$, so the total number of bi-angles in the coloring is at most $(2m+1)m^2$. The number of triangles in $K_n$ is $n\choose3$. Each nonmonochromatic triangle uses up 2 bi-angles, so the number of nonmonochromatic triangles is at most $(2m+1)m^2/2$. This gives you a lower bound on the number of monochromatic triangles, in the $n$ odd case. A simple modification handles the $n$ even case.