Red-blue coloring of the complete graph $K_n$​ such that there are more red edges than blue edges, and there is no red triangle.

Question

Let us prove that for every $n>1$ there exists a $2$-coloring of a complete graph $K_n$ red and blue, where there are more red edges than blue and there is no monochromatic red triangle, where all the edges are red.

I thought of the pigeonhole principle, because a triangle is basically a $K_3$

For $K_5$ there exists, for less than $5$ it is trivial. I tried making cycles, and using these circles to make the coloring.

Answer 1

A slight re-translation of the question asks that you find a class of graphs with $ >\binom{n}{2}/2 = \frac{n(n-1)}{4}$ edges with no triangles. Hint: what is the maximal number of edges in a triangle-free graph and what are the extremal examples? If you are coming in without any knowledge of extremal graph theory, this theorem is called Mantel's theorem.