Problem: Let $K_n$ be a complete graph with $n$ nodes. Let $n \ge 6$. Prove that if we color this graph with two colors, then there will be at least: ${n\choose 3} / {6\choose 3}$ single color triangles.
Question: How can I prove that? I suppose that I should make use of Ramsey's theorem. But I'm not sure how to prove that.
Consider all $6$-vertex subgraphs of $K_n$. Each of them has at least one monochromatic triangle, so at least a fraction $\frac1{\binom63}$ of its triangles are monochromatic. If you average this fraction over all these subgraphs, the result is again at least $\frac1{\binom63}$. In this average, you count all triangles in $K_n$ the same number of times, so the resulting average fraction of monochromatic triangles is the same as in the entire graph. The entire graph has $\binom n3$ triangles, so at least $\frac{\binom n3}{\binom 63}$ of them are monochromatic.