Prove that there is a red triangle or a blue triangle that is is a sub-graph

Question

If the edges of $K_6$ are coloured blue or red, prove that there is a red triangle or a blue triangle that is a sub-graph.

Well I am having a hard time proving this, I try to prove it by contradiction, Meaning that Assume that there is no red or blue triangle that is a sub-graph and I argue that You can't avoid but to have a red or a blue triangle.

But I can't even start the argument.

That was my attempt,

We consider all cycles of length $3$ in $K_6$ which are the triangles,

Now We start coloring one of the edge in this triangle with one red and the other edge with blue, Now the third vertex can be either red or blue, but without loss of generality we choose blue. So now we have a triangle with two blue edges and one red edge.

But then what can I say ?

Answer 1

Let $K_6 = (V,E)$ and $x \in V$ be arbitrary. $x$ has five incident edges, so by the pigeonhole principle at least three of those edges are the same color (without loss of generality, red). So now we have three red edges, $(x,v_1)$, $(x,v_2)$, $(x,v_3)$. If any of the three edges between the $v_i$ are red, that creates a red triangle. If none of them are red, then $(v_1,v_2)$, $(v_1,v_3)$, $(v_2,v_3)$ constitutes a blue triangle.