Proving $R(3,4)\le 9$ I am trying to prove $R(3,4)\le 9$. This is my approach:

For any $K_9$ we have (WLOG) at least 4 red edges by the pigeonhole
  principle. Consider all of the edges between these 4 red edges, if any
  of them are red, we have a red $K_3$. If this is not the case, all of
  the edges between them are blue, forming a $K_4$.

Is this correct? Can anyone help me neaten up this argument?
 A: another way of looking at it is if any vertex has blue degree greater equal to 6 or red degree greater equal to 4 we are done. since degree of each vertex is 8 the only case that we are left with is each vertex has blue degree 5 in which case the sum of all blue degrees is 45 which is odd and hence a contradiction. So we are done.
A: You'd need to expand on how you decided that there must be $4$ red edges, I think. And I don't really understand how you are using "WLOG" in this context.
If the four red edges are a cycle, or even a path, joining those associated vertices with blue edges will not form a blue $K_4$.

If you can find a vertex that connects with $4$ red edges, then those $4$ vertices either have a red connecting edge somewhere - forming a red $K_3$ - or they are all blue, forming a blue $K_4$.
If you can find a vertex that connects with $6$ blue edges, then the $K_6$ connected to it contains either a red triangle $K_3$ or a blue triangle, which with the initial blue-connected vertex forms a blue $K_4$.
So a vertex that connects with exactly $5$ blue edges and $3$ red edges is not suitable for initial consideration.
However, since $K_9$ has $36$ edges, there must be either $23$ blue edges or $14$ red edges, and each edge connects to $2$ vertices. This means that there are either $46\: (> 5\times 9)$ blue connections or $28\: (> 3\times 9)$ red connections. So there must be a vertex with one of the above required colour connections - $6$ blue edges or $4$ red edges - as required.
