Coloring a Complete Graph in Three Colors, Proving that there is a Complete Subgraph 
Color the edges of a complete graph on $n$ vertices $K_n$ in three
  colors (red,blue,yellow) such that at most $\dfrac{n^2}{k}$ are
  colored red ($k$ is some natural number). 
Prove that $K_n$ contains either blue $K_p$ or yellow $K_p$ (complete
  subgraph on p vertices where all of its edges are colored blue or all
  are colored yellow) with $p=\dfrac{1}{100}\log k$. 
What are the bounds for $p$?

I know that a complete graph $K_n$ has ${n\choose 2}$ edges, thus if $\dfrac{n^2}{k}$ are colored red, we have $\dfrac{n^2}{2}-\dfrac{n}{2}-\dfrac{n^2}{k}$ edges that are colored blue or yellow. Now I thought to use Ramsey theorem, but I don't know how to find the relavant bounds for $p$.
Any help will be very appreciated.
 A: We know that the number of edges in $K_{n}$
  is: $|E(K_{n})|={n \choose 2}=\frac{n(n-1)}{2}=\frac{n^{2}}{2}-\frac{n}{2}$
We are given that at most $n^{2}/k$
  edges are colored Red, and thus at least $n^{2}/2-n/2-n^{2}/k$
  edges are not Red, 
i.e. White or Black (Also, $k$
  should be at least 3, for the question to make sense - so we will have 
$n^{2}/k\leq n^{2}/2-n/2$
 , for large enough $n$
 ). 
Denote by $G$
  the graph we obtain from $K_{n}$
  by removing the Red colored edges. 
Let $t=\sqrt[50]{k}-1$
 . By Turan's theorem, if $G$
  has no copy of $K_{t+1}$
  (regardless of color) then $|E(G)|\leq\frac{n^{2}}{2}(1-\frac{1}{t})$
 .
However, $|E(G)|\geq\frac{n^{2}}{2}-\frac{n}{2}-\frac{n^{2}}{k}=\frac{n^{2}}{2}(1-\frac{1}{n}-\frac{2}{k})\underset{(*)\,k\leq n^{2}\Rightarrow\sqrt{k}\leq n}{\geq}\frac{n^{2}}{2}(1-\frac{1}{\sqrt{k}}-\frac{2}{\sqrt{k}})=\frac{n^{2}}{2}(1-\frac{3}{\sqrt{k}})$
 .
And we have: $\frac{n^{2}}{2}(1-\frac{3}{\sqrt{k}})\geq\frac{n^{2}}{2}(1-\frac{1}{\sqrt[50]{k}-1})$
 , where $t=\sqrt[50]{k}-1$
 , as $\frac{3}{\sqrt{k}}<\frac{1}{\sqrt[50]{k}-1}$
  for $k>1$
 .
(easy to check as $\sqrt{k}$
 , $\sqrt[50]{k}$
  are mono. increasing and $\sqrt{k}>\sqrt[50]{k}$
 ).
$\Longrightarrow|E(G)|>\frac{n^{2}}{2}(1-\frac{1}{t})$
 , and so by Turan's theorem,$ G$
  has a copy of $K_{t+1}$
 .
Let $p=\frac{1}{100}\log_{2}k$
 . We know from class the following upper bound on Ramsey's numbers:$R(m,m)\leq4^{m}=2^{2m}$
And so we have:
$R(p,p)\leq2^{\frac{1}{50}\log_{2}k}=k^{\frac{1}{50}}=\sqrt[50]{k}$
As $t=\sqrt[50]{k}-1$
  and $G$
  has a copy of $K_{t+1}=K_{\sqrt[50]{k}}$
 , then every coloring of edges of $G$
  in 2 colors (White and Black)
is also a 2-coloring of this $K_{t+1}$,
  and by Ramsey's theorem, will contain either White $K_{p}$
  or Black $K_{p}$
 , as required.
