I am trying to find a lower-bound for R(k,k,k) , which is defined as "the smallest N such that every red/blue/green coloring of the pairs in $N\choose 2$ contains some set of k elements where every pair receives the same color."
I want to use the Erdos method that is used in proving the 2-color version, but I am not sure where to start. Any helps?
On
If you want to use the probabilistic method you have that $P(c(\{v_i, v_j\})=k) = \frac13$ for each edge $\{v_i, v_j\}$ of the graph and $c: V \to \{r,b,g\}$ is a colouring and each $k \in \{r,b,g\}$. There are $\binom{n}{k}$ many $k$-cliques and the probability that a given one of them is monochrome is $3 \cdot \frac{1}{3^{\binom{k}{2}}}$ by the previous (times 3 for the three colours), and we have $\binom{k}{2}$ pairs of edges that independently all must have the same colour.
So we get that there is at least one such clique as upperbounded by $$u(n,k):=\binom{n}{k}\frac{1}{3^{\binom{k}{2}-1}}$$ as in page 11 of "your" slides, which I take as a model.
Now ask yourself for what small values of $n$ we have that $u(n,k) < 1$, which ensures that the complementary event has positive probability, which will correspond to graphs/colourings as hinted in my other answer ?
To find a lower bound you need a specific graph $G$ and colouring by three colours than has no monochrome $k$-clique. Then you have shown that $R(k,k,k) > |V(G)|$. Presumably you need to find a family of graphs parametrised in $k$..