Prove $$R(k,l) < R(k-1, l) + R(k, l-1)$$ when $R(k-1,l)$ and $R(k,l-1)$ are even. I have not made much progress as I don't really understand the intutiton as to why this would hold.
I understand why $R(k,l) \leq R(k-1, l) + R(k, l-1)$ this is true and the general inductive proof is clear to me.
When I start thinking about the strict less than, I know I want to show that for $K_n$ with $n=R(k-1, l) + R(k, l-1)-1$ is guaranteed to contain a monochromatic $K_k$ or $K_l$.
Let $n=R(k-1,\ell)+R(k,\ell-1)-1.$ Suppose each edge of $K_n$ is colored blue or red, with no monochromatic blue $K_k$ and no monochromatic red $K_\ell.$
From the proof of the inequality $R(k,\ell)\le R(k-1,\ell)+R(k,\ell-1),$ it follows that each vertex of $K_n$ must be incident with exactly $R(k-1,\ell)-1$ blue edges and exactly $R(k,\ell-1)-1$ red edges.
In other words, the spanning subgraph of $K_n$ formed by the blue edges is an
$(R(k-1,\ell)-1)$-regular graph of order $n.$ But $n$ and $R(k-1,\ell)-1$ are odd numbers.