For all $k,l\in \mathbb{N}^{+}$ $R(k,l)\leq R(k-1,l)+R(k,l-1)$ whenever the [[Ramsey Number on Two Colours|Ramsey numbers]] exist. ###### Proof: Let $G=(V,E)$ be a graph with $R(k-1,l)+R(k,l-1)$ vertices. Then for all $v\in V,$ $|N(v)|\geq R(k-1,l)$ or $|V\setminus N(v)|\geq R(k,l-1)$: BWOC suppose there exists $v\in V$ such that $|N(v)|<R(k-1,l)$ and $|V\setminus N(v)|<R(k,l-1)$ then $|V|\leq1+R(k-1,l)-1+R(k,l-1)-1<R(k-1,l)+R(k,l-1)$ since $V=\{ v \}\cup N(v) \cup V\setminus N(v)$ - a contradiction. If $|N(v)|\geq R(k-1,l)$ then there exist a $(k-1)$-clique or independent set of $l$ vertices in $G[N(v)]$ which implies the existence of a $k$-clique or independent set of $l$ vertices in $G$ since we can add $v$ to the $(k-1)$-clique in $G[N(v)].$ Similarly, if $|V\setminus N(v)|\geq R(k,l-1)$ there exists a $k$-clique or independent set of $(l-1)$ vertices in $G[N(v)]$ which implies the existence of a $k$-clique or independent set of $l-1$ vertices in $G.$ This completes the proof of the claim.