$R(p,q)\leq R(R(p − 1, q; r),R(p, q − 1; r); r − 1) + 1.$ it possible to prove? is true? Has anyone seen this inequality of Ramsey's numbers? where? or is it possible to prove?, I found it in some notes and I don't know if it's true
$$R(p,q)\leq R(R(p − 1, q; r),R(p, q − 1; r); r − 1) + 1.$$
I know some properties like
$1- R(q_1, q_2, \ldots, q_k; 1) = q_1 + q_2 + \cdots + q_k − k + 1$
$2- R(p, r; r) = R(r, p; r) = p$
$3- R(p, q)\leq (p − 1, q) + R(p, q − 1)$
but the first inequality I don't know if it's true
 A: Two corrections : 


*

*Property 3 should read $$R(p,q)\leq R(p-1,q)+R(p,q-1)$$

*I think you want to prove that
$$R(p,q;r)\leq R(R(p − 1, q; r),R(p, q − 1; r); r − 1) + 1.$$
Whitout this $r$ the results is obvious by monotonicity over $r$.


If I'm correct, lets proceed.
Define $p_1 = R(p − 1, q; r)$ and $q_1 = R(p, q − 1; r)$. Let S be a set with $n$ elements, where $n=R(p_1, q_1; r − 1)+1$, and color the $r$-subsets of $S$ with two colors, say red and blue. For simplicity, assume $S=\{1,\ldots,n\}$.

We want to prove that $S$ must include either a $p$-subsets whose all $r$-subsets are red, or a $q$-subsets whose all $r$-subsets are blue.

Let $S'=S\setminus\{n\}=\{1,\ldots,n-1\}$. We define a coloring
of the $(r − 1)$-subsets of $S'$ by giving $X \subseteq S'$ the same color as $X \cup \{n\}$. 
By hypothesis $$\vert S'\vert=n-1=R(p_1, q_1; r − 1)$$ hence $S'$ either contains a subset $A$ of size $p_1$ such that all its $(r − 1)$-subsets are red or a subset $B$ of size $q_1$ such that all its $(r − 1)$-subsets are colored blue.
Without loss of generality, suppose that the first situation occurs. Since $A$ has $R(p − 1, q; r)$ elements, there are two possibilities. 


*

*$A$ has a subset of $q$ elements with all its $r$-subsets blue, in which case we are done.

*$A$ has a subset $A'$ of $p − 1$ elements with all its $r$-subsets red. Therefore because $A'\subseteq A$, we know that all $(r − 1)$-subsets and all $r$-subsets of $A'$ are red. Combining this two properties by definition of the coloring on $S'$ from the coloring on $S$, the set $A'\cup \{n\}$ of size $p$ with all $(r)$-subsets red.


Hence $n\geq R(p,q;r)$
