Let $2 \leq p \leq q$ and $2 \leq r \leq s$. Prove that $R(p,r) \leq R(q,s)$ and that equality holds if and only if $p=q$ and $r=s$.
The equality part is clear, cause we will have $R(p,r) = R(p,r)$, which clearly is the same thing. But why it is true only then? I know also that things are equal due to symmetry too, though I don't think it is relevant here. On the other hand, how can I prove the inequality?
Assuming $p\lt q$ and $r\le s,$ here's how you can show that $R(p,r)\lt R(q,s).$
Let $n=R(p,r).$ Take a complete graph $K_{n-1}$; by the definition of the Ramsey number $R(p,r),$ we can color the edges of this graph with two colors, call them color $1$ and color $2,$ so that there is no monochromatic $K_p$ of color $1$ and no monochromatic $K_r$ of color $2.$
Now create a complete graph $K_n$ by adding a new vertex and edges joining the new vertex to all the old vertices. Color all the new edges with color $1.$ In this coloring of $K_n$ there is no monochromatic $K_{p+1}$ of color $1$ and no monochromatic $K_r$ of color $2.$ This shows that $n\lt R(p+1,r),$ i.e., if $p\lt q$ and $r\le s$ then $$R(p,r)=n\lt R(p+1,r)\le R(q,r)\le R(q,s).$$
To prove that $R(x,y)$ is strictly increasing in the second variable, do the same but color the new edges with color $2.$