 Solution

The Ramsey number $R(s,t)$ is the least integer $N$ such that every graph on $N$ vertices contains either a clique of size $s$ or an independent set of size $t$. Equivalently, every red-blue colouring of the edges of $K_N$ contains a red $K_s$ or a blue $K_t$.

Put $a=R(s-1,t)$, $b=R(s,t-1)$ and consider a graph on $a+b$ vertices. Fix a vertex $v$. If it has at least $a$ neighbours, their induced graph contains either a clique of size $s-1$, which together with $v$ gives a clique of size $s$, or an independent set of size $t$. Otherwise $v$ has at least $b$ nonneighbours. Their induced graph contains either a clique of size $s$, or an independent set of size $t-1$, which can be enlarged by $v$. ThusSince $R(2,3)=R(3,2)=3$, this gives $R(3,3)\le 6$. For the reverse bound, the five-cycle has no triangle and no independent set of size three: its complement is again a five-cycle, so it too is triangle-free. Hence the property fails on five vertices and