Can the counting density of the multiples of a finite set ever double beyond $\max(A)$? (Erdős #488)

Let $A$ be a finite set of positive integers and let $$B=\{ n \geq 1 : a\mid n\textrm{ for some }a\in A\}$$ be its set of multiples. Is it true that, for every $m>n\geq \max(A)$, $$\frac{\lvert B\cap [1,m]\rvert }{m}< 2\,\frac{\lvert B\cap [1,n]\rvert}{n}\,?$$ In words: once past $\max(A)$, the running density of $B$ can never be twice as large at a later checkpoint as at an earlier one. The constant $2$ cannot be lowered: taking $A=\{a\}$, $n=2a-1$, $m=2a$ gives density ratio $(2a-1)/a\to 2$. A single explicit triple $(A,n,m)$ violating the inequality would disprove the conjecture.

Acceptance. FULLY RESOLVES (disproof): an explicit finite set $A$ and integers $m>n\geq\max(A)$ with $\lvert B\cap[1,m]\rvert/m \geq 2\lvert B\cap[1,n]\rvert/n$, together with the two exact counts and code that recomputes them by direct sieving — fully machine-checkable. FULLY RESOLVES (proof): a proof of the inequality for all finite $A$ and all $m>n\geq\max(A)$; machine-checkable (Lean, against the existing formal-conjectures statement) preferred, else a complete written proof. ADVANCES: (a) a certified exhaustive verification for all $A\subseteq\{1,\ldots,K\}$ for a stated $K$ — using rigorous tail bounds so that only finitely many $(n,m)$ need checking per $A$ — delivered with code and the exhaustiveness argument; (b) an explicit family of triples $(A,n,m)$ with density ratio provably exceeding a stated constant strictly greater than the trivial benchmarks in the background (approaching 2 from below in a new way, quantifying how tight the constant is); or (c) a proof of the inequality for a natural infinite class of sets $A$ (e.g. all $A$ with $|A|\leq 2$, or sets of pairwise coprime integers), with proof. Deliver the counterexample + checker, or search code + certified range, or the proof file.