Let $x(a)\geq1$ and $y(b)\geq1$. I have a relation $x(a) \leq k(a,b)y(b)$ for all $k(a\geq b) \geq 1$ and $x(a)=y(b)$ when $k(a=b)=1$. Can we conclude that $x(a)\geq y(b)$ ?
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No, it most definitely does not imply that. Indeed, try solving for both $x< k y$ and $x \le y$ at the same time for $x \ge 1 \land y \ge 1$.
For $k=2$, $ x = 3$ and $y = 5$ clearly satisfy both: $ 3 < 2 \times 5$ and $3 \le 5$.