Bounds in NP-completeness proofs
Note that you got it wrong in the question. What you have is that if $Y \leq^p_m X$ and $Y$ is NP-hard, then $X$ is also NP-hard. This is true by definition, unconditional on any other lower bounds known for $X$ or $Y$ (though, as Ryan points out, such lower bounds could be very interesting).

If $X$ is in NP and has a lower bound on its computational complexity that is super-polynomial, it would mean that $P \neq NP$.