Given
$$\mathrm{\#3SAT} = \{ (w, y) \mid w\text{ is a $\mathrm{3SAT}$ instance with at least $y$ satisfying assignments}\}\,,$$
prove that $\mathrm{\#3SAT}$ is NP-Hard.
I am currently stuck with this one.
Basically $\mathrm{\#3SAT}$ is the counting version of $\mathrm{3SAT}$ and to me it is seems clear that establishing membership in the language is more complex than establish the classic $\mathrm{3SAT}$ membership.
I'm trying to reason through the prover-verifier paradigm: let's say you have a $\mathrm{3SAT}$ instance, now suppose the certificate for that instance is $n$ bits. It is obvious that the certificate for the same instance but translated into $\mathrm{\#3SAT}$ with let's say $y=10$ would be at least $10n$ bits. So if the first can be examined in time $t$ then the second will take at least $10t$ and so on. However I am not satisfied with this reasoning, it seems incomplete and superficial. I would greatly appreciate your advice on how to proceed with this proof. Thanks in advance.