Is it sufficient to show that we cannot calculate the inverse of the function because it is a many to one function and not a one-one function.
It would depend on what you need your function for, but by the standard definition of pre-image resistant, no, it's not sufficient. In the standard definition, it must be infeasible, given $x$, to find any $y$ such that $H(y) = x$; it need not be the specific $y$ that you had in mind.
I have to prove that the function is pre-image resistant that is given an output we cannot get back to the input.
It depends on what you mean by proof; however if you mean "proof it mathematically without making any unproven assumptions about the difficulty of an underlying problem", we currently cannot prove that any specific function is pre-image resistant. In fact, we can't even prove that there exists a pre-image resistant efficiently computable function. So, we either:
Rely on a function that we can prove is secure assuming some underlying problem (such as factoring) is hard, or
Rely on a function that we have no proof for, but has been studied by clever people, and they couldn't find anything.
(For hash functions, #2 is far more common).
BTW: if you were able to prove that every output has multiple possible preimages; that implies some mathematical symmetry to the function; is it possible that that symmetry could be used to find preimages?