A proof would be something like:
(Using E as the symbol for set membership, and V for "for all", R is the set of real numbers)
Define: d(x) = a/(x + b)^c, where a,b,c are constants E R and where x E R.
Hypothesis: d /= 0
Proof:
Let a = 0, then d(x) = 0
Also
Take limits as x -> +- infinity
d(x) = lim x->+-inf a/(x+b)^c = 0.
Hence by contradiction d is not non-zero for some a,b,c,x E R.
However if constraints are imposed on a and x d(x) can be non-zero.
Let a E R/{0}, and x be finite:
d(x) = a/(x+b)^c
by observation d(x) can only = 0 if the numerator = 0 or the denominator = +- infinity, as a is non-zero and x is finite and b,c are constants those conditions cannot be satisfied. The condition x+b /= 0 must hold in order for the function to be defined.
And so d(x) /=0 V a,b,c,x E R under those constraints.
That was my attempt at a proof. Its probably not very acceptable at most levels.
(I've noticed some typos please don't execute me for those.)