"Proof" that the chemical potential is less than zero for irreversible material changes
Q1. You won't need the chemical potential terms $\sum_{i,\alpha}\mu_i^\alpha\text{d}n_i^\alpha$ (because it equals zero) in using the differential relation $\text{d}G = -S\,\text{d}T+V\,\text{d}P$ in this case. Nevertheless, the chemical potentials are still relevant in establishing criteria for phase equilibrium: $\mu_i^\alpha = \mu_i^\beta$, etc.


  Why should $\sum_{i,\alpha}\mu_i^\alpha\text{d}n_i^\alpha = 0$?


Reversible changes in composition imply that the process is in equilibrium, so that $\text{d}G = 0$. Assuming thermal and mechanical equilibrium and constant $T$ and $P$ we have $\text{d}T = \text{d}P = 0$, and the result follows by considering the (full) differential relation satisfied by $\text{d}G$.

Q2. Yes, you do need a reversible path. The expression on the right, if you follow its derivation carefully, does represent a reversible path. The inequality occurs because the chemical potential terms have not been accounted for. So the cancellation works out.

It seems, however, that we can derive the same result much more simply. Consider a system at thermal and mechanical equilibrium with constant $T$ and $P$, but not chemical equilibrium. Then $\text{d}G &lt; 0$, since the Gibbs free energy is minimized at equilibrium. If $T$ and $P$ are constant $\text{d}T = \text{d}P = 0$, so all that's left in $\text{d}G$ are the chemical potential terms, which must thus be negative.