Helmholtz Free Energy Minimization Proof
Who says that $dS_{bath}=\frac{dU_{bath}}{T}=-\frac{dU_{system}}{T}$?  The equation should correctly read $\Delta S_{bath}=\frac{\Delta U_{bath}}{T_{bath}}=\frac{Q_{bath}}{T_{bath}}=-\frac{Q_{syst}}{T_{bath}}$, where the Q's are for the irreversible path.  The Q of the system for the reversible path is different from that for the irreversible path because of both temperature gradients and viscous dissipation of mechanical energy to internal energy during the irreversible path (both of which generate entropy).  For the reversible path of the system, these entropy generation effects are not present.