This answer is too long for a comment, and it is for your question following @Themis's answer below.
Denote by $T^0,S^0$ the reservoir's parameters and by $\delta W^0$ the externally controlled work, then for any process connecting two infinitesimally close equilibrium states $$dU=TdS+\delta W=T^0dS^0+\delta W^0 \tag{1}\label{1}$$
For an isothermal process $T=T^0$ and then
$$T(dS-dS^0)=\delta W^0-\delta W=T\sigma \ge 0 \tag{2}\label{2}$$ where $$\sigma = dS-dS^0 \ge 0\tag{3}\label{3}$$ is the internally generated entropy that is never negative, and is zero only for a reversible process.
Therefore $dF=d(U-TS)=dU-TdS-SdT=-SdT + \delta W$ that can be written for an isothermal process, $dT=0$, as
$$ dF|_T = \delta W = - T\sigma +\delta W^0 \tag{4}\label{4}$$ which in the special case of no work done, $\delta W^0=0$, is just $$ dF|_{T,\delta W^0=0} = - T\sigma \le 0 \tag{5}\label{5}$$ with equality only in a reversible process.
The above considerations are for infinitesimal changes between two equilibrium states. One might think that we could integrate $\eqref{4}$ or $\eqref{5}$ and obtain a similar inequality for finite, not infinitesimal, changes. Unfortunately, that would not work in general because in the $\eqref{4}$ we have assumed that each infinitesimal step is bracketed with an equilibrium states, that is the finite path connecting one equilibrium state with another not close to it is almost reversible.
It is possible to generalize our result but we must assume that the internal energy of the system undergoing the irreversible process between two non-neighboring equilibrium states satisfies the Gibbs equation $$U=TS+W\tag{6}\label{6}$$ where $\delta W=\sum_{k=1}^K Y_k dX_k$ is the work function. In other words the internal energy is a 1st order homogeneous function of the entropy.
That being the case assume that the system is connected to a single thermal reservoir at temperature $T^0$ and to various work reservoirs with which it can interchange entropy $dS^0$ and work, resp. that is $\delta W^0 = \sum_{k=1}^N Y_k^0 dX_k^0$ with $dX_k=dX_k^0; k=1,2,..,K$ but $dS-dS^0 \ge 0$.
Let the initial and final states be denoted by $\mathcal A$ and $\mathcal B$. Since the end states are assumed to be in equilibrium, both the initial and final temperature must equal to that of the reservoir, $T^0$.
Now calculate the change in the internal energy over the process
$$\Delta U = U(\mathcal A)-U(\mathcal B)\\=(TS)|_{\mathcal A}+W(\mathcal A)-(TS)|_{\mathcal B} - W(\mathcal B)\\
=T^0(S({\mathcal A})-S({\mathcal B}))+W(\mathcal A) - W(\mathcal B) \\
=T^0 \Delta S + \Delta W \tag{7}\label{7}$$
By definition $F=U-TS$, $\Delta F=\Delta (U-TS)$ but at the endpoints $T=T^0$ is fixed, therefore we can write:
$$\Delta U - T^0 \Delta S =\Delta (U - T^0 S) \tag{8}\label{8}.$$
$$\Delta F = \Delta W \tag{9}\label{9}.$$
By the conservation of energy $\Delta U = T^0 \Delta S^0 + \Delta W^0$ where $\Delta W^0$ is the "experimental", ie., external, work done by the environment on the system and $\Delta S^0$ is the entropy absorbed by the system from the thermal reservoir whose temperature is $T^0$ throughout. But from the 2nd law we know that $\Delta S \ge \Delta S^0$ therefore for some $\sigma = \Delta S-\Delta S^0 \ge 0$:
$$\Delta U - T^0 \Delta S = \Delta F = \Delta W \\
=T^0\Delta S^0 +\Delta W^0 -T^0 (\delta S^0 +\sigma)
=\Delta W^0 +T^0\sigma$$ or
$$\Delta F + T^0\sigma = \Delta W^0 \tag{10}\label{10}.$$
This $\eqref {10}$ is the general equation for the finite variation of the free energy. It shows that since the internal dissipation $T^0\sigma$ is never negative and is always positive for an irreversible process, it reduces the amount the free energy can be increased by the external, experimental, work. Or, if written in the "active mode", $\eqref{11}$
$$-\Delta W^0 + T^0\sigma = - \Delta F \tag{11}\label{11}$$
it shows how much the work done on the environment is reduced by the dissipation for a given change in the free energy.
Important: for this result to be true it is not needed that the system go through an isothermal process throughout. Instead, all is needed is that the beginning and end equlibrium states are at the same temperature and the reservoir with which thermal energy (entropy) is exchanged stay at the same temperature.
