Minimizing Gibbs free energy with Helmholtz free energy expressions If I have an expression for the Helmholtz free energy (from statistical associating fluid theory for a polymer system), can I still find the Gibbs free energy minimum under constant $T,P$ conditions? 

Seems that since 

$\mu_i = \left(\frac{\partial A}{\partial n_i}\right)_{T,V,n_j} = \left(\frac{\partial G}{\partial n_i}\right)_{T,P,n_j}$

where $A$ is the Helmholtz free energy and $G$ is the Gibbs free energy and $\mu_i$ is the chemical potential of component $i$.

If I can just find the expressions for $\mu_i=\left(\frac{\partial A}{\partial n_i}\right)_{T,V,n_j}$ I should be able to use this in the expression for $G = \sum_i \mu_i n_i$ and find its minimum, without having to worry about an equation of state to relate change in volume ($dV$) with respect to constant pressure ($P$)?

Edit I should clarify that I wonder about the Gibbs free energy because I'm interested in the constant $T,P$ (isothermal, isobaric case), and in the case especially for liquids. If you needed to introduce an equation of state to parameterize the change in V as a function of $T,P$ this would not be too difficult for an ideal gas, but for liquids it would be a challenge. So in this case, I wonder how to handle the $PV$ term (for constant $P$) in the Helmholtz free energy which does not appear in the Gibbs.
I don't think you can use the approach you are outlining without an equation of state.

$G = \sum_i \mu_i n_i$ is valid only at constant $T$ and $P$.

$A = \sum_i \mu_i n_i$ is valid only at constant $T$ and $V$.

If $P$ and $V$ are both constant, then you can equate the two and use $G=A= \sum_i \mu_i n_i$.  Actually, that assumption is probably not horrible for most liquids at "normal" pressures: I think it amounts to assuming the liquid is incompressible.