Entropy of subsystem maximized? Let $A$ and $B$ be 2 subsystems of a quantum mechanical system, so a state of the whole system is a vector in $A \otimes B$. As far as I understand, a  density operator $ \rho $ in general can't be written as a tensor product of the density operators of its subsystems. If $\rho_A$ and $\rho_B$ are the density operators of two independent systems, then you can write:
$$
\rho = \rho_A \otimes \rho_B.
$$
Otherwise (the two systems are entangled), to get an expression that contains the same information as $\rho_A$, you would have to take the partial trace of the operator over the subspace $B$:
 $$
\tilde{\rho}_A = \mathrm{tr}_B \rho.
$$
Now we know that in thermodynamic equilibrium, the von Neumann entropy of the system (of $\rho$) is at its maximum. Can we derive from that that the von Neumann entropy of the reduced density matrix $\tilde{\rho}_A$ is also at its maximum?
In the case of no entanglement, 
$$
S[\rho_A \otimes \rho_B] = S[ \rho] = S[\rho_A] + S[\rho_B]
$$
holds, and since all expressions are bigger than zero, one could argue that for $S[\rho]$ to me maximized, one needs also to maximize $S[\rho_A]$ and $S[\rho_B]$. This doesn't work anymore for an entangled system, because here we don't have additivity of the entropy, but instead only subadditivity for $\tilde{\rho}_A = \mathrm{tr}_B \rho$ and $\tilde{\rho}_B = \mathrm{tr}_A \rho$. Is $S[\mathrm{tr}_B \rho]$ still maximized?
Edit: To give a reason, why I'm asking this. The question I was thinking about originally was: If a quantum system is in thermodynamic equilibrium, are the sub systems also in thermodynamic equilibrium? My naive answer to that is "Yes, they should be", but I'm not sure about that, and I can't give a proper reason, why they should. 
 A: Assuming you are considering non-interacting subsystems, let us take two of them, $S_1$ and $S_2$, at given total energy $E = E_1 + E_2$, and seek a maximum entropy state. 
As you already noticed, subadditivity of entropy in the presence of entanglement, $S ≤ S_1 + S_2$, rules out entangled states, and means that entropy necessarily attains its maximum on the set of unentangled states compatible with the given $E$. 
Now sort the latter set according to the energy of one subsystem, say $E_1 = \epsilon$. For each $E_1 = \epsilon$, the state of maximum entropy will be the direct product $\rho_1^0(\epsilon) \otimes \rho_2^0(E- \epsilon)$ of maximum entropy subsystem states $\rho_1^0(\epsilon)$, $\rho_2^0(E- \epsilon)$ corresponding to energies $\epsilon$, $E-\epsilon$. The total entropy is $S^0(\epsilon) = S_1^0(\epsilon)+S_2^0(E-\epsilon)$, and the problem is reduced to maximizing $S^0(\epsilon)$. That is, we need $\epsilon$ such that
$$
\frac{dS^0}{d\epsilon} = \frac{dS_1^0}{d\epsilon}(\epsilon) -  \frac{dS_2^0}{d\epsilon}(E- \epsilon) = 0
$$
From here a standard argument yields that the desired value of $\epsilon$ is that for which $S_1$ and $S_2$ are in mutual thermal equilibrium (common temperature) for the total energy $E$. 
Does this mean that each subsystem is in its own maximum entropy state? 
Relative to its other states of identical energy $E_i$, yes. Across the whole set $\{\rho_i^0(E_i)\}_{E_i}$ compatible with given $E$, no. 
The reason is that the entropy of equilibrium states $\rho_i^0(E_i)$ increases with the energy $E_i$, and so for each subsystem the entropy $S_i^0(E_i)$ attains its maximum for maximum $E_i$. But when this happens the complementary subsystem has minimum energy $E-E_i$, hence minimum entropy, qed. 
