Max. Entropy = Min. Energy? If a closed system has a submaximum entropy, then we can assume that in theory we can extract energy and thus the system goes to the maximum entropy state and no free energy is available. On the other hand, every physical system tends to be in a minimum energy state in the sense of internal energy.
Therefore, can we say that the principle of maximum entropy equals the principle of minimum energy?
 A: Max entropy = min energy if and only if one adds the right information about the physical conditions.
What can be proved for every thermodynamic system is that the condition of maximum entropy at fixed internal energy, and at fixed  remaining extensive variables (e.g. volume and number of particles, for a typical fluid system), is equivalent to the condition that internal energy is minimum at fixed entropy (and at fixed  remaining extensive variables).
Probably the proof makes easier to understand this duality. It can be found in every good thermodynamics textbook like Callen's Thermodynamics and an Introduction to Thermostatistic. I'll summarize it below.
First of all, it should be clear that maximum or minimum should be intended as conditions with respect to one or more variables expressing the internal constraints on the system. Let $X$ indicate a single constraint variable.
If entropy is maximum with respect to $X$, it is an extremum and the condition
$$
\left.\frac{\partial{S}}{\partial{X}} \right|_U = 0
$$
holds. But a well known identity of partial derivatives implies that $U$ must be extremum wrt $X$ at fixed $S$:
$$
Q=\left.\frac{\partial{U}}{\partial{X}}\right|_S = -\frac{\left.\frac{\partial{S}}{\partial{X}}\right|_U}{\left.\frac{\partial{S}}{\partial{U}}\right|_X}=0,~~~~~~[1]
$$
since $\left.\frac{\partial{S}}{\partial{U}}\right|_X=\frac{1}{T}$ can never be zero.
In order to show that $U$ is a minimum at fixed $S$, if $S$ is a maximum at fixed $U$, we need to derive an intermediate result.
We have
$$
\left.\frac{\partial{Q}}{\partial{X}}\right|_S = \left.\frac{\partial{Q}}{\partial{X}}\right|_U
$$
when $\left.\frac{\partial{U}}{\partial{X}}\right|_S=0$. This is a consequence of the identity  $\left.\frac{\partial{Q}}{\partial{X}}\right|_S =
\left.\frac{\partial{Q}}{\partial{U}}\right|_X \left.\frac{\partial{U}}{\partial{X}}\right|_S  + \left.\frac{\partial{Q}}{\partial{X}}\right|_U $.
Therefore,
$$
\left.\frac{\partial^2{U}}{\partial{X}^2}\right|_S = \left.\frac{\partial{Q}}{\partial{X}}\right|_S = \left.\frac{\partial{Q}}{\partial{X}}\right|_U.
$$
We need to evaluate the partial derivative wrt $X$, at constant $U$ of the last ratio in equation [$1$]. The term with the derivative of the denominator vanishes because it is multiplied by $\left.\frac{\partial{S}}{\partial{X}} \right|_U $ which is zero (extemum condition). Thus, we remin with
$$
\left.\frac{\partial^2{U}}{\partial{X}^2}\right|_S =
-\frac{\left.\frac{\partial^2{S}}{\partial{X}^2}\right|_U}{\left.\frac{\partial{S}}{\partial{U}}\right|_X}= -T\left.\frac{\partial^2{S}}{\partial{X}^2}\right|_U,
$$
which is positive if the extremum of $S$ is a maximum.
I would add that a similar max/min duality holds for the corresponding legendre transforms of energy/entropy. For example (although almost trivial), the  principle of minimum Helmholtz free energy would correspond to the maximum of the Legendre transform of entropy with respect to the energy, $\tilde S(1/T,V,N) = S -\frac{U}{T}$.
A: A closed system will stay at the same energy, while its entropy will as you said rise to its maximum. Maximum entropy is therefore minimal free energy, but total energy won't be affected.
