If perfect maximal entanglement is never true, does a remainder invalidate the monogamy of entanglement? If something is only very nearly (and/or observed to be) maximally entangled, does that remainder allow for a menage trois of hybrid correlation (as it relates to AMPS)?
 A: Simple answer: Yes it does.
The monogamy of entanglement is often stated as "when two particles are maximally entangled, they cannot be entangled by a third particle", while really it means that the more two particles are entangled, the less they can be entangled with a third particle. However, since "more" and "less" are ill-defined, phrasing it in the first way makes it a correct statement, which show you the essence of what's going on. 
Now, making the statement precise requires measures of entanglement. Some such measures are various entropies - not all of which satisfy monogamy in a strict sense. The concurrence (for qubits) does, so does the squashed entanglement. See here for one in the long list of papers related to the problem: http://arxiv.org/abs/quant-ph/0502176
A: Maximal Entanglement theoretically cannot be achieved on particular $n$-qu$d$it systems for a specific $(n,d)$ pair. For example, the paper demonstrates this taking the case of a four-qubit system and extends Concurrence to pure states of arbitrary dimensions (aka multi-qudit pure states) in a very intuitive way by looking at the geometry of the tensor products with increasing dimensionality. A pedagogic explanation to reach the generalization from the basic notion of separability may be found in our recent article here: arXiv:1607.00164 (2016).
