Do qutrits violate monogamy of entanglement? According to monogamy of entanglement, if you have three particles $A,B\;\&\; C$, then if $A$ is maximally entangled with $B$, then $C$ is not entangled with either $A$ or $B$.
However one could imagine a state of qutrits like this
$|\psi\rangle=\frac{1}{\sqrt{3}}(|012\rangle+|120\rangle+|201\rangle)$
and clearly measuring any of the particles will tell you the state of the other two. What is wrong here?
 A: In this example $A$ and $B$ are not entangled at all. Parties $A$ and $B$ being (maximally) entangled means that the reduced state on $\mathcal H_A\otimes\mathcal H_B$ is (maximally) entangled. In this case, you have
$$\rho_{AB} \equiv \operatorname{Tr}_C(\mathbb{P}_\psi)
= \frac13(\mathbb{P}_{01} + \mathbb{P}_{12} + \mathbb{P}_{20}),
\qquad \mathbb{P}_\psi\equiv|\psi\rangle\!\langle\psi|.$$
Such $\rho_{AB}$ corresponds to full correlation between the two parties, but is a separable state.
More generally, you're implicitly assuming that "correlation between particles means entanglement", which is not the case. You can have maximal correlation between two parties (meaning measurement outcome on one determines measurement outcome on the other) with no entanglement at all. Though notice that such "maximal correlation" is still a weaker form of correlation compared to what entanglement gives, because entangled states can produce such correlations in different measurement bases, in a way separable states cannot.
