<Entanglement monogamy> says that maximal entanglement with one independent system excludes entanglement with another. For three qubits this is quantified by the <Coffman--Kundu--Wootters inequality>
$$
C_{A:BC}^2\geq C_{AB}^2+C_{AC}^2.
$$
For example, if $AB$ is a <maximally entangled state>, then $C_{AB}=1$ and the global state factorizes as $|\psi^-\rangle_{AB}\otimes|\phi\rangle_C$ up to local unitaries, so $C_{AC}=C_{BC}=0$. The three-qubit GHZ state instead has entanglement across every one-versus-two cut but no pairwise concurrence after the third qubit is traced out.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 343 / 2 / a / Solution

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{title2=$C_{A:BC}^2\geq C_{AB}^2+C_{AC}^2$}
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For a pure state of three qubits, the squared concurrence obeys $C_{A:BC}^2\geq C_{AB}^2+C_{AC}^2$. In particular, if $A$ and $B$ form a maximally entangled pair, then $A$ has no entanglement with $C$.


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 Incoming links: Coffman--Kundu--Wootters inequality