For bipartite <density operator>[density operators],
$$
H(A|B)_{p\rho+(1-p)\sigma}\geq pH(A|B)_\rho+(1-p)H(A|B)_\sigma.
$$
This is a consequence of .

Concavity of quantum conditional entropy

The other convex decomposition is $\omega=q\sigma+(1-q)\Delta$. Applying the <concavity of quantum conditional entropy>, which follows from the , yields
$$
\boxed{H(A|B)_\omega\geq
\frac{H(A|B)_\sigma+\varepsilon H(A|B)_\Delta}{1+\varepsilon}}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 323 / 4 / c / iv / Solution

{c}
{wiki=Strong_subadditivity_of_quantum_entropy}

For a tripartite <density operator> $\rho_{ABC}$,
$$
S(\rho_{ABC})+S(\rho_B)\leq S(\rho_{AB})+S(\rho_{BC}).
$$
Equivalently, conditioning on an additional quantum system cannot increase <quantum conditional entropy>.

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 Incoming links: Strong subadditivity of Von Neumann entropy