Subadditivity of Von Neumann entropy

For a bipartite density operator ${\rho }_{AB}$,Equality holds exactly when ${\rho }_{AB}={\rho }_A\otimes {\rho }_B$.

For a bipartite density operator ${\rho }_{AB}$,

For a tripartite density operator ${\rho }_{ABC}$,Equivalently, conditioning on an additional quantum system cannot increase quantum conditional entropy.

Apply data-processing inequality for quantum relative entropy to ${\rho }_{ABC}$ and ${\rho }_A\otimes {\rho }_{BC}$, tracing out $C$. The resulting comparison is $D({\rho }_{ABC}\Vert {\rho }_A\otimes {\rho }_{BC})\ge D({\rho }_{AB}\Vert {\rho }_A\otimes {\rho }_B)$. Expand the logarithms of the product marginals to obtain $S(AB)+S(BC)-S(B)-S(ABC)\ge 0$, exactly Strong subadditivity of Von Neumann entropy.

Every tripartite density operator satisfies $S(AC)+S(BC)\ge S(A)+S(B)$. Purify to $ABCD$ and apply Strong subadditivity of Von Neumann entropy to $ACD$, conditioned on $A$. The Schmidt decomposition gives $S(AD)=S(BC)$ and $S(ACD)=S(B)$, proving the inequality. The original tripartite state need not be pure.