Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 35 / 4 / ii

False. Take the Bell stateThe joint state is pure, so $S(\rho )=0$. Taking either partial trace kills the off-diagonal terms and gives ${\rho }_1={\rho }_2=I_2/2$, whose two eigenvalues are $1/2$. Thus $S({\rho }_1)=S({\rho }_2)=1$, contradicting the proposed inequality $1\le 0$. Quantum entanglement allows a pure joint state to have mixed subsystem states.