Solution
= Solution
\b[False.] Take the <Bell state>
$$
|\Phi^+\rangle=\frac{|00\rangle+|11\rangle}{\sqrt2},\qquad\rho=|\Phi^+\rangle\langle\Phi^+|.
$$
The 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 $1\le0$. <Quantum entanglement> allows a pure joint state to have mixed subsystem states.