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The fidelity case was already worked in the other answer. Here is an idea for the trace distance one. The trace distance between $\rho$ and some $|\psi\rangle\!\langle\psi|$ is $$\|\rho - |\psi\rangle\!\langle\psi|\|_1 = \operatorname{Tr}\lvert \,\rho - |\psi\rangle\!\langle\psi|\,\rvert,$$ which is equal to the sum of the singular values of $\rho-|\psi\... 5 Recall that for any Hermitian operator$A$and any unit vector$|\psi\rangle$the real number$\langle \psi|A|\psi\rangle$, known as the Rayleigh quotient, is bounded by the largest eigenvalue$\lambda_{max}$of$A$$$\langle \psi|A|\psi\rangle \le \lambda_{max}.$$ Moreover, the maximum is achieved when$|\psi\rangle$is the unit norm eigenvector of$A\$ ...