= Solution
Write $x_i=\langle\psi_i|X|\psi_i\rangle$, which is real because $X$ is a <Hermitian operator>. Set
$$
T=\sum_i\operatorname{sgn}(x_i)|\psi_i\rangle\langle\psi_i|,
$$
with $\operatorname{sgn}(0)=0$. This <Hermitian operator> satisfies $-I\leq T\leq I$. The <trace-norm variational principle for Hermitian operators> gives the <diagonal absolute-sum bound for the trace norm>:
$$
\boxed{\|X\|_1\geq\operatorname{Tr}(XT)=\sum_i|\langle\psi_i|X|\psi_i\rangle|.}
$$
For $X=\rho-P$, with $P=|\psi\rangle\langle\psi|$, extend $|\psi\rangle$ to an <orthonormal basis> $|\psi_1\rangle=|\psi\rangle,|\psi_2\rangle,\ldots$. Put $r=\langle\psi|\rho|\psi\rangle$. The first diagonal entry is $r-1\leq0$, and all the others are nonnegative. Their sum is $1-r$, because $\operatorname{Tr}\rho=1$. The bound therefore yields $\|\rho-P\|_1\geq2(1-r)$. Using the definitions of <trace distance> and <quantum fidelity>,
$$
\boxed{D(\rho,P)\geq1-r=1-F(\rho,P)^2.}
$$
This <pure-target lower bound on trace distance> is attained whenever $\rho$ has no coherence between $|\psi\rangle$ and its orthogonal complement. The proof used the <trace-norm variational principle for Hermitian operators>, together with positivity and normalization of a <density operator>.
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