= Solution
Let $\Delta=\rho-\sigma=\Delta_+-\Delta_-$ be the <positive-negative decomposition of a Hermitian operator>. Since $\operatorname{Tr}\Delta=0$,
$$
\operatorname{Tr}\Delta_+
=\operatorname{Tr}\Delta_-
=\frac12\|\Delta\|_1.
$$
For $0\leq P\leq I$,
$$
\operatorname{Tr}(P\Delta)
\leq\operatorname{Tr}(P\Delta_+)
\leq\operatorname{Tr}\Delta_+.
$$
Equality is attained by the projector onto the positive eigenspace of $\Delta$. This proves the <variational characterization of trace distance>
$$
\boxed{D(\rho,\sigma)
=\max_{0\leq P\leq I}\operatorname{Tr}[P(\rho-\sigma)]}.
$$
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