= Solution
By <Donald's identity>, the objective equals
$$
\sum_jp_jD(\omega_j\|\rho)=\sum_jp_jD(\omega_j\|\overline\omega)+D(\overline\omega\|\rho).
$$
The <nonnegativity of quantum relative entropy> makes the last term at least zero, with equality exactly when $\rho=\overline\omega$. Hence the unique minimizing <density operator> is the ensemble average and
$$
\min_\rho\sum_jp_jD(\omega_j\|\rho)
=\sum_jp_jD(\omega_j\|\overline\omega)
=S(\overline\omega)-\sum_jp_jS(\omega_j).
$$
For the <classical-quantum state> $\sigma_{XB}$, its marginal on $X$ has entropy $H(p)$, its marginal on $B$ is $\overline\omega$, and its joint entropy is $H(p)+\sum_jp_jS(\omega_j)$. Substituting in the definition of <quantum mutual information> gives
$$
\boxed{\min_\rho\sum_jp_jD(\omega_j\|\rho)
=\chi(\{p_j,\omega_j\})=I(X:B)_\sigma,\qquad\rho_{\min}=\overline\omega}.
$$
This is the <relative-entropy barycenter of a quantum ensemble>: the average state minimizes mean distinguishability from the ensemble. Its minimal value is exactly the <Holevo quantity>.
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