= Solution
For the <quantum state ensemble> $\mathcal E=\{p_i,\rho_i\}$ with average $\overline\rho=\sum_ip_i\rho_i$, the <Holevo quantity> is
$$
\boxed{\chi(\mathcal E)=S(\overline\rho)-\sum_ip_iS(\rho_i)=\sum_ip_iD(\rho_i\Vert\overline\rho).}
$$
The second equality follows by expanding the trace definition: $\sum_ip_i\operatorname{Tr}\rho_i\log\overline\rho=\operatorname{Tr}\overline\rho\log\overline\rho$. All positive-weight states are supported within $\overline\rho$, so the expression is finite.
For a <CPTP map> $\Lambda$, the output average is $\Lambda(\overline\rho)$, with unchanged probabilities $p_i$. The <data-processing inequality for quantum relative entropy> gives
$$
\begin{aligned}
\chi(\Lambda\mathcal E)
&=\sum_ip_iD(\Lambda(\rho_i)\Vert\Lambda(\overline\rho))\\
&\leq\sum_ip_iD(\rho_i\Vert\overline\rho)
=\chi(\mathcal E).
\end{aligned}
$$
Thus \b[the Holevo quantity cannot increase under a quantum channel].
The underlying monotonicity can also be seen directly from <Strong subadditivity of Von Neumann entropy>. Attach the classical label $X$, and realize $\Lambda$ by a <Stinespring representation of a completely positive map> with output $B$ and discarded environment $E$. The isometry preserves $I(X:BE)$, whose input value is $\chi(\mathcal E)$. The difference from the output value is
$$
I(X:BE)-I(X:B)=S(XB)+S(BE)-S(B)-S(XBE)\geq0,
$$
which is precisely strong subadditivity. This explains why discarding an environment loses accessible ensemble information.
Here “operation” means the unconditional channel, including the classical outcome when a measurement is retained. Conditioning on a selected outcome can increase the Holevo quantity. For instance, take equally likely vectors $\sqrt a|0\rangle\pm\sqrt{1-a}|1\rangle$, with $1/2<a<1$. Their input Holevo quantity is $h_2(a)<1$. The filter $K=\operatorname{diag}(\sqrt{(1-a)/a},1)$ succeeds on either state with probability $2(1-a)$ and produces the orthogonal states $|+\rangle,|-\rangle$ after normalization. The success-conditioned quantity is $1$. This does not violate the unconditional statement.
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