= Solution
Write $\overline\rho=\sum_i p_i\rho_i$. The <Holevo quantity> is
$$
\boxed{\chi(\mathcal E)=S(\overline\rho)-\sum_i p_iS(\rho_i),}
$$
where $S$ is <Von Neumann entropy>, with a consistent logarithm base. Attach an orthogonal classical label register $X$ to form the <classical-quantum state>
$$
\omega_{XB}=\sum_i p_i|i\rangle\langle i|_X\otimes\rho_i.
$$
The <entropy of an orthogonal quantum mixture> gives
$$
S(X)=H(p),\quad S(XB)=H(p)+\sum_i p_iS(\rho_i),\quad S(B)=S(\overline\rho).
$$
Here $H(p)$ is the <Shannon entropy> of the classical label. Hence $I(X:B)=\chi(\mathcal E)$, where $I$ denotes <quantum mutual information>.
Let $\Phi:B\to C$ be the proposed <quantum channel>. Using its <Kraus representation>, define an <linear isometry> $V=\sum_\ell A_\ell\otimes|\ell\rangle_E$, since $V^\dagger V=I$. This is a <Stinespring representation>. Apply it to $B$, obtaining a state $\eta_{XCE}$; tracing out $E$ applies $\Phi$ to every ensemble member. An <linear isometry> preserves nonzero <eigenvalues> and therefore <Von Neumann entropy>, so
$$
I(X:CE)_\eta=I(X:B)_\omega=\chi(\mathcal E),
\qquad
I(X:C)_\eta=\chi(\{p_i,\Phi(\rho_i)\}).
$$
Their difference is
$$
I(X:CE)-I(X:C)=S(XC)+S(CE)-S(C)-S(XCE).
$$
The <Strong subadditivity of Von Neumann entropy> states that this expression is nonnegative for every tripartite state. Consequently
$$
\boxed{\chi(\{p_i,\Phi(\rho_i)\})\leq\chi(\mathcal E).}
$$
Thus the <Holevo quantity under a quantum channel> decreases because discarding the dilation environment cannot improve correlations with the classical label. The probabilities remain unchanged throughout.
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