= Solution
For an ensemble $\mathcal E=\{p_i,\rho_i\}$, write its channel-output <Holevo quantity> as
$$
\chi(\Phi,\mathcal E)=S\!\left(\sum_i p_i\Phi(\rho_i)\right)-\sum_i p_iS(\Phi(\rho_i)),
\qquad
\chi^*(\Phi)=\sup_{\mathcal E}\chi(\Phi,\mathcal E).
$$
Take ensembles $\mathcal E_1=\{p_i,\rho_i\}$ and $\mathcal E_2=\{q_j,\sigma_j\}$ independently, giving the product ensemble $\{p_iq_j,\rho_i\otimes\sigma_j\}$. Its average output factors:
$$
\sum_{i,j}p_iq_j\,\Phi_1(\rho_i)\otimes\Phi_2(\sigma_j)
=\left(\sum_i p_i\Phi_1(\rho_i)\right)\otimes
\left(\sum_j q_j\Phi_2(\sigma_j)\right).
$$
The <Von Neumann entropy> of a tensor product is the sum of its entropies. Applying this both to the average and to every labelled output gives
$$
\chi(\Phi_1\otimes\Phi_2,\mathcal E_1\otimes\mathcal E_2)
=\chi(\Phi_1,\mathcal E_1)+\chi(\Phi_2,\mathcal E_2).
$$
Choose each ensemble within $\varepsilon$ of its supremum. The supremum for the joint channel includes this product ensemble, so its value is at least $\chi^*(\Phi_1)+\chi^*(\Phi_2)-2\varepsilon$. Let $\varepsilon\downarrow0$. The <superadditivity of Holevo capacity> follows:
$$
\boxed{\chi^*(\Phi_1\otimes\Phi_2)\geq\chi^*(\Phi_1)+\chi^*(\Phi_2)}.
$$
This argument does not restrict the joint supremum to product ensembles and does not require the individual suprema to be attained.
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