= Solution
The construction is a <convex roof extension>. First establish its <convexity>, without assuming that the minimizing <quantum state ensemble> exists. If $\tau=\sum_jw_j\tau_j$, choose a <pure state> ensemble for each $\tau_j$ whose average value of $\mu$ is within $\varepsilon$ of $F(\tau_j)$. Combining these ensembles, with their weights multiplied by $w_j$, gives a <pure state> ensemble for $\tau$. Taking arbitrarily small $\varepsilon>0$ proves
$$
F\left(\sum_jw_j\tau_j\right)\leq\sum_jw_jF(\tau_j).
$$
Also, \b[$F(\pi)=\mu(\pi)$ for a pure state $\pi$]. Indeed, if $\pi=\sum_jp_j\pi_j$ and $v$ is orthogonal to its one-dimensional support, then $0=\sum_jp_j\langle v|\pi_j|v\rangle$. All terms are nonnegative, so every positive-weight $\pi_j$ has that same support and equals $\pi$.
Represent outcome $k$ of the operation by a linear unnormalized map $\mathcal I_k$, as for a <quantum instrument>. Normalization is performed only after applying this map: $\mathcal I_k(\rho)=P_k\rho_k$. For any preparation ensemble $\rho=\sum_jp_j\pi_j$, put
$$
\mathcal I_k(\pi_j)=q_{k|j}\sigma_{jk},\qquad
P_k=\sum_jp_jq_{k|j}.
$$
Linearity gives $P_k\rho_k=\sum_jp_jq_{k|j}\sigma_{jk}$. For $P_k>0$, <Bayes' theorem> identifies the conditional preparation weights
$$
w_{j|k}=\frac{p_jq_{k|j}}{P_k},\qquad \rho_k=\sum_jw_{j|k}\sigma_{jk}.
$$
Outcomes with zero <probability> contribute nothing; normalized states on such branches need not be defined. The conditional states $\sigma_{jk}$ may be mixed, so convexity of the <convex roof> is essential here.
Applying that convexity separately for each recorded outcome, and then the assumed average monotonicity for each pure preparation, yields
$$
\begin{aligned}
\sum_kP_kF(\rho_k)
&\leq\sum_{j,k}p_jq_{k|j}F(\sigma_{jk})\\
&\leq\sum_jp_jF(\pi_j)
=\sum_jp_j\mu(\pi_j).
\end{aligned}
$$
The left side depends only on the initial <density operator> and the operation, not on the chosen preparation ensemble. Taking the <infimum> of the last expression over all its <pure state> ensembles proves
$$
\boxed{F(\rho)\geq\sum_kP_kF(\rho_k).}
$$
Equivalently, choose an ensemble within $\varepsilon$ of the <infimum> and then let $\varepsilon\downarrow0$. This is the <monotonicity of a convex roof under a quantum instrument>. It uses linearity of the unnormalized branches, not a false assumption that normalized conditional states depend linearly on the input. As usual, the averages are assumed well defined; finite-valued roofs and finitely many outcomes suffice. If an input roof is $-\infty$, taking ensembles with averages tending to $-\infty$ gives the same conclusion in the extended-real interpretation whenever the output average is defined.
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