= Solution
For each $x$, choose purifications $|u_x\rangle$ of $\rho_x$ and $|v_x\rangle$ of $\sigma_x$ in a common reference space. By <Uhlmann's theorem>, the second can be chosen, including its overall phase, so that
$$
\langle u_x|v_x\rangle=F(\rho_x,\sigma_x)\geq0.
$$
Construct the two <flagged purifications of a quantum ensemble>
$$
|U\rangle=\sum_x\sqrt{p_x}|u_x\rangle|x\rangle,\qquad
|V\rangle=\sum_x\sqrt{p_x}|v_x\rangle|x\rangle.
$$
Their reduced states are the respective mixtures, and orthogonality of the flags gives $\langle U|V\rangle=\sum_xp_xF(\rho_x,\sigma_x)$. A particular purification overlap cannot exceed the maximizing overlap in <Uhlmann's theorem>. Hence
$$
\boxed{F\left(\sum_xp_x\rho_x,\sum_xp_x\sigma_x\right)\geq\sum_xp_xF(\rho_x,\sigma_x).}
$$
This proves <joint concavity of quantum fidelity>. Choosing each overlap nonnegative prevents cancellation of different phases; the same probability weights in the two mixtures yield $p_x$ rather than distinct square-root weights.
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