= Solution
Use the unsquared convention for <quantum fidelity>:
$$
F(\tau,\omega)=\operatorname{Tr}\sqrt{\sqrt\tau\,\omega\sqrt\tau}.
$$
Choose a <purification of a density operator> $|\Psi\rangle_{RA}$ of $\rho$, and put $\tau_{RA}=(\operatorname{id}_R\otimes\Lambda)(|\Psi\rangle\langle\Psi|)$. By definition,
$$
F_e(\rho,\Lambda)=\langle\Psi|\tau_{RA}|\Psi\rangle
=F(|\Psi\rangle\langle\Psi|,\tau_{RA})^2.
$$
The <monotonicity of quantum fidelity under partial trace> follows from <Uhlmann's theorem>: maximizing purifications of two joint states are also candidate purifications of their marginals, with the discarded subsystem included in the purifying system. The optimization for the marginals therefore cannot give a smaller overlap. Here those marginals are $\rho$ and $\Lambda(\rho)$. Consequently
$$
\boxed{F_e(\rho,\Lambda)\leq F(\rho,\Lambda(\rho))^2.}
$$
This <entanglement fidelity bound by state fidelity> distinguishes preserving a reference entanglement from merely reproducing the input marginal.
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