Suppose $\rho'_{AB} \approx_\varepsilon \rho_{AB}$ in trace distance. Is there an explicit construction of some state $\tilde{\rho}_{AB}$ using $\rho'_{AB}, \rho'_A, \rho'_B, \rho_A$ and $\rho_B$ (but not $\rho_{AB}$) such that
$$\tilde{\rho}_A = \rho_A, \tilde{\rho}_B = \rho_B$$
while ensuring that $\tilde{\rho}_{AB}$ is as close in trace distance to $\rho'_{AB}$ as possible?
To get some obvious answers out of the way, we have the trivial solution of $\tilde{\rho} = \rho_A\otimes\rho_B$. This satisfies the marginal condition but probably is not optimal in terms of the trace distance between $\tilde{\rho}$ and $\rho'$.
A much cleverer attempt is shown in this answer to my previous question. Take
$$\tilde{\rho}_{AB} = \rho'_{AB} + (\rho_A - \rho'_A) \otimes \rho'_B + \rho'_A \otimes (\rho_B - \rho'_B) $$
but this construction is not guaranteed to be positive semidefinite for entangled pure states. Apart from that, it satisfies the requirement on the marginals and keeps $\tilde{\rho}$ close to $\rho'$.
Is there anything else one can try?