In the answer of this question, the last paragraph says that
If you know one decomposition which is optimal for Entanglement of Formation for a given state $\rho$, you can obtain the optimal decomposition for other states by simply shifting the weights $q_i$ in the optimal decomposition.
I'd like to know how to prove this or is there any references?
We will prove the following statement:
Let $\rho=\sum p_i \rho_i$ ($p_i>0$) be a decomposition of a given state $\rho$ which minimizes the cost function $C(\{p_i,\rho_i\})=\sum p_i f(\rho_i)$ for that state. Then, $\rho'= \sum p_i' \rho_i$ is a decomposition which minimizes $C$ for the state $\rho'$.
Note. In the case of entanglement of formation, $f(\rho_i)=S(\mathrm{tr}_B\,\rho_i)$.
Proof. We will prove this by contradiction. Assume that there exists a decomposition $\rho'=\sum q_j \sigma_j$ for which $$\sum q_j\,f(\sigma_j)<\sum p_i'\,f(\rho_i)\ .$$ Choose a $\lambda>0$ s.th. $r_i:=p_i-\lambda p_i'\ge0$ for all $i$. Then, \begin{align} C(\{p_i,\rho_i\}) & = \sum_i p_i f(\rho_i)\\ &=\lambda\left[\sum_i p_i'\, f(\rho_i)\right] + \sum_i r_i\, f(\rho_i)\\ &>\lambda\left[\sum_j q_j\,f(\sigma_j)\right]+ \sum_i r_i\, f(\rho_i)\\ &=C(\{w_k,\tau_k\}) \end{align} with the ensemble $\{w_k,\tau_k\}$ the union of the ensembles $\{\lambda q_j,\sigma_j\}$ and $\{r_i,\rho_i\}$, which is in contradiction to the assumption that $\rho=\sum p_i\rho_i$ is the optimal decomposition for $\rho$. $\Box$
As far as I am aware, this is a well-known result, but I don't know where this is written down.
Late edit: I just stumbled across the result: It is proven in Sec. IV B of K.~G. H. Vollbrecht and R. F. Werner, Phys. Rev. A 64, 062307 (2001) (also quant-ph/0010095).
