# Why is the entanglement of formation upper bounded by the Schmidt number?

I have read many times in several articles (such as https://arxiv.org/abs/1609.05033) that the entanglement of formation EoF puts a lower bound on entanglement dimensionality $$d$$ (i.e., the Schmidt number):

$$\log_2d\geq E_{\rm{oF}}.$$

However, I was not able to find any proof of this fact. How would one go about proving it?

For a pure state $$\psi_{AB}$$ with Schmidt rank $$r$$ the maximum of $$E_{oF}(\psi_{AB})=H({\rm tr}_B(|\psi_{AB}\rangle\langle\psi_{AB}|))=-\sum_{k=1}^r\lambda_k\log_2\lambda_k$$ is known to equal $$\log_2(r)$$, achieved on a maximally entangled state (i.e. $$\lambda_k=\frac1r$$). For mixed states $$E_{oF}(\rho_{AB}):=\min \sum_k\lambda_kE_{oF}(\psi_{AB}^k)$$ where the minimum is taken over all pure state decompositions (and this minimum is achieved somewhere!) meaning the above bound carries over: $$\sum_k\lambda_kE_{oF}(\psi_{AB}^k)\leq \Big(\sum_k\lambda_k\Big)\log_2(\max_k r_k)=\log_2(\max_k r_k)\leq\log_2(\min\{\dim\mathcal H_A,\dim\mathcal H_B\})$$ For further reading refer, e.g., to Chapter 9.1.1.1 in the book of Khatri & Wilde.