# How to prove that the distillable entanglement satisfies $E_D(|\psi_d\rangle\!\langle\psi_d|)\ge \log d$?

If I have a pure state vector: $$\left|\psi_d\right\rangle=\frac{1}{\sqrt{d}}\sum^{d}_{i=1}\left|ii\right\rangle$$ then the distillable entanglement satisfies: $$E_{D}(\left|\psi_d\right\rangle\langle\psi_d|)\geq\log d.$$ I am not seeing how to prove this as I thought if I have a pure state then the distillable Entanglement is equal to the entanglement entropy which is $$\leq\log d$$.

The distillable entanglement of this state is equal $$\log d$$ by definition. Thus, both $$E_D(\lvert \psi_d\rangle\langle\psi_d\rvert)\ge \log d$$ and $$E_D(\lvert \psi_d\rangle\langle\psi_d\rvert)\le \log d$$ holds true.
More generally, the distillable entanglement of a pure state is equal to the Shannon entropy $$-\sum p_k\log p_k$$ of its Schmidt coefficients $$p_k$$ (equivalently, the von Neumann entropy of either reduced density matrix) by definition. In the case of the state you give, this entropy equals $$\log d$$.