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If $\epsilon =\Vert {\rho }_{AB}-{\sigma }_{AB}{\Vert }_1/2$ and $d_A=\dim {\mathcal{H}}_A$, thenThe proof couples the two states through their positive and negative differences, then combines concavity of quantum conditional entropy with the entropy bound for a binary mixture.

Because the two flags $E_C$ and $F_C$ are orthogonal projections, the flagged state is block diagonal. If $h(\lambda )$ denotes the binary entropy, thenandIts unflagged marginals are ${\rho }_{AB}=\lambda {\rho }_{AB}^{\prime }+(1-\lambda ){\rho }_{AB}^{\prime \prime }$ and ${\rho }_B=\lambda {\rho }_B^{\prime }+(1-\lambda ){\rho }_B^{\prime \prime }$. Substitution in Strong subadditivity of Von Neumann entropy cancels the two binary-entropy terms and givesThis is exactly the concavity of quantum conditional entropy.

The other convex decomposition is $\omega =q\sigma +(1-q)\Delta$. Applying the concavity of quantum conditional entropy, which follows from the Strong subadditivity of Von Neumann entropy, yields