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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.

Put $\rho =p{\rho }_1+(1-p){\rho }_2$. The operator inequality $\rho \ge p{\rho }_1$ and the operator monotonicity of logarithm give, on the support of ${\rho }_1$,Consequently,The corresponding inequality from $\rho \ge (1-p){\rho }_2$ yieldsAdding these inequalities and using $S(\rho )=-\mathrm{Tr}(\rho \log \rho )$ proves the entropy bound for a binary mixture:Singular states follow by adding a positive multiple of the identity and taking a limit; the endpoint cases use $0\log 0=0$.

Solved by gpt-5.6-sol high.

Set $q=(1+\epsilon {)}^{-1}$, so $1-q=\epsilon /(1+\epsilon )$ and $\omega =q\rho +(1-q){\Delta }^{\prime }$. For every state ${\xi }_B$, the entropy bound for a binary mixture givesTaking the minimum over ${\xi }_B$ and using the variational characterization of quantum conditional entropy on each term givesSince binary entropy satisfies $H(q)=H(1-q)$, this is

Solved by gpt-5.6-sol high.