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# Concavity of Conditional Quantum Entropy

Let's say I have a bipartite density operator $$\gamma_{12} = (1 - \epsilon) \rho_{12} + \epsilon\sigma_{12}$$, for $$0 \le \epsilon \le 1$$, i.e., a convex combination of $$\rho_{12}$$ and $$\sigma_{12}$$. I want to show that ($$S$$ represents Von Neumann entropy):

$$S(\gamma_{12} | \gamma_2) \ge (1 - \epsilon) S(\rho_{12} | \rho_2) + \epsilon S( \sigma_{12} | \sigma_2).$$ The note that I am following says that this is due to the concavity of conditional entropy, which is not immediately obvious to me. I tried to derive it in the following way:

\begin{align} S(\gamma_{12} | \gamma_2) &= S(\gamma_{12}) - S(\gamma_2) \\ &= S((1 - \epsilon) \rho_{12} + \epsilon\sigma_{12}) - S((1 - \epsilon)\rho_2 + \epsilon\sigma_2) \;\; \text{[definition of \gamma_{12} and using partial trace] } \\ &\ge (1 - \epsilon) S(\rho_{12}) + \epsilon S(\sigma_{12}) - S((1 - \epsilon)\rho_2 + \epsilon\sigma_2) \;\; \text{[using concavity in the first S] } \\ &\stackrel{?}{\ge} (1 - \epsilon) S(\rho_{12}) + \epsilon S(\sigma_{12}) - (1 - \epsilon)S(\rho_2) - \epsilon S(\sigma_2). \end{align} This of course, gives me the desired inequality. But how come the last inequality is true? Isn't $$S((1 - \epsilon)\rho_2 + \epsilon\sigma_2) \ge (1 - \epsilon)S(\rho_2) - \epsilon S(\sigma_2)$$ due to concavity? Thanks!