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I want to prove the subadditivity relation $S(\rho_{AB})\le S(\rho_A)+S(\rho_B)$ for the Von Neumann entropy. The tip is to use the Klein inequality $S(\rho_{AB}\Vert \rho_A\otimes \rho_B)\ge 0$: $$S(\rho_{AB}\Vert \rho_A\otimes \rho_B)=\text{tr}[\rho_{AB}\log \rho_{AB}]-\text{tr}[\rho_{AB}\log(\rho_A\otimes\rho_B)]\ge 0$$ implies that $$ S(\rho_{AB})\le -\text{tr}[\rho_{AB}\log(\rho_A\otimes \rho_B)].$$ (the trace is intended on $AB$). Introducing the spectral decompositions $\rho_A=\sum_l \lambda_l|l\rangle\langle l|_A$ and $\rho_B=\sum_k \chi_k|k\rangle\langle k|_B$ I calculated \begin{align} \log(\rho_A\otimes\rho_B)&=\sum_{lk}\log(\lambda_l \chi_k)|l\rangle\langle l|_A\otimes |k\rangle\langle k|_B \\ &=\sum_{lk}\log\lambda_l|l\rangle\langle l|_A\otimes |k\rangle\langle k|_B+\sum_{lk}\log\chi_k|l\rangle\langle l|_A\otimes |k\rangle\langle k|_B \\ &= \log\rho_A\otimes 1_B+1_A\otimes\log\rho_B \end{align} where I've made use of the completeness relations on $A$ and $B$. Now, if $\rho_{AB}=\rho_A\otimes\rho_B$ then we have simply $$ S(\rho_{AB})\le -\text{tr}(\rho_A\log\rho_A\otimes\rho_B)-\text{tr}(\rho_A\otimes\rho_B\log\rho_B)=S(\rho_A)+S(\rho_B).$$ But what if $\rho_{AB}$ is not separable?

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You are very close.

$$ S(\rho_{AB}\Vert \rho_A\otimes \rho_B)=\text{tr}[\rho_{AB}\log \rho_{AB}]-\text{tr}[\rho_{AB}\log(\rho_A\otimes\rho_B)]\ge 0 $$

Let's continue from here. From this link:

The second term on the left side of the inequality can be written as:

$$ \text{Tr}[\rho_{AB}(\log \rho_A \otimes \mathbb{I} + \mathbb{I} \otimes \log \rho_B)] \\ = \text{Tr}[\rho_A \log \rho_A] + \text{Tr}[\rho_B \log \rho_B] \\ = -S(\rho_A) - S(\rho_B) $$

You can see why the first equality follows from this link (see Watrous's answer). Then from your quoted equation:

$$ -S(\rho_{AB}) + S(A) + S(B) \ge 0 \\ \implies S(\rho_{AB}) \le S(A) + S(B). $$

There is no longer any assumption on $\rho_{AB}$, whether it is separable or not.

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