# How to prove the positivity of the conditional quantum mutual information, $I(A;B|C)\ge0$?

I was reading Wilde's 'Quantum Information Theory' and saw the following theorem at chapter 11 $$(11.7.2)$$:

$$I(A; B | C) \ge 0,$$ where, $$I(A;B|C) := H(A|C) + H(B | C) - H(AB|C).$$

I know that the mutual information is non-negative, i.e. $$I(A;B) \ge 0,$$ where, $$I(A;B) := H(A) + H(B) - H(AB).$$ Now if we have access to an additional subsystem $$C$$, this can't decrease the mutual information of $$A$$ and $$B$$. But I was looking for sort of formal proof of this. I was trying to apply the non-negativity of mutual info. into this, but not sure how to proceed. Thanks in advance!

Here's a relatively simple proof just based on the data processing inequality (DPI) for the relative entropy $$D(\rho\|\sigma) = \mathrm{tr}[\rho (\log \rho - \log \sigma)]$$ -- if you're willing to accept the DPI as a basis for a formal proof. Recall that the DPI says that for any channel $$\Phi$$ we have $$D(\rho \|\sigma) \geq D(\Phi(\rho)\|\Phi(\sigma)).$$
Now \begin{aligned} I(A:B|C) &= H(A|C) + H(B|C) - H(AB|C) \\ &= H(AC) - H(C) + H(BC) - H(C) - H(ABC) + H(C) \\ &= H(AC) + H(BC) - H(ABC) - H(C) \\ &= H(A|C) - H(A|BC) \\ &= -D(\rho_{AC} \| I_A \otimes \rho_C) + D(\rho_{ABC} \| I_A \otimes \rho_{BC}). \end{aligned} Thus $$I(A:B|C) \geq 0$$ is equivalent to $$D(\rho_{ABC} \|I_A \otimes \rho_{BC}) \geq D(\rho_{AC} \|I_A \otimes \rho_C),$$ but this follows immediately from the DPI by taking the channel $$\Phi$$ to be the partial trace over the $$B$$ system.
To expand on @Purva Tharke's comment, the strong subadditivity inequality states: $$H(ABC)+H(C) \le H(AC) + H(BC)$$ $$=H(ABC)+H(C) +H(C) -H(C) \le H(AC) + H(BC)$$ $$=H(AB|C) \le H(A|C) + H(B|C)$$ $$=0\le H(A|C) + H(B|C) - H(AB|C)=H(A;B|C)$$