# Can't understand how $D_{A,B} + D_{A,C} \ge E_{A,B}+E_{A,C}$ is proved used the subadditivity of entropy

I am reading Monogamy properties of quantum and classical correlations. Eq.10 states that $$D_{A,B} + D_{A,C} \ge E_{A,B}+E_{A,C},$$ where $$D_{i,j}$$ is the quantum discord, and $$E_{A,B}$$ is the entanglement of formation.

Now by Eq.3 $$S(\rho_{i|k})=E(\rho_{i,l}),$$ where $$S(\rho_{i|k})=min_{E_{j}^{k}}S(i|\{E_{j}^{k}\})$$ Now I assume that the omission in the paper, going forward, of $$\rho$$ in the entanglement of formation is just to save time. So they then state that, when the state becomes mixed, $$S(\rho_{i|k})\ge E(\rho_{i,l})$$. They do this for tripartite systems. i and k are single subsystems, and l in this case is every other subsystem apart from i. So really it should be (I think) $$S(\rho_{A|B})\ge E_{A,BC}$$

So they get that $$D_{i,k}=S(\rho_{k})-S(\rho_{l})+S(\rho_{i|k}) \ge S(\rho_{k})-S(\rho_{l})+E(\rho_{i,l}),$$

so

$$D_{A,B} + D_{A,C} = S(\rho_{B})-S(\rho_{AB})+S(\rho_{C})-S(\rho_{AC})+S(\rho_{A|B})+S(\rho_{A|C}) \\ \ge S(\rho_{B})-S(\rho_{AB})+S(\rho_{C})-S(\rho_{AC})+ E_{A,B}+E_{A,C}.$$

Then they state that, by using the subadditivity of the entropy, $$D_{A,B} + D_{A,C} \ge E_{A,B}+E_{A,C}.$$

I just don't see how this is the case at all. Strong subadditivity has

$$S(\rho_{B})-S(\rho_{AB})+S(\rho_{C})-S(\rho_{AC})\le 0,$$

which means that $$S(\rho_{A|B})+S(\rho_{A|C})$$ have to be at least equivalent to $$-S(\rho_{B})+S(\rho_{AB})-S(\rho_{C})+S(\rho_{AC})+ E_{A,B}+E_{A,C}$$, so that after the subtraction has occurred, the value left $$\ge E_{A,B}+E_{A,C}$$. But I don't see how this is achieved via subadditivity. Moreover, is $$E_{A,BC}\ge E_{A,B}$$, as I know $$E_{A,BC}\ge E_{A,B}+E_{A,C}$$ doesn't necessarily hold? As this would need to be the case for them even to arrive at the inequality they given that applying subadditivity to apparently proves the first one.

I feel like I am missing something very obvious here, but I think I have pidgeon holed my thought process right now.