# Proving $S(A:B)+S(A:C) \le 2S(A)$ using subadditivity

I am fairly new to this, but my attempt below involves using a fourth system R to purify.

$$S(A,B,C)+S(B) \le S(A,B)+S(B,C),$$ introducing a system R which purifies ABC, we get $$S(A,B,R)+S(B) \le S(A,B)+S(B,R).$$

$$S(C)+S(B) \le S(A,B)+S(A,C)\\ \Longrightarrow S(A|C)+S(A|B) \ge0 \\\Longrightarrow -S(A|C)-S(A|B) \le 0$$

$$S(A:C)-S(A)+S(A:B)-S(A) \le 0 \to S(A:C)+S(A:B) \le 2S(A)$$

I'm not sure if my reasoning here is correct. Is is valid to do what I did by subbing in the system R? I am still getting used to proving things in this manner. S is the Von Neumann Entropy, and the individual systems are d-dimensional quantum systems.

• What kind of thing are $A$, $B$, and $C$? What does $S$ stand for? What do you mean by system? Aug 11, 2020 at 11:21
• Added info in edit. S is the Von Neumann Entropy, A, B and C are quantum systems. Aug 11, 2020 at 14:13
• Not sure about your question. Your inequality follows directly from $S(B)+S(C) \le S(A,B) + S(A,C)$, which is one well-known statement of the strong subadditivity property of von Neumann entropy (see en.wikipedia.org/wiki/Von_Neumann_entropy#Strong_subadditivity). If you use this inequality, there is no need to introduce another system $R$ Aug 18, 2020 at 17:45
• Yeah I used that inequality in my proof. But the way the question was phrased to me, it sounded like they wanted you to start from $S(A,B,C)+S(B) \le S(A,B)+S(B,C)$. I think I got fixated on showing it from this point. I take it that using R is valid, but un-needed as I can just go straight from the 3rd inequality? Aug 19, 2020 at 11:03