# Conditional version of the triangle inequality for Von Neumann entropy

I'm trying to solve problem 11.3 in Nielsen Chuang:

(3) Prove the conditional version of the triangle inequality: $$S(A,B|C)\geq S(A|C)-S(B|C)$$

But the inequality seems incorrect. For example, let $$|\psi\rangle = \frac{|0\rangle|0\rangle|00\rangle+|0\rangle|1\rangle|01\rangle+|1\rangle|0\rangle|10\rangle+|1\rangle|1\rangle|11\rangle}{2},\\ \rho^{ABC} = |\psi\rangle\langle\psi|.$$ Then: $$\rho^{C}=\frac{I}{4},\\ \rho^{AC}=\rho^{BC}=\frac{I}{2},\\ S(A,B|C)=S(A,B,C)-S(C)=0-2,\\ S(A|C)=S(B|C)=1-2=-1,\\ S(A,B|C)=-2<0=S(A|C)-S(B|C).$$

Am I doing something wrong, or did the problem mean to say $$S(A,B)\geq S(A|C)-S(B|C)$$ or something?

You've computed $$\rho^{AC},\rho^{BC}$$ incorrectly $$-$$ they are matrices of size $$8\times 8$$, also they are not equal to $$I/8$$. Nevertheless their entropy is equal, i.e. $$S(A,C) = S(B,C)$$. So $$|\psi \rangle$$ is indeed a counterexample to the statement.
The inequality $$S(A,B)\geq S(A|C)-S(B|C)$$, which is equivalent to $$S(A,B) + S(B,C) \geq S(A,C)$$ is probably what they had in mind.