The triangle inequality or Araki-Lieb inequality of the von Neumann entropy is $$ S(A,B)\ge|S(A)-S(B)| $$ this is proven by introducing a system $R$ which purifies systems $A$ and $B$. Applying subadditivity obtains $S(R)+S(A)\ge S(A,R)$.
$ABR$ is in a pure state $\implies $ $S(A,R)=S(B)$ and $S(R)=S(A,B)$. Therefore,
\begin{align} S(R)+S(A)&=S(A,B)+S(A)\ge S(A,R)=S(B)\\ &\implies S(A,B)+S(A)\ge S(B)\\ &\implies S(A,B)\ge S(B)-S(A) \end{align}
In Page 516, Quantum Computation and Quantum Information by Nielsen and Chuang, Exercise 11.16 is given as
Let $ρ^{AB}=\sum_i λ_i|i\rangle\langle i|$ is a spectral decomposition for $ρ^{AB}$. Show that $S(A,B)=S(B)−S(A)$ if and only if the operators $ρ^A_i≡\text{tr}_B(|i\rangle\langle i|)$ have a common eigenbasis, and the $ρ^B_i≡\text{tr}_A(|i\rangle\langle i|)$ have orthogonal support.
How do I approach the problem, and are there any physical explanations for the equality conditions?
My Attempt
Klein inequality:$$S(\rho||\sigma)=-S(\rho)-\text{tr}(\rho\log\sigma)= 0\implies \rho=\sigma\,.$$ Setting $\rho=\rho^{AR}$ and $\sigma=\rho^A\otimes\rho^R$ then \begin{align} S(\rho^{AR}||\rho^A\otimes\rho^R)&=-S(\rho^{AR})-\text{tr}(\rho^{AR}(\log\rho^A\otimes I_R+I_A\otimes\log\rho^R))\\ &=-S(\rho^{AR})-\text{tr}((\rho^{AR}(\log\rho^A\otimes I_R)+-\text{tr}((I_A\otimes\log\rho^R))\\ &=-S(\rho^{AR})-\text{tr}(\rho^A\log\rho^A)-\text{tr}(\rho^R\log\rho^R)\\ &=-S(\rho^{AR})+S(\rho^A)+S(\rho^R)\\ &=-S(AR)+S(A)+S(R)\\ \end{align} We used $$\log(\rho^A\otimes\rho^R)=\log\rho^A\otimes I_R+I_A\otimes\log\rho^R$$ $$\text{tr}(\rho^{AR}(\sigma^A\otimes I))=tr(\rho^A\sigma^A)$$ Please check Partial trace over a product of matrices - one factor is in tensor product form for proof.
Therefore, $$S(\rho^{AR}||\rho^A\otimes\rho^R)=0\implies S(A,R)= S(A)+S(R)\implies \rho^{AR}=\rho^A\otimes\rho^R$$
$$ S(A,R)=S(A)+S(R)\implies S(A,B)=S(B)−S(A)\implies ρ^{AR}=ρ^A\otimes ρ^R\\ \rho^A=tr_B(\rho^{AB})=\sum_i λ_itr_B(|i\rangle\langle i|)\\ \rho^B=tr_A(\rho^{AB})=\sum_i λ_itr_A(|i\rangle\langle i|)\\ $$
