= Solution
Let a <density operator> on a <Hilbert space> $\mathcal H_S$ have the <spectral decomposition>
$$
\rho_S=\sum_{j=1}^r\lambda_j|j\rangle_S\langle j|,
\qquad\lambda_j>0,\qquad\sum_j\lambda_j=1.
$$
Introduce an auxiliary <Hilbert space> $\mathcal H_R$ with an <orthonormal basis> containing $r$ vectors $|j\rangle_R$, and set
$$
|\Omega\rangle_{SR}=\sum_{j=1}^r\sqrt{\lambda_j}|j\rangle_S|j\rangle_R.
$$
This vector has norm one. Its <partial trace> is
$$
\operatorname{Tr}_R|\Omega\rangle\langle\Omega|
=\sum_{j,k}\sqrt{\lambda_j\lambda_k}|j\rangle_S\langle k|\langle k|j\rangle_R
=\rho_S.
$$
This explicitly constructs a <purification of a density operator>. The same construction works for an arbitrary trace-class <density operator>: its nonzero <eigenvalues> form a finite or countable summable family, and the displayed vector converges in the <Hilbert space> norm. The entropy inequality below is understood with finite <Von Neumann entropies>, avoiding undefined differences of infinities.
Apply this construction with $S=AB$ to obtain a joint <pure state> on $ABC$. The <Schmidt decomposition> across $AB:C$, $A:BC$ and $B:AC$ respectively gives
$$
S(AB)=S(C),\qquad S(A)=S(BC),\qquad S(B)=S(AC).
$$
Now apply <Subadditivity of Von Neumann entropy> to the <reduced density matrices> on $BC$ and $AC$:
$$
S(A)=S(BC)\leq S(B)+S(C),\qquad
S(B)=S(AC)\leq S(A)+S(C).
$$
These inequalities say $S(A)-S(B)\leq S(C)$ and $S(B)-S(A)\leq S(C)$. Replacing $S(C)$ by $S(AB)$ proves the <Araki–Lieb inequality>:
$$
\boxed{|S(A)-S(B)|\leq S(A,B).}
$$
The subadditivity used here concerns the actual reduced <density operators>, rather than statistical independence. In particular it follows from <nonnegativity of quantum relative entropy>, since
$$
D(\rho_{BC}\Vert\rho_B\otimes\rho_C)
=S(\rho_B)+S(\rho_C)-S(\rho_{BC})\geq0,
$$
and similarly for $AC$. Thus no independence assumption on the <purification of a density operator> is needed.
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