Choose the <trace-distance-preserving binary measurement> from the previous part. Let $p,q$ be its output probability distributions, so $\lVert p-q\rVert_1=2D(\rho,\sigma)$. The <data-processing inequality for quantum relative entropy> gives
$$
D(\rho\|\sigma)\geq D(\Phi(\rho)\|\Phi(\sigma))=D(p\|q).
$$
For diagonal <density operators>, the quantum expression is the classical <Kullback-Leibler divergence>. Applying the supplied <Pinsker's inequality> gives
$$
D(p\|q)\geq\frac1{2\ln2}\lVert p-q\rVert_1^2
=\frac2{\ln2}D(\rho,\sigma)^2.
$$
Consequently the <quantum Pinsker inequality> is
$$
\boxed{D(\rho\|\sigma)\geq\frac2{\ln2}D(\rho,\sigma)^2}.
$$
The $\ln2$ factor converts natural-logarithm relative entropy to bits. If the relative entropy is infinite because its support condition fails, the inequality holds automatically.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 323 / 4 / iv / Solution

In bits, the quantum Pinsker inequality is
$$
D(\rho\|\sigma)\geq\frac{2}{\ln2}D(\rho,\sigma)^2.
$$
The <trace-distance-preserving binary measurement> converts the <trace distance> into classical <total variation distance> without loss. Apply the <data-processing inequality for quantum relative entropy> to that measurement, then classical <Pinsker's inequality>. The factor $\ln2$ is omitted when relative entropy uses natural <logarithms>.

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