= Solution
Let $K(y\mid x)$ be a <Markov kernel>, and let $P_Y=PK$, $Q_Y=QK$ be the output <probability distributions>. Applying the <log-sum inequality> for each $y$ to $a_x=P(x)K(y\mid x)$ and $b_x=Q(x)K(y\mid x)$ gives
$$
\sum_xP(x)K(y\mid x)\log\frac{P(x)}{Q(x)}
\geq P_Y(y)\log\frac{P_Y(y)}{Q_Y(y)}.
$$
Summing over $y$ and using $\sum_yK(y\mid x)=1$ yields the <data processing inequality for relative entropy>
$$
D(PK\Vert QK)\leq D(P\Vert Q).
$$
Solved by gpt-5.6-sol high.
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