The chain rule, now separating $X_i$ first, gives
$$
D(Q\Vert P)
=D(Q_{X_i}\Vert P_{X_i})
+D(Q_{X^{(i)}\mid X_i}\Vert P_{X^{(i)}}\mid Q_{X_i}),
$$
where the second term denotes the conditional divergence averaged over $Q_{X_i}$. Summing over $i$ yields
$$
\sum_iD(Q_{X^{(i)}\mid X_i}\Vert P_{X^{(i)}}\mid Q_{X_i})
=nD(Q\Vert P)-\sum_iD(Q_{X_i}\Vert P_{X_i}).
$$
Repeated use of the chain rule and convexity gives the <tensorization lower bound for relative entropy>
$$
\sum_iD(Q_{X_i}\Vert P_{X_i})\leq D(Q\Vert P).
$$
Therefore the preceding sum is at least $(n-1)D(Q\Vert P)$, which is equivalent to the required inequality.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 208 / 2 / c / Solution

For a product reference $P=\bigotimes_iP_i$ and arbitrary $Q$,
$$
D(Q\Vert P)\geq\sum_iD(Q_i\Vert P_i).
$$
It follows from the <chain rule for relative entropy> and convexity of relative entropy in its first argument.

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