Yes. Write $D_S=D(Q_{X_S}\Vert P_{X_S})$ and $D=D_{\{1,\ldots,n\}}$. The bounds from b and d are respectively
$$
B_{\mathrm{old}}=nD-\sum_iD_{X^{(i)}},
\qquad
B_{\mathrm{new}}=\frac{nD-\sum_iD_{\{i\}}}{n-1}.
$$
The , obtained by repeated <entropy submodularity>, states
$$
(n-1)\sum_iH(X^{(i)})
\geq n(n-2)H(X_{1:n})+\sum_iH(X_i).
$$
After replacing entropies by divergences from the product reference, whose cross-entropy terms cancel, this is exactly $B_{\mathrm{new}}\leq B_{\mathrm{old}}$. Equality holds for product $Q$ and when $n=2$; dependence can make the new bound strictly smaller.

Solution

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

{c}

The strong form
$$
(n-1)\sum_iH(X^{(i)})
\geq n(n-2)H(X_{1:n})+\sum_iH(X_i)
$$
follows by iterating <entropy submodularity>.

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