The elementary <logarithm inequality> $\log_2u\leq(\log_2e)(u-1)$ gives
$$
D(P\Vert Q)
\leq(\log e)\sum_xP(x)\left(\frac{P(x)}{Q(x)}-1\right).
$$
Since $\sum_x(P(x)-Q(x))=0$,
$$
\sum_xP(x)\frac{P(x)-Q(x)}{Q(x)}
=\sum_x\frac{(P(x)-Q(x))^2}{Q(x)}
=\chi^2(P\Vert Q).
$$
This proves $D(P\Vert Q)\leq(\log e)\chi^2(P\Vert Q)$, relating <relative entropy> to <chi-squared divergence>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 224 / 3 / c / Solution

Put $L=L^*(X)$ and $\mu=\mathbb E L$. Since $L$ is a function of $X$, the <chain rule for information entropy> and part (c) give
$$
H(X)=H(L)+H(X\mid L)\leq H(L)+\mu.
$$
Part (a), applied to the nonnegative integer-valued random variable $L$, yields
$$
H(L)\leq(1+\mu)h\left(\frac1{1+\mu}\right)
=\log(1+\mu)+\mu\log\left(1+\frac1\mu\right).
$$
The <logarithm inequality> $\log(1+t)\leq t\log e$ implies $\mu\log(1+1/\mu)\leq\log e$. Thus
$$
H(X)\leq\mu+\log(1+\mu)+\log e.
$$
Part (c) also gives $\mu\leq H(X)$, so monotonicity of the <logarithm> lets us replace $\log(1+\mu)$ by $\log(H(X)+1)$. Rearranging proves
$$
\mathbb E[L^*(X)]
\geq H(X)-\log[H(X)+1]-\log e.
$$

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 224 / 4 / d / Solution

For $x>0$, <concavity> of the <natural logarithm> gives
$$
\ln x\leq x-1.
$$
Equivalently, $\log_bx\leq(\log_be)(x-1)$ for $b>1$.

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