Write $S_1=H(X)+H(Y)+H(Z)$ and $S_2=H(X,Y)+H(Y,Z)+H(Z,X)$. By the definition of <mutual information>,
$$
I(X;Y)+I(Y;Z)+I(Z;X)=2S_1-S_2,
$$
so the required right-hand side is $(2S_2-S_1)/3$.

Apply <entropy submodularity> to the pairs $(X,Y),(X,Z)$ and then cyclically permute the variables:
$$
\begin{aligned}
H(X,Y)+H(X,Z)&\geq H(X)+H(X,Y,Z),\\
H(Y,Z)+H(Y,X)&\geq H(Y)+H(X,Y,Z),\\
H(Z,X)+H(Z,Y)&\geq H(Z)+H(X,Y,Z).
\end{aligned}
$$
Adding gives $2S_2\geq S_1+3H(X,Y,Z)$, which is exactly
$$
H(X,Y,Z)\leq\frac12S_2-\frac16(2S_1-S_2).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 164 / 1 / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-164/1/ii/solution