The Markov property gives $I(X_1;X_4\mid X_3)=I(X_2;X_4\mid X_3)=0$. The chain rule therefore yields
$$
I(X_1;X_3)-I(X_1;X_4)=I(X_1;X_3\mid X_4)
$$
and
$$
I(X_2;X_3)-I(X_2;X_4)=I(X_2;X_3\mid X_4).
$$
Conditional on $X_4$, the factorization of the chain still gives $X_1\to X_2\to X_3$. The <conditional data-processing inequality> thus gives
$$
I(X_1;X_3\mid X_4)\leq I(X_2;X_3\mid X_4).
$$
Rearranging proves
$$
I(X_1;X_3)+I(X_2;X_4)
\leq I(X_1;X_4)+I(X_2;X_3).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 224 / 1 / d / Solution

If $X\to Y\to Z$ is a Markov chain conditionally on $W$, then
$$
I(X;Z\mid W)\leq I(X;Y\mid W).
$$
It follows by applying ordinary data processing inside every conditional distribution and averaging over $W$.


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 Incoming links: Conditional data-processing inequality