Solution
= Solution
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
$$
I(X_1;X_3)+I(X_2;X_4)
\leq I(X_1;X_4)+I(X_2;X_3).
$$