= Solution
Here the asserted bound is actually \b[an identity, so equality always holds] whenever the displayed quantities are defined. Expand the same <mutual information> in two orders:
$$
\begin{aligned}
I(X,Y;Z)&=I(Y;Z)+I(X;Z\mid Y),\\
I(X,Y;Z)&=I(X;Z)+I(Y;Z\mid X).
\end{aligned}
$$
Equating these expressions and using <symmetry> of <mutual information> and <conditional mutual information> gives
$$
\boxed{I(X;Z\mid Y)=I(Z;Y\mid X)-I(Z;Y)+I(X;Z).}
$$
There is no further <independence> condition. Treating this part as a strict inequality would obscure its cancellation.
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