The rate is information plus mismatch, for any encoder

Insert $q(\vec z)$ into the log-ratio: $\log\frac{q(\vec z\mid n)}{p(\vec z)} =\log\frac{q(\vec z\mid n)}{q(\vec z)}+\log\frac{q(\vec z)}{p(\vec z)}$.

The identity. Average the split in the hint over $q(n,\vec z)=q(n)\,q(\vec z\mid n)$:

In the first term, $q(\vec z\mid n)/q(\vec z)=q(n,\vec z)/\big(q(n)\,q(\vec z)\big)$, so it is $\KL(q(n,\vec z)\,\|\,q(n)q(\vec z))=I_q(n;\vec z)$ (Mutual information · 0.IT.04). The second term depends on $\vec z$ alone, and the marginal of $q(n,\vec z)$ over $\vec z$ is $q(\vec z)$, so it is $\KL(q(\vec z)\,\|\,p(\vec z))$. $\blacksquare$

Zero rate means no information. Both terms on the right are non-negative. If the left side is zero, both are zero. Then $I_q(n;\vec z)=0$: the code is independent of the example, so it tells the decoder nothing about which input it came from. Equivalently, each $\KL(q(\vec z\mid n)\,\|\,p)$ is non-negative and their average is zero, so each is zero and every $q(\vec z\mid n)$ is the prior. That is posterior collapse.

The rate bounds the information. Dropping the non-negative second term gives $I_q(n;\vec z)\leq\frac1N\sum_n\KL(q(\vec z\mid n)\,\|\,p)$. The rate is an upper bound on what the code says about the example. Since $n$ takes $N$ values, the information is also at most $\log N$; any rate above that is spent on the mismatch.

The identity holds for every encoder, not only at an optimum. The script checks it for three encoders on the four points, including the collapsed one, where both sides are zero.