= Solution
The <mean> of a shape-rate <gamma distribution> is shape divided by rate. Hence the <posterior mean> is
$$
\mathbb E(\theta\mid x)=\frac{a+n\overline x}{b+n}
=\frac b{b+n}\frac ab+\frac n{b+n}\overline x.
$$
Both weights are positive and sum to one. This is a <convex combination> of the <prior mean> and the <sample mean>, proving
$$
\boxed{\min\{a/b,\overline x\}\leq\mathbb E(\theta\mid x)\leq\max\{a/b,\overline x\}.}
$$
For $n\geq1$ it lies strictly between them if the two means differ, and equals both if they agree. The rate $b$ acts as the prior weight relative to the $n$ observed exposure units.
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