Step 1: Consider two unbiased estimators of $\tau(\theta)$, one being $\phi$ and the other being $\phi^{*} = E[\phi \mid T]$ built by averaging $\phi$ over the values consistent with a sufficient statistic $T$.
Step 2: Using the variance decomposition $\text{Var}(\phi) = E[\text{Var}(\phi \mid T)] + \text{Var}(E[\phi \mid T])$, the first term on the right is non-negative, so $\text{Var}(\phi) \ge \text{Var}(\phi^{*})$.
Step 3: This inequality holds purely because $T$ is sufficient, sufficiency guarantees $E[\phi \mid T]$ is a computable statistic free of $\theta$.
Step 4: Repeating this conditioning process on any unbiased estimator therefore drives the variance down, and the tool that makes this possible is the sufficient statistic, not mere unbiasedness, completeness or prior efficiency.
\[\boxed{\text{A sufficient statistic}}\]