Bayes risk is at most minimax risk

Remarks

For each fixed rule $\delta$, an average of $R(\cdot, \delta)$ is at most its supremum, so $\int R(\theta, \delta) \, \pi(d\theta) \le \sup_\theta R(\theta, \delta)$. Taking the infimum over $\delta$ on both sides gives the claim. Hence also $\sup_\pi r(\pi) \le R^*$, where $R^*$ is the minimax risk.

This inequality is the standard way to prove lower bounds for the minimax risk. If a rule $\delta^*$ has $\sup_\theta R(\theta, \delta^*) = r(\pi)$ for some prior $\pi$, then $\delta^*$ is minimax. For instance, a Bayes rule with constant risk is minimax, as for the binomial proportion with the beta prior of parameters $\sqrt{n}/2$ and $\sqrt{n}/2$.

Going deeper

For a student

To show that no estimator can do better than some value in the worst case, it is enough to find a prior whose Bayes risk equals that value. This is the Bayesian route to minimax lower bounds, often used with a sequence of priors whose Bayes risks increase to the candidate value. For a finite parameter set with randomized rules allowed, minimax theorems show that the two sides are equal when the supremum over priors is taken.