Without loss of generality take $\sigma^2 = 1$. The general case follows the same argument with $1/n$ replaced by $\sigma^2/n$. The risk of the MLE is constant: $R(\hat{\theta}_{\mathrm{MLE}}, \theta) = \operatorname{Var}_\theta(\bar{X}_n) = 1/n$ for all $\theta \in \mathbb{R}$. For any decision rule $\delta$, the bias-variance decomposition gives \begin{align*} R(\delta, \theta) = B(\theta)^2 + \operatorname{Var}_\theta(\delta(X)), \end{align*} where $B(\theta) = \mathbb{E}_\theta[\delta(X)] - \theta$ is the bias. The Cramér–Rao lower bound applies to unbiased estimators, but the proof of the bound works for any sufficiently regular estimator and yields \begin{align*} \operatorname{Var}_\theta(\delta) \geq \frac{(1 + B'(\theta))^2}{n}, \end{align*} since the Fisher information of the $N(\theta,1)$ model is $I(\theta) = 1$ and $\mathbb{E}_\theta[\delta(X)] = \theta + B(\theta)$ differentiates to give derivative $1 + B'(\theta)$. If $\delta$ dominates $\bar{X}_n$, then $R(\delta, \theta) \leq 1/n$ for all $\theta \in \mathbb{R}$, which combines with the above to give \begin{align*} B(\theta)^2 + \frac{(1 + B'(\theta))^2}{n} \leq \frac{1}{n} \qquad \text{for all } \theta. \end{align*} This is the key inequality $(\dagger)$. It immediately bounds $B(\theta)$ and forces $B'(\theta) \leq 0$, so $B$ is nonincreasing. If $B'$ were bounded away from $0$ for large $|\theta|$, then $B$ would be unbounded, contradicting its boundedness from $(\dagger)$. Hence there exist sequences $\theta_n \to -\infty$ and $\theta_n' \to +\infty$ along which $B'(\theta_n) \to 0$ and $B'(\theta_n') \to 0$. Substituting into $(\dagger)$ forces $B(\theta_n)^2 \to 0$ along both sequences. Since $B$ is nonincreasing and vanishes along sequences going to $\pm\infty$, it must be identically zero. With $B \equiv 0$, the Cramér–Rao bound applies in its standard form: $\operatorname{Var}_\theta(\delta) \geq 1/n$, so $R(\delta, \theta) \geq 1/n$. The only way to achieve $R(\delta, \theta) \leq 1/n$ for all $\theta$ is to have equality everywhere. Therefore $\delta$ does not strictly dominate $\bar{X}_n$, which is admissible. Constant risk then implies minimaxity by the constant-risk admissibility criterion from the previous chapter.