I just came through an inequality that looks simple, but I haven't been able to figure out the proof. The problem statement is:
Let $Z$ be a standard normal random variable, and $f: \mathbb{R} \rightarrow \mathbb{R}$ is a function with bounded first derivative. Prove that:
$$\mathbb{E}[f'(Z)]^2 \leq \operatorname{Var}[f(Z)] \leq \mathbb{E}\left[f'(Z)^2\right]$$
Comment: I saw on some online post a similar inequality called Gaussian Poincare inequality, which involves proving the second inequality in the statement.
As stated in the comments, this is the Cramer-Rao lower bound.
Let $\varphi$ be the pdf of the standard normal distribution. Let $f:\mathbb{R}\to \mathbb{R}$ be differentiable with bounded derivative. We assume throughout that all interchanges of derivatives and integrals are valid. Each case can be verified by dominated convergence theorem. Let $Z\sim N(0;1)$. Let $$M(\theta)=\mathbb{E}[f(\theta+Z)].$$ Let $\varphi$ be the standard normal pdf. Note that $$\frac{\partial}{\partial \theta} \bigl[\log \varphi(z-\theta)\bigr]\varphi(z-\theta)\Bigr|_{\theta=0}=z\varphi(z).$$
We calculate \begin{align*} \mathbb{E}[f'(Z)]^2 & = M'(0)^2 = \Bigl[\int f(z)\frac{\partial}{\partial \theta}\varphi(z-\theta)\Bigr|_{\theta=0}dz\Bigr]^2 \\ & = \Bigl[\int f(z)\frac{\partial}{\partial \theta} \bigl[\log \varphi(z-\theta)\bigr]\varphi(z-\theta)\Bigr|_{\theta=0} dz\Bigr]^2 \\ & = \Bigr[\int f(z)z\varphi(z)dz\Bigr]^2=\mathbb{E}\Bigl[f(Z)Z\Bigr]^2 = \text{Cov}(f(Z),Z)^2 \\ & \leqslant \text{Var}[f(Z)]\text{Var}[Z]=\text{Var}[f(Z)].\end{align*}