Let $f: \mathbb R \to \mathbb R$ be square-integrable w.r.t the standard normal distribution $N(0,1)$. For $G \sim N(0, 1)$, let $\sigma_0(f) := E_G[f(G)]$ and $\sigma_1(f) := E[Gf(G)]$ be the first 2 Hermite coefficients of $f$.
Question. What is a characterization of functions $f$ for which $\sigma_0(f^2) \ge \sigma_0(f)^2 + \sigma_1(f)^2$ ?
Observation
Note that for any $f \in L^2(\mathbb R,N(0,1))$, then $$ \sigma_0(f^2) \ge \max(\sigma_0(f)^2,\sigma_1(f)^2). \tag{*} $$
Indeed, by Cauchy-Schwarz, $\sigma_1(f)^2 := E[Gf(G)]^2 \le E[G^2]E[f(G)^2] = E[f(G)^2] = \sigma_0(f^2)$, i.e $\sigma_0(f^2) \ge \sigma_1(f)^2$. Also, since the variance of the r.v $f(G)$ is nonnegative, we have $\sigma_0(f^2) \ge \sigma_0(f)^2$.
Even functions. As noted below by user Clement C., if $f$ is even (meaning $f(-x) = f(x)$ for all $x \in \mathbb R$), then the inequality holds. Indeed, for such $f$, we have for $G \sim N(0,1)$, $$ \sigma_1(f) = E[Gf(G)] = E[Gf(-G)] = E[-Gf(G)] = -\sigma_1(f), $$ and so $\sigma_1(f) = 0$. Thus, the sought-for inequality then follows from (*).
Odd functions. Similarly, i turns out that the inequality holds true if $f$ is odd. Indeed for such $f$, we have $\sigma_0(f) = E[f(G)] = E[-f(-G)] = -E[f(-G)] = -E[f(G)]=-\sigma_0(f)$, and so $\sigma_0(f)=0$. Thus, the sought-for inequality then follows from (*).