Inequality in Gaussian space -- possibly provable by rearrangement?

That is just Cuchy-Schwartz (though pretty well hidden). Writing everything down in terms of the joint density, dropping irrelevant factors, and taking into account that $f(-t)=-f(t)$, we see that the inequality in question is equivalent to $$ \iint_{(0,+\infty)^2}e^{-\frac 12(x^2+y^2)}(e^{\rho xy}-e^{-\rho xy})f(x)f(y)dxdy\le \iint_{(0,+\infty)^2}e^{-\frac 12(x^2+y^2)}(e^{\rho xy}-e^{-\rho xy})f(x)^2 dxdy $$ Now, $$ e^{\rho xy}-e^{-\rho xy}=\sum_n c_n(\rho) x^ny^n $$ with $c_n(\rho)\ge 0$. Thus, it suffices to show that $$ \iint_{(0,+\infty)^2}e^{-\frac 12(x^2+y^2)}x^ny^nf(x)f(y)dxdy\le \iint_{(0,+\infty)^2}e^{-\frac 12(x^2+y^2)}x^ny^n f(x)^2 dxdy $$ But, since the integrands are pure products now, this rewrites as $$ \left[\int_{(0,+\infty)}e^{-\frac 12 x^2}x^n f(x)dx\right]^2\le \left[\int_{(0,+\infty)}e^{-\frac 12 x^2}x^n f(x)^2dx\right]\cdot \left[\int_{(0,+\infty)}e^{-\frac 12 x^2}x^n dx\right], $$ which is pure Cauchy-Schwartz.

Using the antisymmetry of $f$ and $\mathrm{sgn}$ to bring the expectations to integral expressions over $[0, \infty) \times [0, \infty)$, the first expectation takes the form:

$const\times \int f(x) f(y) \exp\left(-\frac{x^2+y^2}{2(1-\rho^2)}\right)\sinh\left(\frac{2\rho xy}{2(1-\rho^2)}\right) dx dy$

while for the second case $f(y)$ is replaced by $f(x)$.

The proof becomes an application of the Cauchy–Schwarz inequality.