Inequality in Gaussian space -- possibly provable by rearrangement? The following problem arose for my collaborators and me when studying the computational complexity of the Maximum-Cut problem.
Let $f : \mathbb{R} \to \mathbb{R}$ be an odd function.  Let $\rho \in [0,1]$.  Let $X$ and $Y$ be standard Gaussians with covariance $\rho$.  Prove that $\mathbf{E}[f(X)f(Y)]$ ≤ $\mathbf{E}[f(X)^2 \mathrm{sgn}(X) \mathrm{sgn}(Y)]$.
The quantity on the left-hand side arises naturally in many contexts; e.g., it is the integral of $f(x)f(y)$ against the Mehler kernel (with parameter $\rho$).  
I have some reason to believe this inequality is true.  For one piece of evidence, suppose $f$'s range is $\pm 1$.  Then the inequality reduces to 
$\mathbf{E}[f(X)f(Y)]$ ≤ $\mathbf{E}[\mathrm{sgn}(X) \mathrm{sgn}(Y)]$;
i.e., it's saying that $\mathrm{sgn}$ is the $\pm 1$-valued odd function maximizing the LHS.  This is indeed true; it follows from a result of Christer Borell ("Geometric bounds on the Ornstein-Uhlenbeck velocity process"), proved by Ehrhard symmetrization.  It was also given a different proof by Beckner, deducing it from a rearrangment inequality on the sphere.  
The second inequality generalizes to the case of functions $f : \mathbb{R}^n \to$ {$-1,1$}, but I believe the first inequality, which I would like to prove, is inherently $1$-dimensional.
Any ideas, or pointers to literature that might help?
Thanks!
 A: 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.
A: 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.
