Whether does the following equation have only one finite zero? Here is a calculus problem which bored me for sometime. Let $a>0$ and $b<0$ be fixed.Define the following function (EDIT: Following the comment by Barry Cipra, you may only consider the case where $a=1$)
$$
W_{a,b} (x)= e^{-2 b x}\left( \Phi^2\left(\frac{a b -x}{\sqrt{a}}\right)-\Phi\left(\frac{2 a b -x}{\sqrt{a}}\right)\right),
$$
where 
$$
\Phi(x): = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^x e^{-y^2/2} d y = \frac{1}{2} \left(Erf\left(\frac{x}{\sqrt{2}}\right)+1\right).
$$
The question is whether the following equation has only three zeors: $x=0$ and $x=\pm\infty$: 
$$
W_{a,b}(x) = W_{a,b}(-x).
$$
Plotting the function $W_{a,b}(x)$ suggests that the answer is right. But to find a proof seems quite hard.
Here are some graphs of the functions: $W_{1,-1}(x)$ (the blue one), $W_{1,-1}(-x)$ (the red one), and $W_{1,-1}(x)-W_{1,-1}(-x)$ (the one crossing the origin).



Here is the original problem. Define
$$
E_{a,b}(x) = e^{-b x}\Phi\left(\frac{ab-x}{\sqrt{a}}\right)+e^{b x}\Phi\left(\frac{ab+x}{\sqrt{a}}\right).
$$
We wish to prove that for $a>0$, $b<0$, 
$$
E_{a,b}^2(x)\ge E_{a,2b}(x),\quad\text{for all $x\in R$.}
$$
If one define 
$$
F_{a,b}(x) =E_{a,b}^2(x)- E_{a,2b}(x),
$$
then
$$
\frac{d F_{a,b}(x)}{d x} = -b \left( W_{a,b}(x) - W_{a,b}(-x)\right).
$$
Hence, this problem reduces to the above question.
Thank you very much for any suggestions!
Anand
 A: I'll try to address the original problem directly.
It is a bit obscure with all that $\Phi$  notation, but, if I deciphered the meaning of it all correctly (please, let me know if I'm wrong), you will be completely satisfied with showing that the ratio
$$
\frac{(Ee^{-b|x-\xi|})^2}{Ee^{-2b|x-\xi|}}\,,
$$
where $b>0$ and $\xi$ is the standard Gaussian random variable on the line, is decreasing in $x$ when $x>0$, so its infimum is attained at infinity, where it equals $e^{-b^2}$.
Taking the (minus) logarithmic derivative with respect to $x$ and shifting the variable by $x$, we can rewrite it as
$$
\frac{\int_0^\infty{e^{-bt}e^{-(x-t)^2/2}}dt-\int_0^\infty{e^{-bt}e^{-(x+t)^2/2}}dt}
{\int_0^\infty{e^{-bt}e^{-(x-t)^2/2}}dt+\int_0^\infty{e^{-bt}e^{-(x+t)^2/2}}dt}
\ge
\frac{\int_0^\infty{e^{-2bt}e^{-(x-t)^2/2}}dt-\int_0^\infty{e^{-2bt}e^{-(x+t)^2/2}}dt}
{\int_0^\infty{e^{-2bt}e^{-(x-t)^2/2}}dt+\int_0^\infty{e^{-2bt}e^{-(x+t)^2/2}}dt}\,,
$$
which can be restated as
$$
\frac{\int_0^\infty{e^{-bt}e^{-(x+t)^2/2}}dt}
{\int_0^\infty{e^{-bt}e^{-(x-t)^2/2}}dt}
\le
\frac{\int_0^\infty{e^{-2bt}e^{-(x+t)^2/2}}dt}
{\int_0^\infty{e^{-2bt}e^{-(x-t)^2/2}}dt}\,,
$$ 
or, equivalently,
$$
\frac{\int_0^\infty{e^{-bt}e^{-(x+t)^2/2}}dt}
{\int_0^\infty{e^{-2bt}e^{-(x+t)^2/2}}dt}
\le
\frac{\int_0^\infty{e^{-bt}e^{-(x-t)^2/2}}dt}
{\int_0^\infty{e^{-2bt}e^{-(x-t)^2/2}}dt}\,.
$$ 
Now, for any measure $\mu$ on $(0,+\infty)$, we have
$$
\log\left(\int e^{-bt}d\mu(t)\right)-\log\left(\int e^{-2bt}d\mu(t)\right)=
\int_b^{2b}\frac{\int te^{-ct}\,d\mu(t)}{\int e^{-ct}\,d\mu(t)}\,dc\,.
$$
Thus it suffices to show that the ratio $\frac{\int t\,d\nu(t)}{\int 1\,d\nu(t)}$ is less for $\nu_+$ than for $\nu_-$ where $d\nu_{\pm}(t)=e^{-ct}e^{-(x\pm t)^2/2}$. To this end, it suffices to show that for every $y>0$, we have
$$
\frac{\nu_+([0,y])}{\nu_+([0,+\infty))}\ge \frac{\nu_-([0,y])}{\nu_-([0,+\infty))} 
$$
or, equivalently,
$$
\frac{\nu_+([0,y])}{\nu_+([y,+\infty))}\ge \frac{\nu_-([0,y])}{\nu_-([y,+\infty))} 
$$
But this is obvious because the density of $\nu_-$ is just $e^{2xt}$ times that of $\nu_+$, so $\nu_-([0,y])\le e^{2xy}\nu_+([0,y])$ while  $\nu_-([y,\infty])\ge e^{2xy}\nu_+([y,\infty])$.
