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 $\nu_-([y,\infty])\ge e^{2xy}\nu_+([y,\infty])$.