I have spent a couple of days on this without success, so I really hope for help here.
Let $\phi_\mu(s) = \frac{1}{\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(s-\mu)^2}{2\sigma^2}\right)$ be the density for a $N(\mu,\sigma)$ random variable and let $\bar{\Phi}_\mu(t) = \int_t^\infty \phi_\mu(s)ds$ be the survivor function.
I know that $\frac{\phi_0(s)}{\phi_1(s)}$ and $\frac{\bar{\Phi}_0(s)}{\bar{\Phi}_1(s)}$ are decreasing. This follows since the normal distribution has the monotone likelihood ratio property.
I conjecture (and hope to be true) that $h(s)=2\frac{\phi_0(s)}{\phi_1(s)} - \frac{\bar{\Phi}_0(s)}{\bar{\Phi}_1(s)}$ is also decreasing.
I have not been able to find a contradiction numerically.
Some observations:
$h(s)>0,s \in R$
$\lim_{s \rightarrow -\infty} h(s) = \infty$
$\lim_{s \rightarrow \infty} h(s)= 0$
$h^{\prime }(s)=-\frac{2}{\sigma ^{2}}\frac{\phi _{0}(s)}{\phi _{1}(s)}+\frac{\phi _{0}(s)% }{\bar{\Phi}_{1}(s)}-\frac{\phi _{1}(s)\bar{\Phi}_{0}(s)}{\bar{\Phi}_{1}^{2}(s)}$
$\lim_{s \rightarrow -\infty} h'(s) = -\infty$
$\lim_{s \rightarrow \infty} h'(s)= 0$.
Generalisation of the result, if true, beyond the normal distribution would be fantastic.
The desired property will be useful for proving uniqueness of equilibrium in a model that involves random signals.
Here is a plot of $h$ (in $\color{blue}{\text{blue}}$) and $h'$ (in $\color{red}{\text{red}}$) for $\sigma = 2$:
I have found a partial answer myself. It works for $\sigma=1$. I am still hoping for extension to any $\sigma>0.$ Any help I could get on that would be most appreciated.
Define $\lambda_\mu(s) = \frac{\phi_\mu(s)}{\bar{\Phi}_\mu(s)}$ and note that $\lambda_\mu(s) = \lambda(s-\mu)$.
Lemma (Adapted from Feller (1957)) $$ \frac{s}{\sigma ^{2}}<\lambda _{0}\left( s\right) <\frac{s^{3}}{\sigma ^{2}s^{2}-\sigma ^{4}},s>0. $$ Proof: Observe by differentiation that for $s>0$ \begin{eqnarray*} \phi _{0}\left( s\right) \frac{\sigma ^{2}}{s} &=&\int_{s}^{\infty }\phi _{0}\left( t\right) \left[ 1+\frac{\sigma ^{2}}{t^{2}}\right] dt>\bar{\Phi}% \left( s\right) \\ \phi _{0}\left( s\right) \left[ \frac{\sigma ^{2}}{s}-\frac{\sigma ^{4}}{% s^{3}}\right] &=&\int_{s}^{\infty }\phi _{0}\left( t\right) \left[ 1-\frac{% 3\sigma ^{4}}{t^{4}}\right] dt<\bar{\Phi}\left( s\right) . \end{eqnarray*} The conclusion follows.QED.
Proof that $h\left( s\right) =2\frac{\phi _{0}\left( s\right) }{\phi _{1}\left( s\right) }-\frac{\bar{\Phi}_{0}\left( s\right) }{\bar{\Phi}_{1}\left( s\right) } $ is decreasing when $\sigma=1$.
Note that $\lambda_\mu(s)$ is positive, increasing in $s$ and therefore decreasing in $\mu $.
The derivative of $h$ is \begin{eqnarray*} h^{\prime } &=&-2\frac{\phi _{0}}{\phi _{1}}+\frac{\phi _{0}}{\bar{\Phi}_{1}}% -\frac{\phi _{1}\bar{\Phi}_{0}}{\bar{\Phi}_{1}^{2}} \\ &=&-2\frac{\phi _{0}}{\phi _{1}}+\left[ \frac{\phi _{0}}{\bar{\Phi}_{0}}-% \frac{\phi _{1}}{\bar{\Phi}_{1}}\right] \frac{\bar{\Phi}_{0}}{\bar{\Phi}_{1}} \\ &=&\left[ -2\frac{\lambda _{0}}{\lambda _{1}}+\lambda _{0}-\lambda _{1}% \right] \frac{\bar{\Phi}_{0}}{\bar{\Phi}_{1}} \end{eqnarray*} Then $h^{\prime }<0$ if and only if $\lambda _{0}\left( \lambda _{1}-2\right) -\lambda _{1}^{2}<0$. This holds trivially if $\lambda _{1}\leq 2$. But $\lambda _{1}\left( 2.5\right) <2$ so we may restrict attention to $s>2.5$.
Now use the Lemma to find that for $s>2.5$, \begin{eqnarray} \lambda _{0}-\lambda _{1}-2\frac{\lambda _{0}}{\lambda _{1}} &<&\frac{1}{% \frac{1}{s}-\frac{1}{s^{3}}}-s+1-2s\left( \frac{1}{s-1}-\frac{1}{\left( s-1\right) ^{3}}\right) \nonumber \\ &=&-\frac{s^{4}-s^{3}-2s^{2}-3s+1}{\left( s-1\right) ^{3}\left( s+1\right) } \label{upperbound} \end{eqnarray}
The upper bound for $h'$ becomes negative for $s$ larger than the largest root of the denominator. The real roots of the numerator are $% \left\{ 0.277\,04,2.331\,1\right\} $, which means that the upper bound is smaller than zero when $s>2.5$. QED