Is $\Phi(-a(x+b))+\Phi(-a(x-b))$ log-concave in $x$ over the interval $x \in [0, \infty)$? Let $f(x)=\Phi(-a(x+b))+\Phi(-a(x-b))$, where $\Phi(\cdot)$ is the c.d.f of the standard normal, and $a>0$.
I would like to know if $\partial^2 \ln(f(x))/\partial x^2<0$  over the interval $x \in [0, \infty)$. I believe this is true, but I am having difficulty proving it. I would be immensely thankful to anyone who might help.
 A: $\newcommand\num{\operatorname{num}}\newcommand\den{\operatorname{den}}$This is not true in general. E.g.,
$$\frac{\partial^2 \ln f(x)}{\partial x^2}=0.16522\ldots>0$$
at $(a,b,x)=(1,-5,-4)$.

The OP has changed the question, by adding the condition $x\ge0$, thus invalidating the answer above.
After the change, the answer becomes positive.
Indeed, by a horizontal rescaling of the graph of $f$, without loss of generality (wlog) $a=1$.  Also, wlog $b>0$, since $f$ is even in $b$.
It is enough to show that
\begin{equation}
    r:=L'=\frac\num\den
\end{equation}
is decreasing on $[0,\infty)$, where
\begin{equation}
    L:=\ln f,
\end{equation}
\begin{equation}
    \num(x):=-h(-b-x)-h(b-x),\quad\den(x):=H(-b-x)+H(b-x),
\end{equation}
and $h$ and $H$ denote, respectively, the pdf and the cdf of the standard normal distribution.
Consider the "derivative ratio"
\begin{equation}
    \rho(x):=\frac{\num'(x)}{\den'(x)}= \frac{b \left(e^{2 b x}-1\right)}{e^{2 b x}+1}-x. 
\end{equation}
Note that
\begin{equation}
    \rho''(x)=-\frac{8 b^3 e^{2 b x} \left(e^{2 b x}-1\right)}{\left(e^{2 b
   x}+1\right)^3}<0 
\end{equation}
(for $x>0$). So, the function $\rho$ is concave. Also, $\rho(0)=0$. Hence, $\rho$ is up-down on $[0,\infty)$ -- that is, for some $c\in[0,\infty]$ the function $\rho$ is increasing on $[0,c]$ and decreasing on $[c,\infty)$. Note also that $\num(\infty-)=0=\den(\infty-)$.
The crucial point of this proof is the use of the l'Hospital-type rule for monotonicity given in line 3 of Table 4.1, which now implies that, just as the "derivative ratio" $\rho$, the function $r$ is up-down on $[0,\infty)$. But
\begin{equation}
    r'(0)=-\frac{4h(b)^2}{(H(-b)+H(b))^2}<0. 
\end{equation}
Thus, $r$ decreasing on $[0,\infty)$. $\quad\Box$
