Let \begin{align} g_c(x)= (x+c)e^{-\frac{(x+c)^2}{2}}+(x-c)e^{-\frac{(x-c)^2}{2}} \end{align} where $x$ and $c$ are non-negative.
We are interested in finding
\begin{align}
C= \sup\{ c: g_c(x)\ge 0, \forall x\ge 0\}
\end{align}
In other words, we want to find the largest $c$ such that $g_c(x)$ is non-negative for all $x$.
My attempt:
Since, we want $g_c(x) \ge 0$
\begin{align}
(x+c)\ge -(x-c)e^{-2cx}
\end{align}
This emiditly implies that we have to look at $0\le x \le c$ and solve
\begin{align}
\frac{c+x}{c-x}-e^{-2cx} \ge 0
\end{align}
This is as far as I got. Not sure how to proceed. The numerical simulations suggest that $\bar{c}=1$.
Thanks in advance.
I won’t write here every detail, only the main stream of the proof.
Be $f(x):=x e^{-\frac{1}{2}x^2}$ and therefore $g_c(x):=f(c+x)-f(c-x)$.
$x>0 => f(x)>0$, $f(0)=0$, $x<0 => f(x)<0$
The maximum of $f(x)$ is $\max(f(x))=f(1)$.
Be $x\geq 0$, $c\geq 0$.
(1) $\,0\leq c<x$: $f(c-x)<0<f(c+x)$
(2) $\,0\leq c\leq 1$ and $0<x \leq c$: $f(c-x)<f(c+x)$
(3) $\,c>1$ and $0<x \leq c-1$: $f(c+x)<f(c-x)$
$\,\,\,\,\,\,\,\,$ because $f(c-x)=f(1)$ means maximum for $x=c-1$
$\,\,\,\,\,\,\,\,$ Therefore $g_c(x)<0$.
Result:
(1) and (2) means $g_c(x)\geq 0$ for $c\leq 1$
(1) and (3) means $g_c(x)<0$ exists for $c>1$
=> $\max(c)=1$