Let $a$,$b$ be positive real numbers such that $a0$ where
$f(x)= \phi(a-x)-\phi(b+x)$ where $\phi$ is the standard normal PDF.
So far I have that $f(0)>0$ and that
$f'(x)=(a-x)\phi(a-x)+(b+x)\phi(b+x)$
and then I get stuck
Let $a$,$b$ be positive real numbers such that $a0$ where
$f(x)= \phi(a-x)-\phi(b+x)$ where $\phi$ is the standard normal PDF.
So far I have that $f(0)>0$ and that
$f'(x)=(a-x)\phi(a-x)+(b+x)\phi(b+x)$
and then I get stuck
Since $a0$ we have $-b-x < a - x < b + x$, or $|a-x| < |b+x|$, so that $-(b+x)^2/2 < -(a-x)^2/2$, and hence that
$$ \sqrt{2\pi} \phi(b+x) = e^{-(b+x)^2/2} < e^{-(a-x)^2/2} = \sqrt{2\pi} \phi(a-x). $$
Thus $f(x) = \phi(a-x) - \phi(b+x) > 0$.