Is this function decreasing in $x$? Consider
$$x \mapsto \frac{\int_{x-b}^{x+b} e^{-\frac{z^2}{2} }\text d z}{\int_{x-c}^{x+c} e^{-\frac{z^2}{2} }\text d z}$$
decreasing in $x\in [0,\infty)$, if $c > b > 0$ ?
What I tried:
taking the derivative in $x$ yields:
$$\frac{\left( e^{-\frac{(x+b)^2}{2}} - e^{-\frac{(x-b)^2}{2}} \right)\int_{x-c}^{x+c} e^{-\frac{z^2}{2}}\text d z - \left( e^{-\frac{(x+c)^2}{2}} - e^{-\frac{(x-c)^2}{2}} \right)\int_{x-b}^{x+b} e^{-\frac{z^2}{2}}\text d z}{\left( \int_{x-c}^{x+c} e^{-\frac{z^2}{2}}\text d z \right)^2} $$
where the sign is determined by the numerator, so the question is if
$$\frac{e^{-\frac{(x+b)^2}{2}} - e^{-\frac{(x-b)^2}{2}}}{e^{-\frac{(x+c)^2}{2}} - e^{-\frac{(x-c)^2}{2}}} - \frac{\int_{x-b}^{x+b} e^{-\frac{z^2}{2} }\text d z}{\int_{x-c}^{x+c} e^{-\frac{z^2}{2} }\text d z} \geq 0 \ ?$$
 A: Let me just show you a few tricks so the next time you'll be able to figure such things out without pen or paper, not to mention MSE.
First, drop the common part. We need to show that 
$$
x\mapsto \frac{\int_{[x-b,x+b]}e^{-z^2/2}dz}{\int_{[x-c,x+c]\setminus[x-b,x+b]}e^{-z^2/2}dz}
$$ 
is decreasing for $x\ge 0$. 
Next, use symmetry and the fact that the integral of a family of decreasing functions over the parameter is still a decreasing function. Thus, it suffices to show that for every $0<s<b$,
$$
x\mapsto \frac{e^{-(x-s)^2/2}+e^{-(x+s)^2/2}}{\int_{[x-c,x+c]\setminus[x-b,x+b]}e^{-z^2/2}\,dz}
$$ 
is decreasing for $x\ge 0$.
Turn the fraction upside down and use the same logic to conclude that it suffices to show that for all $0<s<b<t$, 
$$
x\mapsto \frac{e^{-(x-t)^2/2}+e^{-(x+t)^2/2}}{e^{-(x-s)^2/2}+e^{-(x+s)^2/2}}
$$ 
is increasing for $x\ge 0$.
Now open parentheses and cancel inessential stuff. We just need to show that for $0<s<t$,
$$
x\mapsto \frac{e^{xt}+e^{-xt}}{e^{xs}+e^{-xs}}
$$ 
is increasing for $x\ge 0$.
This can be done by hand, but here is a general principle that makes it a no-brainer. Let $f(x)=\sum_{k\ge 0}a_k x^k$ and $g(x)=\sum_{k\ge 0}b_k x^k$. If $a_k,b_k\ge 0$ and $a_k/b_k$ increases with $k$, then $f(x)/g(x)$ increases for $x\ge 0$. Indeed, let $\mu=f(x)/g(x)$ for some $x$. Then $0=f(x)-\mu g(x)=\sum_{k\ge 0}(a_k-\mu b_k)x^k$. Note that the coefficients $a_k-\mu b_k$  must change sign just once from $-$ to $+$ as $k$ increases. Let them change sign between $K$ and $K+1$. Then, if we replace $x$ by $x'>x$, all negative terms will be multiplied by at most $(x'/x)^K$ while all positive terms will be multiplied by at least $(x'/x)^{K+1}$, so the difference $f(x')-\mu g(x')$ will be positive.
The rest should be clear. 
