Suppose that $c>0, b>1,$ and $y-c>0$. Show that:
$\frac{erfc[\frac{x}{\sqrt{2}}]}{erfc[\frac{y}{\sqrt{2}}]}<\frac{erfc[\frac{x-c}{b\sqrt{2}}]}{erfc[\frac{y-c}{b\sqrt{2}}]}$
where $x>y$.
What I have done:
I have confirmed that given my assumptions this inequality holds with the help of Mathematica.
The inequality does not hold for every convex function (note that, since $y-c>0$, we are always on the convex part of the complementary error function). My guess is that there is some property of the complementary error function that makes it converge sufficiently quickly to zero so that the inequality holds.
(edit 1) I drew a graph for an easier understanding of the problem.
(edit 2) I believe that by proving the following key property the rest will follow. I should show that
$\forall x>0$ and $|dx|<x$,
$erfc[x]^2>erfc[x+dx]erfc[x-dx]$
Any solution/idea on how this can be solved will be highly appreciated!
Thank you in advance.
I found the 'edit 2' statement challenging to prove and hope you have success with using it in the rest of your proof. I'll sketch a proof of the inequality in edit 2: $$ \text{erfc}^2(x) \ge \text{erfc}(x(1+d))\cdot \text{erfc}(x(1-d)), \quad |d|<1 $$
In a paper about incomplete gamma functions there is the formula $$\gamma(a,t)\gamma(b,s) = \int_0^{\tfrac{t}{t+s}}\gamma(a+b,s/(1-w))w^{a-1}(1-w)^{b-1}dw \quad + $$ $$+ \int_{\tfrac{t}{t+s}}^1\gamma(a+b,t/w)w^{a-1}(1-w)^{b-1} dw$$ This is relevant because for first argument = 1/2, the incomplete gamma function reduces to an error function. Thus we have a product relation represented as an integral, and we can hope to derive the inequality by looking at the integrand. I don't use exactly the prior formula because we want erfc(), not erf(). Using the fact that $\gamma(1,x)=1-e^{-x},$ the '1' from the addition of the two '1/2', it is not difficult to show $$\pi\,\text{erfc}(\sqrt{t}) \text{erfc}(\sqrt{s})= \int_0^{\tfrac{s}{t+s}} e^{-s/w}\frac{dw}{\sqrt{w(1-w)}} +\int_0^{\tfrac{t}{t+s}} e^{-t/w}\frac{dw}{\sqrt{w(1-w)}}$$
Obviously we will now stick in $t=s=x^2,$ and $t=x^2(1+d)^2, s=x^2(1-d)^2.$ Doing this, and by scaling the integrals and some algebra we get the tractable expression
$$\pi\,\big(\text{erfc}^2(x) -\text{erfc}(x(1+d))\cdot \text{erfc}(x(1-d)) \big)$$ $$= \int_{0}^{1/2}\dfrac{e^{-x^2/w}}{\sqrt{w}}\Big(\frac{2}{\sqrt{1-w}} - e^{-(x\,d)^2/w} \Big( \big(\frac{1+d^2}{(1+d)^2} - w\big)^{-1/2} + \big(\frac{1+d^2}{(1-d)^2} - w\big)^{-1/2} \Big) dw $$ $$= 4\int_{0}^{1/\sqrt{2}}e^{-x^2/u^2}\underbrace{\Big(\frac{1}{\sqrt{1-u^2}} - \dfrac{e^{-(x\,d)^2/u^2}}{2} \Big( \big(\frac{1+d^2}{(1+d)^2} - u^2\big)^{-1/2} + \big(\frac{1+d^2}{(1-d)^2} - u^2 \big)^{-1/2} \Big)}_{} du $$ $$ \quad\quad\quad\quad\quad :=h(u; x,d) $$ The object is to show $h(u; x,d)>0$ for $x\ge 0$, $|d|<1$, $u<0<1/\sqrt{2}.$ The plot below has four curves as a function of $u$ and for $d=0.4.$ The blue one is for $1/\sqrt{1-u^2}$ and the red one for $$g(u; d)=\frac{1}{2}\Big( \big(\frac{1+d^2}{(1+d)^2} - u^2\big)^{-1/2} + \big(\frac{1+d^2}{(1-d)^2} - u^2 \big)^{-1/2} \Big) $$ When examining the red and blue alone, it is as if $x=0.$ The integral is identically 0 and that means that the area between the blue and red curves is identically zero. Of course since the red curve is below the blue curve until about $u=1/2,$ it must exceed it past that point. The dotted curves are what occur when the $x$ parameter is put into play. In both cases there is a dip at the origin but the dotted curves approach the red curve on the right. The black dotted curve is for a smaller $x$ than for the green dotted curve. As $x \to 0,$ the depression at the origin becomes a narrow spike and for more of the range of $u$ it will align more closely to the red curve.
It is apparent by eye that the crossing of the dotted curves with the blue curve are to the right of the intersection of red and blue curve. That's good news because then there is no way to have more'negative area' than what happened in the blue-red analysis. Since there will always be a depression at the origin for $x>0$ that means there will be a positive net area. Thus to prove the inequality we need only prove what we can see by eye in the case $x \to 0.$
The intersection between red and blue can be approximated by (use $w=u^2$) $$\frac{1}{\sqrt{1-w} } = g(\sqrt{w};d) = \frac{1}{\sqrt{1-w}} + \frac{(4w-1)d^2}{2(1-w)^{5/2}} \, + \, ...$$ where an expansion is made since $d<1.$ I actually take the next term too, get a cubic equation, solve it explicitly, and expand the appropriate root for small $d,$ obtaining, $$ x=0 \text{ intersection: } w_0 \sim \frac{1}{4} + \frac{d^2}{9} $$ Doing the same procedure for small $x$ by using the linear approximation from the Taylor series of $\exp{(-(x\,d)^2/w)},$ one can derive an equation involving $x.$ The root for small $x$ and $d$ is $$ x>0 \text{ intersection: } w_0 \sim \frac{1}{4} + \frac{d^2}{9} +\big(\frac{9}{8}d + \frac{d^2}{4}\big)x^2$$ The last term is positive, so we've proven what we expected on visual inspection to hold as $x \to 0.$