I am stuck with proving the following. Let $A\geq0
, B>0$, and $\alpha \in (0,1)$. I would like to show that
$$
\text{erf}\left(\frac{\alpha A+x}{\sqrt{2} \alpha B}\right)+\text{erf}\left(\frac{A-x}{\sqrt{2} B}\right)-\text{erf}\left(\frac{\alpha A-x}{\sqrt{2} \alpha B}\right)-\text{erf}\left(\frac{A+x}{\sqrt{2} B}\right)\geq 0,\quad \mbox{for all } x\geq0,$$ where erf($\cdot$) denotes the error function.
In my numerical experiments I have not found a single counterexample for this inequality, but I have not been able yet to prove it (e.g., via known bounds for the error function). Anyone a clue?
Best Answer
Let $C = \frac{A}{B\sqrt{2}}$, $y = \frac{x}{B \sqrt{2}}$ and $u = \frac{1}{\alpha} > 1$. We need to prove that $$\mathrm{erf}(C + uy) + \mathrm{erf}(C - y) - \mathrm{erf}(C - uy) - \mathrm{erf}(C + y) \ge 0, \ \forall y \ge 0.$$
Let $$f(u) = \mathrm{erf}(C + uy) + \mathrm{erf}(C - y) - \mathrm{erf}(C - uy) - \mathrm{erf}(C + y).$$ We have $$f'(u) = \frac{2y}{\sqrt{\pi}} \mathrm{e}^{-(C+uy)^2} + \frac{2y}{\sqrt{\pi}} \mathrm{e}^{-(C - uy)^2} \ge 0.$$ Also, $f(1) = 0$. Thus, $f(u)\ge 0$ for all $u \ge 1$.
We are done.