Inequality of sum of error functions

Question

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?

Answer 1

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.

Answer 2

@Yves Daoust gave the good hint.

Keeping your notations, at least small values of the arguments, we have for the expression $$f(x)=2 \sqrt{\frac{2}{\pi }}\,\frac {(1-\alpha)}{\alpha B}\,e^{-\frac{A^2}{2 B^2}}\, \Bigg[1+\sum_{n=1}^\infty\frac{ a_n}{(2 n+1)! \,\left(\alpha B^2\right)^{2 n} }\, x^{2n}\Bigg]\,x$$ with $$a_1=\left(\alpha ^2+\alpha +1\right) (A^2-B^2) $$ $$a_2=\left(\alpha ^4+\alpha ^3+\alpha ^2+\alpha +1\right) \left(A^4-6 A^2 B^2+3 B^4\right)$$ $$a_2=\left(\alpha ^6+\alpha ^5+\alpha ^4+\alpha ^3+\alpha ^2+\alpha +1\right) \left(A^6-15 A^4 B^2+45 A^2 B^4-15 B^6\right)$$