Inequality of sum of error functions 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?

@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
$$a_1=\left(\alpha ^2+\alpha +1\right) (A^2-B^2) $$
$$a_2=\left(\alpha ^4+\alpha ^3+\alpha ^2+\alpha +1\right) \left(A^4-6 A^2 B^2+3 B^4\right)$$
$$a_2=\left(\alpha ^6+\alpha ^5+\alpha ^4+\alpha ^3+\alpha ^2+\alpha +1\right) \left(A^6-15 A^4 B^2+45 A^2 B^4-15 B^6\right)$$
