I have a system of equations: 


$$1/2 + {\rm Erf}(x) - {\rm Erf}(\frac{x+y}{2})=0$$
$$-1/2 + {\rm Erf}(y) - {\rm Erf}(\frac{x+y}{2})=0,$$
Where $x \le y$ and ${\rm Erf}$ is the Error Function.


By ansatz I know that one solution is $(x,y)=(-{\rm Erf}^{-1}(1/2),{\rm Erf}^{-1}(1/2))$, but I can not manage to prove that this is the only solution, even though I strongly suspect it. My question therefore is how to prove uniquness. 


One re-arrangement I have tried results in:
${\rm Erf}(\frac{x+y}{2}) = \frac{{\rm Erf}(x)+{\rm Erf}(y)}{2}$, which I am pretty sure is equivalent to $y=-x$ (apparent by implicit plot), but again I can not prove this. Proving that $y=-x$ for all solutions would imply the desired uniqueness. 


Any advice on manipulation tactics for the error function would be much appreciated..! 


Tagged as probability theory because of the relation to the Normal distribution CDF.



Reducing system of equations involving Erf, Error Function
This can be solved as a problem in inequalities.
The desired implication
$$
{\rm Erf}\Bigl(\frac12(x+y)\Bigr) = \frac12\bigl({\rm Erf}(x) + {\rm Erf}(y)\bigr)
\Longleftrightarrow x + y = 0
$$
is not quite correct, because the hypothesis holds also for $x=y$;
but these are the only possibilities, which is enough to yield
the desired result. Because the error function is odd, it is enough to prove:


**Proposition.** *If $x$ and $y$ are real numbers such that $x+y > 0$ then
$$
{\rm Erf}\Bigl(\frac12(x+y)\Bigr) \geq \frac12\bigl({\rm Erf}(x) + {\rm Erf}(y)\bigr),
$$
with equality if and only if $x=y$.*


*Proof*: Let $a = (x+y)/2$ and $b = (x-y)/2$. Then $a>0$ and $x,y = a \pm b$,
and we claim
$$
\bigl({\rm Erf}(a+b) + {\rm Erf}(a-b)\bigr) - 2\phantom.{\rm Erf}(a) \leq 0,
$$
with equality **iff** $b=0$. By symmetry we may assume $b\geq 0$.
Equality clearly holds for $b=0$. We claim that the left-hand side (LHS)
is a decreasing function of $b$ for $b \geq 0$, which will prove that
it is negative for all $b>0$. Indeed derivative of the LHS is
$$
\frac2{\sqrt{\pi}} \left( e^{-(a+b)^2} - e^{-(a-b)^2} \right)
= \frac2{\sqrt{\pi}} e^{-(a^2+b^2)} (e^{-2ab} - e^{+2ab}) < 0
$$
for $b>0$, **QED**.


(This uses the normalization of ${\rm Erf}(t)$ that has derivative
$(2/\sqrt\pi) \phantom. e^{-t^2}$. There are other normalizations, but the proof
works in the same way.)


