I'm investigating if and how the positivity or negativity of a multivariable function can be proved. Consider $y\_{1},y\_{2},y\_{3}\in\mathbb{R}$ and the following function
$$f\left(y\_{1},y\_{2},y\_{3}\right)=y\_{1}\mbox{artanh}(y\_{1}-y\_{3})+y\_{2}\mbox{artanh}(y\_{2}-y\_{1})+y\_{3}\mbox{artanh}(y\_{3}-y\_{2})$$
where $\mbox{artanh}(\cdot)$ is the inverse hyperbolic tangent function. How can I prove whether $f(\cdot)$ is positive or negative? I have tried to find the critical points of this function by computing its gradient given by
$$\Delta{f}\left(y\_{1},y\_{2},y\_{3}\right)=\begin{bmatrix}\mbox{artanh}(y\_{1}-y\_{3})-\frac{y\_{1}}{(y\_{1}-y\_{3})^2-1}+\frac{y\_{2}}{(y\_{1}-y\_{2})^2-1}\\\frac{y\_{3}}{(y\_{2}-y\_{3})^2-1}-\frac{y\_{2}}{(y\_{1}-y\_{2})^2-1}-\mbox{artanh}(y\_{1}-y\_{2})\\
\frac{y\_{1}}{(y\_{1}-y\_{3})^2-1}-\mbox{artanh}(y\_{2}-y\_{3})-\frac{y\_{3}}{(y\_{2}-y\_{3})^2-1}
\end{bmatrix}^{T}$$ 
which becomes equal to zero when $y\_{1}=y\_{2}=y\_{3}=y\_{\ast}$ where $y\_{\ast}\in\mathbb{R}$ is an arbitrary real number. The Hessian of the function computed at these points turned out to be a positive semi definite matrix. However, I'm not sure whether this solution is unique. Is this a good way to approach the problem or is there another way to show whether the function $f(\cdot)$ is positive or negative? Thank you.


Kind regards,



investigating positivity/negativity of a function
This function can be either positive or negative, even if
$y\_1,y\_2,y\_3$ are close enough that the hyperbolic tangents are real
(see **Robert Israel**'s comment). Fix $\eta\_1,\eta\_2,\eta\_3$ such that
$$
\phi(\eta\_1,\eta\_2,\eta\_3) :=
 {\mathop{\rm arctanh}} (\eta\_1 - \eta\_3)
+ {\mathop{\rm arctanh}} (\eta\_2 - \eta\_1)
+ {\mathop{\rm arctanh}} (\eta\_3 - \eta\_2)
\neq 0.
$$
(Almost any choice works as long as $0 < |\eta\_i-\eta\_j| < 1$
for all $i \neq j$.) Then 
$$
f(t+\eta\_1,t+\eta\_2,t+\eta\_3) 
 = \phi(\eta\_1,\eta\_2,\eta\_3) t + f(\eta\_1,\eta\_2,\eta\_3)
$$
has one sign for $t \rightarrow +\infty$
and the opposite sign for $t \rightarrow -\infty$.


[*added later* $-$ Note that this used almost nothing about the
arctanh function: the same argument works for any function
of the form
$$
f(y\_1,y\_2,y\_3) = y\_1 \, a(y\_1-y\_3) + y\_2 \, a(y\_2-y\_1) + y\_3 \, a(y\_3-y\_2)
$$
as long as the real-valued function $a$ is not linear.]


