$n\geq 3$: Is positivity implied by an $n$ variables constraint equation? In pursuing a more extended project I came across the following
constraint problem—where I'm stuck now. The obstacle can easily be formulated in an isolated way:

We consider real numbers $x_1,x_2,\dots,x_n\,$ satisfying
  $$\prod^n_{k=1}\left(1-x_k^2\right)\:=\:\prod^n_{k=1}\,(2x_k+1)\,.$$
  If $\,-0.5 < x_1,\dots,x_n < 1\,$ (then all the preceding factors are positive), does it follow that
  $$\sum^n_{k=1}\,x_k\,\geqslant\,0\;\;?$$

For $\,n\,$ fixed we denote this statement by $S(n)$. 


*

*If $\,S(N)\,$ holds true, then $S(n)$ is true for every $n<N,\,$ just set
$\,x_{n+1}, x_{n+2},\dots, x_N=0$.  

*If $\,S(M)\,$ is wrong, then $S(m)$ cannot hold for any $m>M\,$ because if $(x_1, x_2,\dots, x_M)$ does not satisfy $S(M)$, then it can be padded with zeros to yield a counter-example to $S(m)$ for any $m>M$.


Here is my current status:
The instances $n=1,2$ are true. 
$S(3)$ would be a great nice-to-have! An ansatz to prove it by contradiction is shown below, but I fail to finalise.
I consider it very possible that the validity gap lies between $S(3)$ and $S(4)$.
$\checkmark\;S(1)$ is simple:
$$1-x^2\:=\:2x+1\;\Rightarrow\:x=0$$
$\checkmark\;S(2)$ is true: Let 
$$\left(1-x^2\right)\left(1-y^2\right)\:=\:(2x +1)(2y+1)$$
and assume $x+y<0$, thus $\,1-y>1+x\,$ and $\,1-x>1+y$,
and obtain the contradiction
$$\left(1-x^2\right)\left(1-y^2\right) = 
(1-y)(1-x)(1+x)(1+y) \:>\: (1+x)^2(1+y)^2\geq (2x +1)(2y+1)\,.$$
? $\;S(3)$ attempt: Let 
$$\left(1-x^2\right)\left(1-y^2\right)\left(1-z^2\right)\:=\:
(2x +1)(2y+1)(2z+1)$$
and assume $x+y+z<0\,$. Then $\,1-x > 1+y+z\,$ and cyclically, hence 
$$\begin{align}\left(1-x^2\right)\left(1-y^2\right)\left(1-y^2\right) &\:>\:
(1+x)(1+x+y)(1+y)(1+y+z)(1+z)(1+z+x) \\[2ex]
 & \stackrel{?}{\:\geq\:} (2x +1)(2y+1)(2z+1)
\end{align}$$
Observation for $S(n\geq 4)\,$:
The assumption $\sum^n_{k=1}\,x_k<0$ implies
$1-x_i > 1+\sum\limits_{k\ne i}x_k\,$ but the RHS cannot be (any longer) assumed to be positive.
 A: The conjectured validity gap turns out to be correct.
$S(4)\,$ is wrong, a counterexample is
$$x_1 = -\frac14,\, x_2=x_3 = -\frac38,\,
x_4 = \frac{\sqrt{1970156929}-2048}{45375}\approx 0.933 \\[2ex]
\text{then }\;\sum^4_{k=1}x_k\approx -0.067$$
presented by
Iosif Pinelis in this linked MO post .
In that same post he shows that $S(3)\,$ is true.

There is a close relationship to a 3-variables inequality on this site:
Write the constraint as
$$\prod_{k=1}^n\frac{1-x_k^2}{2x_k+1}\:=\: 1$$
and introduce each factor as a new variable. By means of the bijection
$$\begin{align}\left(-0.5,1\right) \; & \longleftrightarrow\; (0,\infty ) \\[1ex]
x \; & \longmapsto\;\:\frac{1-x^2}{2x+1} \\
\text{having the inverse}\quad\sqrt{a^2-a+1} -a\;\; & \longleftarrow\; a
\end{align}$$
we may equivalently formulate the initial problem as
$$a_1a_2\dots a_n =1 \;\;\stackrel{?}{\implies}\;
\sum^n_{k=1}a_k\:\leqslant\:\sum^n_{k=1} \sqrt{a_k^2-a_k+1}$$
The $n=3$ instance of the preceding is the theme of this
contest inequality question .
