Given a polynomial constraint equation in $n$ variables, can one conclude that the sum of the variables is non-negative? Currently I'm stuck as follows;
at least a positive proof if $n=3$ would be a great nice-to-have!

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 $\,-\frac12 < x_1,\dots,x_n < 1\,$ holds (then all the preceding factors are positive), does it follow that
  $$\sum^n_{k=1}\,x_k\,\geqslant\,0\;\;?$$

For fixed $n$ denote this statement by $S(n)$. Three easy observations are


*

*$S(1)$ is true.

*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 $m>M$.


Next look at $S(2)$ which is true as well:
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 one gets the contradiction
$$\begin{align}\left(1-x^2\right)\left(1-y^2\right) & = 
(1-y)(1-x)(1+x)(1+y)\\ & >\:(1+x)^2(1+y)^2\:\geq\: (2x +1)(2y+1)\,.
\end{align}$$
My attempt to analogously prove $S(3)$:
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 
$$\begin{align}(1-x^2)(1-y^2)(1-y^2) &>
(1+x)(1+x+y)(1+y)(1+y+z)(1+z)(1+z+x) \\[1ex]
 & \stackrel{?}{\geqslant} (2x +1)(2y+1)(2z+1)
\end{align}$$
Can this be finalised?
My guess: The 'validity gap' lies between $S(3)$ and $S(4)$.

This is a Cross-post
from math.SE after a decent waiting period
(lasting 21 days, resulting in no comments, no downvotes, no answers, and 59 views).
 A: A counterexample: $n=4$, 
$$(x_1,\dots,x_4)=\left(-\frac{1}{4},-\frac{3}{8},-\frac{3}{8},\frac{\sqrt{1970156929}-2048}{45375}\right). 
$$
So, by what you noted, your conjecture fails to hold for any $n\ge4$. 

The validity gap is indeed between $n=3$ and $n=4$. 
Indeed, for $n=3$ Mathematica confirms your conjecture:

I think this can also be easily checked using other computer algebra packages. 
Alternatively, one can try to do the case $n=3$ manually, for instance, as follows: solve the (quadratic) equation 
$(1-x^2)(1-y^2)(1-z^2)=
(2x +1)(2y+1)(2z+1)$ for (say) $z$, then rewrite the corresponding inequalities as polynomial ones (in $x,y$), and finally use (say) resultants -- as explained and done e.g. in Section 5. 

Here is a simpler manual, almost ingenuity-free solution for $n=3$: Consider the minimization of $x+y+z$ given
\begin{equation*}
(1-x^2)(1-y^2)(1-z^2)=(2x+1)(2y+1)(2z+1) \tag{1} 
\end{equation*}
and the non-strict inequality constraints 
\begin{equation*}
-1/2\le x,y,z\le1. \tag{2} 
\end{equation*}
If one of these non-strict inequalities is an equality, then without loss of generality (wlog) $x=1$ and $y=-1/2$, whence $x+y+z\ge1-1/2-1/2=0$. So, it remains to consider possible interior extrema, with all inequalities (2) being strict. Then (1) can be rewritten as 
\begin{equation*}
 f(x)+f(y)+f(z)=0,
\end{equation*}
where $f(u):=\ln\frac{1-u^2}{2u+1}$. Suppose that $(x,y,z)$ is an interior extremum point. Then, by Lagrange multipliers, 
\begin{equation*}
 f'(x)=f'(y)=f'(z). \tag{3}
\end{equation*}
It is easy to see that 
\begin{equation*}
 f'''(u)=-\frac{4 \left(4 u^6+12 u^5+42 u^4+37 u^3+6 u^2+3 u+4\right)}{(2 u+1)^3 \left(1-u^2\right)^3}<0
\end{equation*}
if $-1/2<u<1$, so that $f'$ is strictly concave on $(-1/2,1)$. Hence, for any real $t$, equation $f'(u)=t$ has at most two real roots $u$. 
So, by (3), wlog $y=x$. 
Solving now (1) (with $y=x$) for $z$, we see that $x+y+z$ becomes 
\begin{equation*}
 s:=\frac{-p+\sqrt d}{(1-x^2)^2},
\end{equation*}
where 
\begin{equation*}
 p:=1 + 2 x + 4 x^2 + 4 x^3 - 2 x^5
\end{equation*}
and 
\begin{equation*}
 d:=p^2+x^2 (1-x^2)^2 (6+16 x+9 x^2-4 x^4)>p^2, 
\end{equation*}
so that $s$ is manifestly positive, as claimed. 
A: A proof for $n=3$.
Let $\frac{1-x_1^2}{1+2x_1}=a$, $\frac{1-x_2^2}{1+2x_2}=b$ and $\frac{1-x_3^2}{1+2x_3}=c$.
Thus, $a$, $b$ and $c$ are positives such that $abc=1$.
Also, $$x_1^2+2ax_1+a-1=0,$$ which gives
$$x_1=-a+\sqrt{a^2-a+1}$$ because $x_1=-a-\sqrt{a^2-a+1}\leq-1.$
Id est, we need to prove that
$$\sum_{cyc}\sqrt{a^2-a+1}\geq a+b+c$$ and see here: 
https://math.stackexchange.com/questions/3323045
In the general case after substitution $\frac{1-x_i^2}{1+2x_i}=e^{a_i}$ we need to prove that $\sum\limits_{k=1}^nf(a_i)\geq0,$ where 
$$f(x)=\sqrt{e^{2x}-e^x+1}-e^x$$ and $\sum\limits_{k=1}^na_k=0.$
Since $f''(x)>0$ for all $x\geq0,$ by Vasc's RCF Theorem it's enough to prove our inequality for equality case of $n-1$ variables, which is wrong for any $n\geq4.$ 
About RCF see here: https://kheavan.files.wordpress.com/2010/06/2007_1_applic.pdf
