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.
