<s>[INST] New Question: Find $x,y,z$ s.t. $x+y+z=1$ and $x^2+y^2+z^2=1$, and $x,y,z \in ( - 1,0) \cup (0,1)$ If in my earlier question [1], we relieve the constraints such that $x,y,z$ can also get values between $(−1,0)$, it seems that the system has a solution. Now, the question is how we can find this solution? 
$$\begin{cases}&x + y + z = 1\\
&{x^2} + {y^2} + {z^2} = 1\\
&x \ne y \ne z, x \ne z\\
&x,y,z \in ( - 1,0) \cup (0,1)
 \end{cases}$$
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[1] Multivariate-Multi-objective Optimization Problem: $x+y+z = 1$ and $x^2+y^2+z^2 = 1$
 [/INST] First observe that exactly one of the $x,y,z$ must be in $(-1,0)$. Without loss assume that $x\in(-1,0)$ and $y,z \in (0,1)$. Then, the first condition gives 
\begin{align}y+z=1-x\implies& y^2+2yz+z^2=1-2x+x^2\\[0.2cm]\overset{(2)}\implies &1-x^2+2yz=1-2x+x^2\\[0.2cm]\implies& y=\frac{x(x-1)}z\end{align} Now, return and plug in the value of $y$ in $(1)$
\begin{align}x+\frac{x(x-1)}{z}+z=1\implies &z^2+z(x-1)+x(x-1)=0\\[0.2cm]\implies &\frac1{(x-1)}z^2+z+x=0\\[0.2cm]\implies& z_{1,2}=\left(-1\pm\sqrt{1-\frac{4x}{(x-1)}}\right)\cdot\frac{x-1}2\end{align} Now, the condition $z>0$ implies that we can drop the one solution (which is obviously negative) and hence obtain $$z=\left(-1+\sqrt{\frac{1+3x}{1-x}}\right)\cdot\frac{x-1}2$$ Since the term in the square root must me non-negative we get the restriction \begin{align}\frac{1+3x}{1-x}\ge0 \iff (1+3x)(1-x)\ge 0 \iff -\frac13\le x\le 1\end{align} which together with $x<0$ implies that $-\frac13\le x<0$. For these values of $x$ the given $z$ exists and is $>0$. With this value of $z$, $y$ becomes $$y=\frac{x(x-1)}{z}=\frac{2x}{-1+\sqrt{\frac{1+3x}{1-x}}}$$ Hence, it remains to find all $x$ such that $$x+\frac{2x}{-1+\sqrt{\frac{1+3x}{1-x}}}+\left(-1+\sqrt{\frac{1+3x}{1-x}}\right)\cdot\frac{x-1}{2}=1$$ 
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