Can't solve non linear system I'm trying to solve this system:
$\begin{cases}(1+y)(1+z) = 2\lambda x \\ (1+x)(1+z) = 2\lambda y \\ (1+x)(1+y)= 2\lambda z \\ x^2+y^2+z^2 - 1 = 0\end{cases}$
We can assume that $\lambda \neq 0$
How would you solve this? No matter in which direction I go it seems unpleasant.
 A: Here are some thought, hope they help.
Subtract the second from the first, you get 
$$(1+z)(y-x)=2\lambda (x-y)$$ 
thus 
$$x=y \ \ \text{or} \ \ z=-2\lambda-1$$ similarly, 
$$x=z \ \ \text{or} \ \ y=-2\lambda-1$$
$$z=y \ \ \text{or} \ \ x=-2\lambda-1$$
This means that either $x=y=z$ or 
$$x=y=-2\lambda-1\ \ \text{and} \ \ z=2 \lambda$$ 
Well, you can study those two cases.
In the first case you get $x=y=z=\pm \frac{1}{\sqrt{3}}$ in the second $\lambda=-\frac{1}{2}, -\frac{1}{6}$.
A: We have 
$$
\left\{
\begin{array}{rcl}
(1+x)(1+y)(1+z) &=&2\lambda x(1+x)\\
(1+x)(1+y)(1+z) &=&2\lambda y(1+y)\\
(1+x)(1+y)(1+z) &=&2\lambda z(1+z)\\
x^2+y^2+z^2&=&1
\end{array}
\right.
$$
Note in this case that if we switch as unknowns $x$, $y$ and $z$ with each other the system does not change. It is suggested that one of the solutions is $x=y=z$. Then $x^2+y^2+z^2=1$ and $x=y=z$ implies $3x^2=1$, $3y^2=1$, $3z^2=1$, that is, $x=\pm\frac{1}{\sqrt{3}}$, $y=\pm\frac{1}{\sqrt{3}}$ and $z=\pm\frac{1}{\sqrt{3}}$.
