Solve the System of Equations in Real $x$,$y$ and $z$ Solve for $x$,$y$ and $z$ $\in $ $\mathbb{R}$ if
$$\begin{align} x^2+x-1=y \\
                   y^2+y-1=z\\
z^2+z-1=x \end{align}$$
My Try:
if  $x=y=z$ then the two triplets $(1,1,1)$ and $(-1,-1,-1)$ are the Solutions.
if $x \ne y \ne z$ Then we have
$$\begin{align} x(x+1)=y+1 \\
                   y(y+1)=z+1\\
z(z+1)=x+1 \end{align}$$
Multiplying all we get  $$xyz=1 \tag{1}$$ and adding all we get
$$x^2+y^2+z^2=3 \tag{2}$$
Now from Original Equations
$$\begin{align} x^2=y+1-x\\
                y^2=z+1-y\\
                z^2=x+1-z \end{align}$$ Multiplying all and Using $(1)$ we get
$$(y+1-x)(z+1-y)(x+1-z)=1 $$    $\implies$
$$xy+yz+zx-3=(x-y)(y-z)(z-x) \tag{3}$$ I am unable to proceed further..   
 A: using the RMS GM inequality 
$$\sqrt{\frac{x^2+y^2+z^2}{3}}\geq\sqrt[3]{xyz}$$
with equality if and only if $x=y=z$, plugging in your values for $xyz$ and$x^2+y^2+z^2$ we get that
$$\sqrt{\frac{3}{3}}\geq\sqrt[3]{1}$$
$$1=1$$
thus the only possible solutons are those you already stated.
A: No solutions exist for real, distinct $(x,y,z)$.
Writing $f(x)=x^2+x-1$, the existence of such a solution would mean that $f$ has a real point of least period 3; that is, $f^3(x)=x$ for some real $x$ with $f(x)\neq1$ i.e. $x\neq \pm 1$. Sarkovsky's theorem then implies that $f$ has points of arbitrary least period. In particular, there would exist $x\in\mathbb{R}-\{-1,1\}$ such that $f^2(x)=x$. But $$f^2(x) - x = (x^2+x-1)^2+(x^2+x-1)-1-x = x^4+2x^3-2x-1 = (x-1)(x+1)^3.$$ and so there are no points of least period 2. Hence $f$ has no real points of least period 3.
P.S.: I'm actually only using the simplest case of Sarkovsky's theorem, namely that 3-cycles imply period 2-cycles. But the full theorem implies that the same conclusion follows for every least period other than one (that is, $f$ has fixed points but not cycles.) This means that we would be no no better off if we increased the number of variables and equations beyond 3: we would still find no nontrivial real solutions. (Though this can be proven much more simply by the same RMS-GM inequality stated by cirpis in his answer...)
A: $\textbf{Hint:}$Note that $\displaystyle x^2-x-1=y$ is equal $\displaystyle (x-\frac{1}{2})^2-\frac{5}{2}=y-\frac{1}{2}$, so if you substitute $x_1=x-\frac{1}{2}$, $y_1=y-\frac{1}{2}$ and $z_1=z-\frac{1}{2}$ you get:
$$x_1^2-\frac{5}{2}=y_1$$
$$y_1^2-\frac{5}{2}=z_1$$
$$z_1^2-\frac{5}{2}=x_1$$
Substitute (1) to (2), next (2) to (3), you get polynomial equation to calculate $x_1$. The same with $y_1$ and $z_1$.
A: $\textbf{1)}$ Substituting Eq. 1 into Eq. 2 gives $xy(x+1)=z+1$, and then substituting Eq. 2 into Eq. 3 gives $xyz(x+1)=x+1$; so $x=-1$ or $xyz=1$.  By symmetry, we have that $y=-1$ or $xyz=1$ and $z=-1$ or $xyz=1$.  Since $x=-1, y=-1, z=-1$ is a solution, all other solutions must satisfy $xyz=1$.
Since $x=1,y=1, z=1$ is clearly a solution, it is left to show that there are no other solutions satisfying $xyz=1$.
$\textbf{2)}$ If $x\ge1$, then $y=x(x+1)-1\ge2-1=1$ and $z=y(y+1)-1\ge2-1=1$; so $xyz=1\implies x=1, y=1, z=1$.  Therefore we can assume $x<1$ and, by symmetry, $y<1$ and $z<1$.
$\textbf{3)}$ If $x, y, z >0$, then $0<x, y, z <1\implies xyz<1$; so two of the variables must be negative, and we can assume that $x<0, y<0, z>0$.  Since $z<1$, $xy=\frac{1}{z}>1$.  However, this is impossible since $xy=x(x^2+x-1)=x^3+x^2-x$, and $g(x)=x^3+x^2-x$ has a maximum for $x<0$ given by $g(-1)=1$ 
since $g^{\prime}(x)=(3x-1)(x+1)$.
