<s>[INST] 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..   
 [/INST] 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.
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