Find the real solutions of the system below: $$y=\frac{x^{3}+12x}{3x^{2}+4},\ z=\frac{y^{3}+12y}{3y^{2}+4},\ x=\frac{z^{3}+12z}{3z^{2}+4}$$
I guessed that, considering the symmetry, $x=y=z$ must hold. In this case $(x,y,z)=(0,0,0),\ (2,2,2),\ (-2,-2,-2)$. Suppose that $x=y$, thus $x=y=z$. So without loss of generality, we can assume $x>y>z$. I couldn't use this in order to lead to contradiction.
Let $f(t):=\dfrac{t^3+12t}{3t^2+4}$ for each $t\in\mathbb{R}$. Then, $$f'(t)=3\,\left(\frac{(t-2)(t+2)}{3t^2+4}\right)^2>0\text{ for almost every }t\in\mathbb{R}\,.$$ That is, $f$ is a strictly increasing function on $\mathbb{R}$. Now, if $(x,y,z)\in\mathbb{R}$ are such that $y=f(x)$, $z=f(y)$, and $x=f(z)$, then $x=y=z$. To show this, without loss of generality, we suppose on the contrary that $x\neq y$. If $x>y$, then $$y=f(x)>f(y)=z\text{ and so }x=f(z)<f(y)=z<y\,,$$ which is a contradiction. Likewise, if $x<y$, then $$y=f(x)<f(y)=z\text{ and so }x=f(z)>f(y)=z>y\,,$$ which is absurd. Hence, we indeed have $x=y=z$. Since $f(t)=t$ iff $t\in\{-2,0,+2\}$, we conclude that the solutions $(x,y,z)$ are $$x=y=z=-2\,,\,\, x=y=z=0\,,\,\,\text{ and }x=y=z=+2\,.$$
Alternatively, we can avoid using calculus to show that $f$ is strictly increasing. Note that $$f(s)-f(t)=(s-t)\,\left(\frac{3s^2t^2+4s^2+4t^2-32st+48}{(3s^2+4)(3t^2+4)}\right)\text{ for all }s,t\in\mathbb{R}\,.$$ Since $$3s^2t^2+4s^2+4t^2-32st+48=3(st-4)^2+4(s-t)^2\geq 0\text{ for all }s,t\in\mathbb{R}$$ and the inequality becomes an equality iff $s=t=-2$ or $s=t=+2$, we conclude that $$f(s)>f(t)\text{ if }s,t\in\mathbb{R}\text{ are such that }s>t\,.$$ That is, $f$ is a strictly increasing function.