<s>[INST] If $x,y,z\in {\mathbb R}$, Solve this system equation: If $x,y,z\in {\mathbb R}$, Solve this system equation:

$$
\left\lbrace\begin{array}{ccccccl}
    x^4 & + & y^2 & + & 4         & = & 5yz
\\[1mm]
y^{4} & + & z^{2} & + & 4 & = &5zx
\\[1mm]
z^{4} & + & x^{2} & + & 4 & = & 5xy
\end{array}\right.
$$

This is an olympiad question in Turkey (not international), which that, I could not solve it.
My idea:
$xy=a \\ yz=b \\ xz=c$
$$
\left\lbrace\begin{array}{ccccccl}
    a^2c^3 & + & b^3a & + & 4b^2c        & = & 5b^3c
\\[1mm]
a^{3}b^2 & + & bc^3 & + & 4c^2a & = &5c^3a
\\[1mm]
b^{3}c^2 & + & a^3c & + & 4a^2b & = & 5a^3b
\end{array}\right.
$$
Yes, I know, this is a stupid idea, because it did not work at all ( last system equation is more difficult).
 [/INST] Your system can be written as follows 

$$
\left\lbrace\begin{array}{clc}\tag{1}
   \left( {x}^{2}-2 \right) ^{2}+ \left( y-z \right) ^{2}+ \left( x-z
 \right)  \left( x+z \right) +3\,({x}^{2}-\,zy)&=&0
\\ \\[1mm]
 \left( {y}^{2}-2 \right) ^{2}+ \left( z-x \right) ^{2}+ \left( y-x
 \right)  \left( y+x \right) +3(\,{y}^{2}-\,zx)&=&0
\\ \\[1mm]
\left( {z}^{2}-2 \right) ^{2}+ \left( x-y \right) ^{2}+ \left( z-y
 \right)  \left( z+y \right) +3(\,{z}^{2}-\,xy) &=&0
\end{array}\right.
$$

Now one solution of the $(1)$ is $x=y=z=\sqrt{2}$  which is mentioned in the first comment.  
</s>