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).
 A: I checked the Possible Solutions thoroughly. The following solution is possible:

$$x^4+y^2+4+y^4+z^2+4+z^4+x^2+4-5yz-5xz-5xy=0 \Rightarrow (x^4-4x^2+4)+(y^4-4y^2+4)+(z^4-4z^2+4)+\left(\frac {5x^2}{2}-5xy+\frac {5y^2}{2} \right)+\left(\frac {5x^2}{2}-5xz+\frac {5z^2}{2} \right)+\left(\frac {5y^2}{2}-5yz+\frac {5z^2}{2} \right)=0\Rightarrow  (x^2-2)^2+(y^2-2)^2+(z^2-2)^2+\left(x \sqrt{\frac 52}-y \sqrt{\frac 52}\right)^2+
\left(x \sqrt{\frac 52}-z \sqrt{\frac 52}\right)^2+
\left(y \sqrt{\frac 52}-z \sqrt{\frac 52}\right)^2=0$$

You can continue from here.
Only solutions are $x=y=z=±\sqrt2$
A: Summing up all the equtions you find:
$$\sum_{cyc} x_i^4 + \sum_{cyc} x_i^2 - 5\sum_{cyc} x_ix_j +12= 0$$
Now manipulate in such way to have the equality as sum of squares!
EG
$$-5xy =\frac52 \left( x-y\right)^2-\frac52x^2-\frac52y^2$$
That is:
$$x^4+y^4+z^4-4x^2-4y^2-4z^2+\frac52 \left( x-y\right)^2+\frac52 \left(y-z\right)^2+\frac52 \left( z-x\right)^2+12=0$$
$$(x^2-2)^2+(y^2-2)^2+(z^2-2)^2+\frac52 \left( x-y\right)^2+\frac52 \left(y-z\right)^2+\frac52 \left( z-x\right)^2=0$$
The system has 2 different solutions:

$$x=y=z=\sqrt 2$$
and
$$x=y=z=-\sqrt 2$$

A: 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.  
A: Another way, adding gives you
$$\sum x^4 + \sum x^2 +4 =5\sum xy$$
where $\sum$ represents cyclic sums. However by AM-GM,
$\sum (x^4+4) \geqslant 4\sum x^2$ and $\sum x^2 \ge \sum xy$. 
So we need equality in both those inequalities, which is possible iff $a=b=c=\pm\sqrt2$. 
