<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] 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
$$x=y=z=-\sqrt 2$$

</s>