<s>[INST] Is the proof I am using, sufficient/ correct for the system of equation? I want to ask MSE to confirm the correctness of the alternate solution and its mistake. 
I know possible solution: https://math.stackexchange.com/a/2557094/456510
If $x,y,z\in {\mathbb R}$, Solve the system equation:

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

I wrote a solution myself (after more work).
My attempts / solution:
It is obvious that, if $x>0,y>0,z>0$ are solutions, $x<0,y<0,z<0$ are also solutions and it is obvious $x≠0,y≠0,z≠0$.
If the equations have a solution, then $ x = y = z $ should be.

Proof:
I will accept $x,y,z\in {\mathbb R^+}$ 
a-1)
Let $x≥z>y$
We can write :
$z^4>y^4 \\ x^2≥z^2 \\ z^4+x^2+4>y^4+z^2+4 \\ 5xy > 5zx \\ y>z$
We get the contradiction : $y>z$
Because, it must be $z>y$
a-2)
Let $x>z≥y$
We can write:
$z^4≥y^4 \\ x^2>z^2 \\ z^4+x^2+4>y^4+z^2+4 \\ 5xy > 5zx \\ y>z$
We get the same contradiction : $y>z$
Because, it must be $z≥y$
b)
$y≥x>z$ 
We can write:
$x^4>z^4 \\ y^2≥x^2 \\ x^4+y^2+4>z^4+x^2+4 \\ 5yz > 5xy \\ z>x$
But, this is contradiction, because it must be $z<x$.
We get the same contradiction for : $y>x≥z$
c)
$y>z≥x$
We can write:
$y^4>z^4 \\ z^2≥x^2 \\ y^4+z^2+4>z^4+x^2+4 \\ 5zx > 5xy \\ z>y$
But, this is contradiction, because it must be $z<y$.
We get the same contradiction for : $y≥z>x$
d)
$z>x≥y$
We can write:
$z^4>x^4 \\ x^2≥y^2 \\ z^4+x^2+4>x^4+y^2+4 \\ 5xy > 5yz \\ x>z$
But, this is contradiction, because it must be $z>x$.
We get the same contradiction for : $z≥x>y$
e)
$z≥y>x$
We can write:
$y^4>x^4 \\ z^2≥y^2 \\ y^4+z^2+4>x^4+y^2+4 \\ 5zx > 5yz \\ x>y$
But, this is contradiction, because it must be $x<y$.
We get the same contradiction for : $z>y≥x$
f)
$x>y≥z$
We can write:
$x^4>y^4 \\ y^2≥z^2 \\ x^4+y^2+4>y^4+z^2+4 \\ 5yz > 5zx \\ y>x$
But, this is contradiction, because it must be $x>y$.
We get the same contradiction for : $x≥y>z$
Then, solution must be $x=y=z$ (if there is a solution).
The proof is completed.

Finally, 

$$x^4+x^2+4-5x^2=0 \Rightarrow x^4-4x^2+4=0 \Rightarrow (x^2-2)^2=0 \Rightarrow x=±\sqrt2\Rightarrow x=y=z=±\sqrt2 .$$

Is my proof/ solution correct?
Thanks. 
 [/INST] Good job. Just too verbose.
You are correct into saying you can assume $x$, $y$ and $z$ all positive (there will be a corresponding solution with their negatives). The case when two are positive and one negative cannot appear, nor can the case of two negative and one positive, because the positivity of the left-hand sides forces positivity of the right-hand sides, so all three numbers must share the sign.
However, there is another simplification, namely, you can also assume $x$ is the maximum solution, because the equations are cyclic. Thus
$$
x\ge y\ge z \qquad\text{or}\qquad x\ge z>y
$$
You have already excluded the second case, so we can concentrate on the first.
In order to show that for a solution you need $x=y=z$, you just have to exclude $x>y$ and $y>z$.
In the case $x>y\ge z$, we have, according to your method,
$$
x^4>y^4
\qquad
y^2\ge z^2
$$
Then
$$
5yz=x^4+y^2+4>y^4+z^2+4=5zx
$$
which implies $y>x$: a contradiction.
In the case $x\ge y>z$ we have
$$
y^2>z^2
\qquad
x^4\ge y^4
$$
which implies
$$
5yz=x^4+y^2+4>y^4+z^2+4=5zx
$$
implying $y>x$, again a contradiction.
We saw that assuming either $x>y$ or $y>z$ leads to a contradiction. Since $x\ge y\ge z$ by assumption and we cannot have neither $x>y$ nor $y>z$, we deduce that $x=y$ and $y=z$.
Now, finding what's the common value is easy: we have
$$
x^4-4x^2+4=0
$$
so $x^2=2$ and $x=\pm\sqrt{2}$. The problem has exactly two solutions.
</s>