Is it possible to shorten the solution for this 2014 RMO question?

I was solving a question from the Regional Math Olympiad (RMO) 2014.

Find all positive real numbers $x,y,z$ such that

$$2x-2y+\frac1z=\frac1{2014},\quad2y-2z+\frac1x=\frac1{2014},\quad2z-2x+\frac1y=\frac1{2014}$$

Here's my solution:

These expressions are cyclic. Therefore all solution sets must be unordered. This implies that $x=y=z$.

Thus, $x=2014$ and the solution is

$$x=2014\quad y=2014\quad z=2014$$

Here's the official solution:

Adding the three equations, we get $$\frac1x+\frac1y+\frac1z=\frac3{2014}$$

We can also write them as $$2xz-2yz+1=\frac z{2014},\quad2xy-2xz+1=\frac x{2014},\quad2yz-2xy+1=\frac y{2014}$$

Adding these, we get $$x+y+z=3\times2014$$

Therefore, $$\left(\frac1x+\frac1y+\frac1z\right)(x+y+z)=9$$

Using $\text{AM-GM}$ inequality, we therefore obtain $$9=\left(\frac1x+\frac1y+\frac1z\right)(x+y+z)\ge9\times(xyz)^{\frac13}\left({1\over xyz}\right)^{\frac13}=9$$

Hence equality holds and we conclude that $x=y=z$.

Thus we conclude $$x=2014\quad y=2014\quad z=2014$$

What I wonder is if there is something wrong with my approach. If yes, what is it? If no, then why is the official solution so long winded?

Consider the system of equations

$xy + z = 1, \quad yz + x = 1, \quad zx + y = 1$

These equations are related by cyclic permutations of $(x,y,z)$, but they are satisified by $(1,1,0)$ (and its cyclic permutations) when $x$, $y$ and $z$ are not all equal.

There are also solutions where $x=y=z=\frac{\pm \sqrt{5}-1}{2}$, but these are not the only solutions.

As Mohammad Riazi-Kermani and gandalf61 show, you cannot conclude $x=y=z$ simply from the cyclic invariance of the system of equations. However, in this case you can make a simple argument that starts with an observation based on cyclic invariance, namely that you may as well assume $x\ge y,z$ (i.e., cycle through $(x,y,z)$, $(y,z,x)$ and $(z,x,y)$ and pick the one that starts with the largest of the three numbers).

If $x\ge y,z$, then $2x-2y\ge0$ while $2z-2x\le0$, so that

$${1\over2014}=2x-2y+{1\over z}\ge{1\over z}\implies z\ge2014$$

and

$${1\over2014}=2z-2x+{1\over y}\le{1\over y}\implies 2014\ge y$$

so we now have $x\ge z\ge 2014\ge y$. But this now tells us $2y-2z\le0$, so that

$${1\over2014}=2y-2z+{1\over x}\le{1\over x}\implies2014\ge x$$

so we now have $2014\ge x\ge z\ge 2014\ge y$, from which we see $x=z=2014\ge y$. The final equality, $2014=y$, comes by sustituting $x=z=2014$ into any of the three equations.

Note, the implication ${1\over2014}\ge{1\over z}\implies z\ge2014$ requires the assumption $z\gt0$.

The equations being cyclic does not necessarily mean the variables are equal.

For example $$ x+y+z=6\\x^2+y^2+z^2=14\\x^3+y^3+z^3=36$$ Solutions are not equal. $$x=1, y=2,z=3$$ is one solutions set.

The general cyclic system is $$ f(x,y,z)=f(y,z,x)=f(z,x,y)=0$$ with some function $f$. There is not the slightest reason to assume that $f(42,\pi,e)\ne 0$.