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

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.