TITLE: Is it possible to shorten the solution for this 2014 RMO question?
QUESTION [15 upvotes]: 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?

REPLY [2 votes]: 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 $f(42,\pi,e)\ne 0$.