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?
Best Answer
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.