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

Question

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

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

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

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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

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

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

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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?

Answer 1

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.

Answer 2

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 $z\gt0$.

Answer 3

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.

Answer 4

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 $f(42,\pi,e)\ne 0$.

Answer 5

First, your use of the word "cyclic" isn't quite consistent with its usual meaning.

Second, a counterexample:
xy+z=0
yz+x = 0
zx+y = 0

(1,1,-1) satisfies these equations.