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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?

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    $\begingroup$ What exactly is a "solution set" and what does it mean for a solution set to be unordered? If a "solution set" is a set, then of course it is "unordered" because all sets are. Why would this imply that $x=y=z$? $\endgroup$ – littleO May 14 '18 at 10:48
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    $\begingroup$ Any solution of $(x-2y)(y-2x)=0$, if permuted, results in another solution. That doesn't mean we can conclude that $x=y$. It just means our solution set is symmetric across the line $x=y$. $\endgroup$ – G Tony Jacobs May 14 '18 at 10:57
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    $\begingroup$ That's still a claim that would require proof. $\endgroup$ – G Tony Jacobs May 14 '18 at 11:05
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    $\begingroup$ Would $x+y+z=4$ be a counterexample to your logic? $\endgroup$ – Bram28 May 14 '18 at 11:21
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    $\begingroup$ see jstor.org/stable/2975573 entitled "Do symmetric problems have symmetric solutions?" $\endgroup$ – Matthew Towers May 14 '18 at 11:48
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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.

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  • $\begingroup$ Ah. Of course. I didn't realize that. However, the question asks for positive reals. Could you provide an example where the solution set is wholly positive and unordered, but where the unknowns aren't equal? $\endgroup$ – MalayTheDynamo May 14 '18 at 11:04
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    $\begingroup$ How about the same left-hand sides, but they're all equal to $3$? Then every permutation of $(1,1,2)$ works. $\endgroup$ – G Tony Jacobs May 14 '18 at 11:07
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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 $z\gt0$.

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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.

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  • $\begingroup$ This isn't cyclic. Cyclic would be $3x^3+2y^2+z=11,3y^3+2z^2+x=11,3z^3+2x^2+y=11$. What I mean by cyclic is that you can just shift the places of the variables one to the left or one to the right to get another one of the given equations. $\endgroup$ – MalayTheDynamo May 14 '18 at 13:46
  • $\begingroup$ @MalayTheDynamo Thanks for the comment. That is quite different from what I was thinking. $\endgroup$ – Mohammad Riazi-Kermani May 14 '18 at 14:09
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    $\begingroup$ @MalayTheDynamo That's not the standard definition of "cyclic". $\endgroup$ – Acccumulation May 14 '18 at 17:40
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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 $f(42,\pi,e)\ne 0$.

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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.

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