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For $x,y,z \in \mathbb{R}_{> 0}$, how can I prove that there are solutions to this set of equations,

$$x+y=xyz \land xy=x+z \land y+z=yz$$ in a legitimate way, not by feeding it to WolframAlpha?

I have found out that

$x = \frac{y}{yz-1} = \frac{z}{y-1}$

$y = \frac{x}{xz-1} = \frac{x+z}{x} = \frac{z}{z-1}$

$z = \frac{x+y}{xy} = xy-x = \frac{y}{y-1}$

but I am not sure how to progress from there.

vlnkaowo
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1 Answers1

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The solutions can be parametrized by $$ x=y - 2, \quad z=y^2 - 3y + 2, $$ where $y$ has to satisfy the equation $$ y^3 - 4y^2 + 4y - 2=0. $$ Every cubic polynomial has a real root, and this one has a positive one satisfying $y>2$. So we are done, since $x,z>0$ for any $y>2$.

Dietrich Burde
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  • it's not a typo, it is probably not supposed to be an integer solution so it's not that easy – vlnkaowo Nov 27 '20 at 13:09
  • OK, then my solution is correct. – Dietrich Burde Nov 27 '20 at 13:09
  • I am still worried that this solution doesn't really eliminate the need for a calculator. I don't think the author of the exercise meant to use the brutal cubic formula which I would probably need here, but thanks regardless. – vlnkaowo Nov 27 '20 at 13:41
  • You don't have to use the "brutal" cubic formula. You can do this much easier to see that the real root is positive. – Dietrich Burde Nov 27 '20 at 13:44
  • positive - sure, but I still have to work with $x=y-2$ for example, where $y > 0$ isn't enough to tell whether x is positive or not. Or am I missing something? – vlnkaowo Nov 27 '20 at 13:55
  • You only need that $y>2$, right? And this is clear by an easy estimate, without a brutal formula. I have edited this. – Dietrich Burde Nov 27 '20 at 13:57