$x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$, prove that $xyz=1$

Question

Let $x>0$, $y>0$ and $z>0$ such that $x\neq y $ , $y\neq z$ and $x\neq z$. If $x+\dfrac{1}{y}=y+\dfrac{1}{z}=z+\dfrac{1}{x}$, prove that $xyz=1$

score 7 · Accepted Answer · answered Oct 19 '16 at 21:56

7

Hint: Notice that \begin{align} x-y &= \frac{y-z}{yz} \\ y-z &= \frac{z-x}{xz} \\ z-x &= \frac{x-y}{xy}. \end{align} Now multiply.

answered Oct 19 '16 at 21:56

Joey Zou

1

nice. but what you get is $(xyz)^2 = 1$, which mean $xyz$ could be $-1$ as well. – Ahmad Bazzi Oct 19 '16 at 22:12
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@ElBazzi by hypothesis $x,y,z>0$. – Joey Zou Oct 19 '16 at 22:13
OH okay.. My bad, i did not see that – Ahmad Bazzi Oct 19 '16 at 22:16
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btw it seems the assumptions of this question cannot be satisfied. Indeed $x,y,z$ cannot all be $1$ so suppose $x\neq1$. Then $1-x=xy(z-1/y)=xy(x-1/x)=y(x^2-1)$, so $1=-y(x+1)<0$. – stewbasic Oct 20 '16 at 00:03

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