If $x,y,z$ are real numbers, each greater than 1, then show that

$\frac{y-x}{y^2-1}$+$\frac{z-y}{z^2-1}$+$\frac{x-z}{x^2-1}\gt 0$

It is not the actual problem,I deducted the actual problem in those stage. Then I did not find any way. Seeing this problem ,I tried to use rearrangement inequality but found nothing good. Please help me.

  • 1
    $\begingroup$ Its not. In fact, I think $\frac {y-x}{y^2-1} + \frac {z-y}{z^2-1} + \frac {z-x}{x^2-1} \le 0$ $\endgroup$ – Doug M Apr 5 '18 at 3:09

First, I think your inequality is "$\leq$" instead of "$>$". The inequality is equivalent to

$$ \frac{x-1}{x^2-1} + \frac{y-1}{y^2-1} + \frac{z-1}{z^2-1} < \frac{z-1}{x^2-1} + \frac{x-1}{y^2-1} + \frac{y-1}{z^2-1}. $$

Then use rearrangement inequality on $x-1, y-1, z-1$ and $(x^2-1)^{-1}, (y^2-1)^{-1}, (z^2-1)^{-1}$.

  • $\begingroup$ Why the $-1$s in numerator? $\endgroup$ – Macavity Apr 5 '18 at 3:20
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    $\begingroup$ Uh, you are right. The $-1$ is redundant (but did not affect the correctness). $\endgroup$ – Hw Chu Apr 5 '18 at 3:28
  • $\begingroup$ @Hw Chu Why my inequality is equivalent to that you have written in the first line.In my inequality there is $(y-x)$ but in your answer there is $(y-1)$ $\endgroup$ – Sufaid Saleel Apr 5 '18 at 6:45

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