Question: Prove that if $x$, $y$, $z$ are positive real numbers then the following inequality holds: $$\frac{x+y}{x^2+y^2}+\frac{y+z}{y^2+z^2}+\frac{z+x}{z^2+x^2}\leq\frac{1}{x}+\frac{1}{y}+\frac{1}{z}.$$I tried thinking of applying QM-AM, but I didn't know what variables to use. I also thought that this could be related to Cauchy's inequality, but I wasn't sure, again, of what variables to use. Any help would be greatly appreciated, thanks!

  • $\begingroup$ Hey @Pakquebchsoflwty, please check my answer! =) $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 6 '14 at 23:23
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    $\begingroup$ @ Jose Arnaldo Dris, already done :D thank you again :). $\endgroup$ – Pakquebchsoflwty Nov 6 '14 at 23:25

We want to show that:


Note that, by the QM-AM Inequality:

$${\left(\frac{x+y}{2}\right)}^2 \leq \frac{x^2 + y^2}{2}.$$


$$\frac{x + y}{x^2 + y^2} \leq \frac{2}{x + y}.$$

Therefore, it remains to show that:

$$\frac{2}{x + y} + \frac{2}{y + z} + \frac{2}{x + z} \leq \frac{1}{x} + \frac{1}{y} + \frac{1}{z}.$$

By the AM-HM Inequality:

$$\frac{2}{\frac{1}{x} + \frac{1}{y}} \leq \frac{x + y}{2}.$$


$$\frac{2}{x + y} \leq \frac{\frac{1}{x} + \frac{1}{y}}{2}$$


$$\frac{2}{y + z} \leq \frac{\frac{1}{y} + \frac{1}{z}}{2},$$


$$\frac{2}{x + z} \leq \frac{\frac{1}{x} + \frac{1}{z}}{2}.$$


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    $\begingroup$ This is a nicely explained solution, thanks. $\endgroup$ – Pakquebchsoflwty Nov 6 '14 at 23:23
  • $\begingroup$ You're welcome! =) $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 6 '14 at 23:24
  • $\begingroup$ @Pakquebchsoflwty Well it's basically my solution posted later and with unnecessarily lenghty explanation. $\endgroup$ – user2345215 Nov 6 '14 at 23:26
  • $\begingroup$ @ user2345215, I realize that, however, his is easier to follow and although he posted it later, it was only by 1 minute, which does not make too much of a difference. $\endgroup$ – Pakquebchsoflwty Nov 6 '14 at 23:30

It's true that $(x+y)^2\le2(x^2+y^2)$ and it can be seen as a consequence of CS, so $$\frac{x+y}{x^2+y^2}+\frac{y+z}{y^2+z^2}+\frac{z+x}{z^2+x^2}\le\frac2{x+y}+\frac2{y+z}+\frac2{z+x}.$$ Additionally from $(x+y)^2\ge4xy$ it can be deduced that $$\frac2{x+y}\le\frac1{2x}+\frac1{2y},$$ hence the inequality follows easily.


after some algebra we get $y^3(x-z)(x^4-z^4)+z^3(x-y)(x^4-y^4)+x^3(y-z)(y^4-z^4)\geq 0$ which is true.This have i got by canceling the denominators and simplifying the given terms and moving all to the left side.

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    $\begingroup$ Could you explain how you got there? $\endgroup$ – Pakquebchsoflwty Nov 6 '14 at 23:17

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