1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ Hey @Pakquebchsoflwty, please check my answer! =) $\endgroup$ – Jose Arnaldo Bebita-Dris Nov 6 '14 at 23:23
  • 1
    $\begingroup$ @ Jose Arnaldo Dris, already done :D thank you again :). $\endgroup$ – Pakquebchsoflwty Nov 6 '14 at 23:25
3
$\begingroup$

We want to show that:

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

Note that, by the QM-AM Inequality:

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

Consequently:

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

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:

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

Therefore:

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

Similarly:

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

and

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

QED.

$\endgroup$
  • 1
    $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Could you explain how you got there? $\endgroup$ – Pakquebchsoflwty Nov 6 '14 at 23:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.