Inequality problem involving QM-AM-GM-HM or Cauchy Schwarz inequality 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!
 A: 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.
A: 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.
A: 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.
