TITLE: How can I prove the inequality $\frac{1}{x} +  \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$?
QUESTION [5 upvotes]: For $x > 0$, $y > 0$, $z > 0$, prove:
$$\frac{1}{x} +  \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z} .$$
I can see that this is true, I also checked it with a few numbers. But I guess that is not enough to prove it. So how can I prove it?
(You don't need to show the whole proof, I think just a hint will be enough. I simply don't know how to start.)

REPLY [5 votes]: We don't need all above. We can use simple inequality 
arithmetic mean ≥ geometric ≥ harmonic mean 
$$
\frac{x+y+z}{3} ≥ \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}} \iff (x+y+z)\left(\frac{1}{x} +  \frac{1}{y} + \frac{1}{z}\right) ≥ 9
$$
hence proved..