How can I prove the inequality $\frac{1}{x} +  \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$? 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.) 

$$\frac{1}{x} +  \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$$
$\leftrightarrow (x+y+z)(\frac{1}{x} +  \frac{1}{y} + \frac{1}{z}) \geq 9$.
Using Cauchy Inequality, we have $x+y+z\geq 3\sqrt[3]{xyz}$ and $\frac{1}{x} +  \frac{1}{y} + \frac{1}{z}\geq 3\sqrt[3]{\frac{1}{xyz}}$
$$\Rightarrow (x+y+z)\left(\frac{1}{x} +  \frac{1}{y} + \frac{1}{z}\right) \geq 3\sqrt[3]{xyz}\cdot 3\sqrt[3]{\frac{1}{xyz}} = 9$$.
Equality occurs when $x=y=z$.
