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.) 

You can also brute-force it along with some careful re-grouping of terms:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}$
$\Leftrightarrow \frac{xy+xz+yz}{xyz}\geq \frac{9}{x+y+z}$
$\Leftrightarrow x^2 y + xyz + x^2 z + x y^2 + xyz + y^2 z + xyz + xz^2 + y z^2 \geq 9xyz$
$\Leftrightarrow x^2y-2xyz + yz^2 + x^2z-2xyz + yz^2 + xz^2-2xyz + xy^2 \geq 0$
$\Leftrightarrow y(x-z)^2 + z(x-y)^2 + x(y-z)^2 \geq 0$.
The last inequality holds (since $x,y,z>0$), therefore the first inequality holds.
