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.) 
 A: $$\frac{1}{x} +  \frac{1}{y} + \frac{1}{z} \geq \frac{9}{x+y+z}$$
$\leftrightarrow (x+y+z)(\frac{1}{x} +  \frac{1}{y} + \frac{1}{z}) \geq 9$.
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 $x=y=z$.
A: 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..
A: 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.
A: In general, the harmonic mean is at most the arithmetic mean, $$((1/n)(x_1^{-1}+\cdots+x_n^{-1}))^{-1}\le(1/n)(x_1+\cdots+x_n)$$ (assuming all quantities positive). Your question is the case $n=3$. 
A: Since $w+\frac{1}{w} \geq 2$ for all $w>0$, we have
$$
\begin{split}
(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right) 
&= 3 + \frac{x}{y}+\frac{y}{x} + \frac{x}{z}+ \frac{z}{x} +\frac{y}{z}+\frac{z}{y} \\&\geq 3+2+2+2 = 9
\end{split}
$$
A: This also follows from the Cauchy-Schwarz inequality:
Let $u=( x,y,z)$ and $v=( \frac{1}{x}, \frac{1}{y}, \frac{1}{z})$.
Then $u \cdot v \leq \|u\| \|v\|$ is exactly your inequality.  
P.S. I personally find the other proofs mentioned above much simpler... And I would use the C-S inequality for numbers not vectors. But from my experience, many undergrad students know the C-S vector inequality and yet they don't know the HM-GM-AM inequality, this is the only reason I mentioned this proof :)
