Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$ Please help me for prove this inequality:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$

Beferore we prove this inequality first prove that the following inequality:
For $a>0$, $b>0$ is true this inequality: $\frac{a}{b}+\frac{b}{a}\geq 2$
$\frac{a}{b}+\frac{b}{a}-2$=$\frac{a^2+b^2-2ab}{ab}$=$\frac{(a-b)^2}{ab}$.
Since $a>0$, $b>0$  $\Rightarrow$  $a-b>0$ $\Rightarrow$ $(a-b)^2>0$,  $ab>0$.
Means, $(a-b)^2>0$, $ab>0$ $\Rightarrow$ $\frac{(a-b)^2}{ab}>0$. 
Since $a=b$ $\Rightarrow$ $\frac{(a-b)^2}{ab}=0$. 
Means, $\frac{(a-b)^2}{ab}\geq 0$ $\Rightarrow$ $\frac{a}{b}+\frac{b}{a}-2$=$\frac{(a-b)^2}{ab}\geq 0$.
$\frac{a}{b}+\frac{b}{a}-2\geq 0$ $\Rightarrow$ $\frac{a}{b}+\frac{b}{a}\geq 2$.
Now turn to prove the first inequality.
$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c}$
Since $a>0$, $b>0$, $c>0$ $\Rightarrow$ $a+b+c\geq 0$, so
$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(a+b+c)$=$(\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}\frac{c}{a})+(\frac{b}{c}+\frac{c}{b})+3\geq 2+2+2+3=9$
