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:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$

Inequality can be written as: 
$$\left(a+b+c\right) \cdot \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) \geq 9 .$$
And now we apply the $AM-GM$ inequality for both of parenthesis. So: 
$\displaystyle \frac{a+b+c}{3} \geq \sqrt[3]{abc} \tag{1}$ and $\displaystyle \frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3} \geq \frac{1}{\sqrt[3]{abc}} \tag{2}.$ Now multiplying relation $(1)$ with relation $(2)$ we obtained that :
$$\left(\frac{a+b+c}{3}\right) \cdot \left(\frac{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}{3}\right) \geq \frac{\sqrt[3]{abc}}{\sqrt[3]{abc}}=1.  $$ So, we obtained our inequality. 
