TITLE: Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$
QUESTION [6 upvotes]: Please help me for prove this inequality:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$

REPLY [5 votes]: There is a basic inequality you should know:
$$(a_{1}+a_{2}+\cdots+a_{n}) \cdot \left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots +\frac{1}{a_{n}}\right) \ge n^2 $$
that is elementarily solved by Cauchy-Schwarz in a single row proof 
$$\left(\sqrt{a_{1}}\frac{1}{\sqrt{a_{1}}}+\cdots +\sqrt{a_{n}}\frac{1}{\sqrt{a_{n}}}\right)^2=n^2\le (a_{1}+\cdots+a_{n}) \cdot \left(\frac{1}{a_{1}}+\cdots +\frac{1}{a_{n}}\right)  $$
Q.E.D.