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:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq \frac{9}{a+b+c} : (a, b, c) > 0$$
 A: Depends on the machinery you are allowed. It is immediate from Cauchy-Schwarz. 
A: 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.
A: It's a convexity inequality: the function $f(x) = \frac{1}{x}$ is convex on $\mathbb{R}^{+*}$ (its second derivative is positive), so for any $a,b,c\in\mathbb{R}^{+*}, \frac{1}{3}(f(a)+f(b)+f(c))\geq f(\frac{a+b+c}{3})$, which is $\frac{1}{3}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq\frac{3}{a+b+c}$, or finally $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{9}{a+b+c}$
A: 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. 
A: This inequality can also be rewritten as 
$$\frac{a+b+c}{3} \geq \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}} \,,$$
which is just the AM-HM inequality.
A more direct proof would be to simply multiply:
$$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=3+ (\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}+\frac{c}{a})+(\frac{c}{b}+\frac{b}{c}) \geq 3+2+2+2 =9$$
A: 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
$(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})(a+b+c)$=$(\frac{a}{b}+\frac{b}{a})+(\frac{a}{c}\frac{c}{a})+(\frac{b}{c}+\frac{c}{b})+3\geq 2+2+2+3=9$
