<s>[INST] Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$ for positive $a,b,c$ 
Show that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq \frac{9}{a+b+c}$, if $a,b,c$ are positive.

Well, I got that $bc(a+b+c)+ac(a+b+c)+ab(a+b+c)\geq9abc$. 
 [/INST] Approach 1: Write the desired inequality as 
$$
(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq 9\tag{i}
$$
and use the AM-GM inequality $x+y+z\geq 3(xyz)^{1/3}$ to each sum in the parentheses above.

Edit: just a few more approaches (among potentially many others). Hope you'll find this useful.
Approach 2:
Multiply out the LHS of (i) as
$$
1+1+1+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)
$$
and use $x+y\geq 2\sqrt{xy}$ 3 times.
Approach 3: Use Cauchy-Schwarz like Macavity suggested below:
$$
\left(\sqrt{a}^2+\sqrt{b}^2+\sqrt{c}^2\right)\left(\sqrt{\frac{1}{a}}^2+\sqrt{\frac{1}{b}}^2+\sqrt{\frac{1}{c}}^2\right)\geq(1+1+1)^2=9
$$
Approach 4: Use Jensen's inequality: the function $f(x)=\frac{1}{x}$ is convex for $x>0$ so:
$$
\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{3}f(a)+\frac{1}{3}f(b)+\frac{1}{3}f(c)\\
\geq f\left(\frac{1}{3}(a+b+c)\right)=\frac{3}{a+b+c}
$$
from which you can rearrange to obtain your claim.
Approach 5: Please also look up the Chebyshev's sum inequality from which your inequality is an immediate consequence.
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