<s>[INST] AM-GM application How to show that 
$$ \displaystyle (a_1 + a_2 + a_3)(\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_3}) \ge 9 $$
$a_1,a_2,a_3$ are all of same algebraic sign.
 [/INST] WLOG we may assume $a_{1}\leq a_{2}\leq a_{3}$.
$$\dfrac{a_{1}}{a_{1}}+\dfrac{a_{2}}{a_{2}}+\dfrac{a_{3}}{a_{3}}=3\tag{1}$$
It is not hard to see that $a_{1},a_{2},a_{3}$ and $\dfrac{1}{a_{1}},\dfrac{1}{a_{2}},\dfrac{1}{a_{3}}$ are oppositely sorted
Using Rearrangement Inequality we find
$$\dfrac{a_{1}}{a_{2}}+\dfrac{a_{2}}{a_{3}}+\dfrac{a_{3}}{a_{1}}\geq \dfrac{a_{1}}{a_{1}}+\dfrac{a_{2}}{a_{2}}+\dfrac{a_{3}}{a_{3}}=3\tag{2}$$ 
$$\dfrac{a_{1}}{a_{3}}+\dfrac{a_{2}}{a_{1}}+\dfrac{a_{3}}{a_{2}}\geq \dfrac{a_{1}}{a_{1}}+\dfrac{a_{2}}{a_{2}}+\dfrac{a_{3}}{a_{3}}=3\tag{3}$$ 
Add the 3 inequalities to get the required result.
Another way would be to apply AM-GM since on expanding the RHS we get the terms of the form $$\dfrac{x}{y}+\dfrac{y}{x}$$ which is obviously $\geq 2$
There will be 3 such pairs + the result from (1) gives us the RHS$\geq 9$
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