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.
 A: Hint: Cauchy-Schwarz
Hint2: Two applications of the AM-GM will also solve it: one is $(x+y+z)/3 \ge \cdots $ and the other is ... (try to fill in the details).
A: One more way is to note that it's just the AM-HM:
$$\frac{a_1 + a_2 + a_3}{3} \ge \frac{3}{\frac{1}{a_1} + \frac{1}{a_2} + \frac{1}{a_2}}$$
In general it's useful to remember the power mean inequality: Suppose that $\alpha \ge \beta$. Then
$$\left(\frac{a_1^\alpha + \dots + a_n^\alpha}{n}\right)^{1/\alpha} \ge \left(\frac{a_1^\beta + \dots + a_n^\beta}{n}\right)^{1/\beta}.$$ The inequality works also in the case that $\alpha$ or $\beta$ is zero, when we take that to mean the geometric mean.
A: HINT: The function   $f(x,y,z)=(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$ has the absolute minimum in $(1,1,1)$
Note that:
$\displaystyle\frac{\partial f}{\partial x}=\frac{1}{y}+\frac{1}{z}-\frac{y+z}{x^2}$
$\displaystyle\frac{\partial f}{\partial y}=\frac{1}{x}+\frac{1}{z}-\frac{x+z}{y^2}$
$\displaystyle\frac{\partial f}{\partial x}=\frac{1}{x}+\frac{1}{y}-\frac{x+y}{z^2}$
Remember that you must resolve:
$$\left\{
    \begin{array}{ll}
      \displaystyle\frac{\partial f}{\partial x}=0  \\
      \displaystyle\frac{\partial f}{\partial y}=0  \\
      \displaystyle\frac{\partial f}{\partial z}=0 
    \end{array}
  \right.$$
The solution of this system is: (1,1,1) (critical point).
A: One way, although not very thoughtfull, is to simply expand the brackets.  You will get a sum of 9 terms on the LHS.  3 of these terms will be 1's and the other 6 will be fractions involving the three "a" terms.  Get a common denominator for the pair of fractions a/b and b/a (three such pairs) then you can use the following to show each of these three fractions is at least 2, producing a total of at least 9.
http://en.wikipedia.org/wiki/Inequality_of_arithmetic_and_geometric_means
Sorry, you can only use this inequality if the numbers are non-negative.
A: 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$\geq 9$
A: Hint: $$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = 9 \cdot\frac{1}{3}(a+b+c)\cdot\frac{1}{3}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)$$
Assuming $a,b,c$ are positive, you can use the AM-GM inequality twice.
