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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.

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    Is this for a class? If so, what class?2010-12-04
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    It's valid only when the numbers $a_1,a_2,a_3$ are all positive or all negative. A simple counter example would be $2,-1000,2$. I shall edit it.2013-03-23

6 Answers 6

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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).

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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.

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    I do think the intention was to prove AM-HM... :)2010-12-05
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    @J.M.: I agree, but I was thinking that it could be helpful to provide some information on how people solve this kind of problems in general.2010-12-05
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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).

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    This is a question tagged algebra-precalculus, so using Lagrange multiplier may not be a good idea.2010-12-04
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    @Soarer For use Lagrange multipliers are needed conditions do not exist here. Is simply to minimize the function, do not multipliers Lagrange .2010-12-04
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    @Soarer: If he wishes to nuke mosquitoes, who are we to prevent him? ;)2010-12-05
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    @J.M. Jajaja sorry, I think I went too far2010-12-05
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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.

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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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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.