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Show that if $ x_1,x_2,x_3 \in \mathbb{R}$ , and $x_1+x_2+x_3=0$ , we can say that:

$$\sum_{i=1}^{3}\frac{1}{x^2_i} = \left({\sum_{i=1}^{3}\frac{1}{x_i}}\right)^2.$$

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You can easly find that $x_3=-x_1-x_2$. In order to simplify rhs you may use formula $(a+b-c)^2=a^2+b^2+c^2+2ab-2ac-2bc$ –  no identity Feb 29 '12 at 11:56
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Do I have to show that, or do you have to show that? If this is a homework question please use the [homework] tag. –  Asaf Karagila Feb 29 '12 at 12:08
    
@Norbert: Let us try to keep (and use) the symmetry of the problem (if possible). –  Did Feb 29 '12 at 14:00
    
I don't insist . –  no identity Feb 29 '12 at 14:02

2 Answers 2

Hint:
What is value of $\frac{1}{x_1.x_2}+\frac{1}{x_2.x_3}+\frac{1}{x_3.x_1}$ ,when $x_1+x_2+x_3=0$.

If you got the value of $\frac{1}{x_1.x_2}+\frac{1}{x_2.x_3}+\frac{1}{x_3.x_1}$, then proceed by expanding $(\sum_{i=1}^3 \frac{1}{x_i})^2$ by using the formula $(a+b+c)^2= a^2+b^2+c^2+ 2(ab+bc+ac)$

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Thanks , I figured it out. –  Felipe Feb 29 '12 at 12:30

Take the equatin $x_1+x_2+x_3=0$, divide by $x_1x_2x_3$, multiply by $2$, add $x_1^{-2}+x_2^{-2}+x_3^{-2}$. This is essentially reverse-engineered from taking the suspected equality, multiplying out the right-hand side, subtracting out the left-hand side and multiplying by $x_1x_2x_3$...

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