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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$2012-02-29
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    @Norbert: Let us try to keep (and use) the symmetry of the problem (if possible).2012-02-29
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    I don't insist .2012-02-29

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