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How to prove that $a^2+b^2+c^2=(a+b+c)^2$ given that $\frac1a+\frac1b+\frac1c=0$?

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closed as off-topic by Davide Giraudo, dfeuer, user1337, Adriano, azimut Aug 28 '13 at 22:00

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It will be better if you show at least some of your work or whatever you have attempted ... that way people would be more likely to answer your questions – user67258 Aug 3 '13 at 2:41

2 Answers 2

Expand the right side:

$$(a+b+c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc.$$

From the supplementary condition we have

$$\frac{1}{a}+\frac{1}{b} + \frac{1}{c} = 0$$


$$\frac{ab+ac+bc}{abc} = 0.$$

Therefore $ab+ac+bc = 0$ and the result follows. (None of $a,b,c$ can be $0$ else their inverses would be undefined and so the supplementary condition would be ill-posed.)

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since $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0$.

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