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

jimjim
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user88784
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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 –  Aug 03 '13 at 02:41

2 Answers2

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Expand the right side:

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

From the supplementary condition we have

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

Or

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

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.)

Cameron Williams
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$(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc=a^{2}+b^{2}+c^{2}+2abc\big(\frac{1}{c}+\frac{1}{b}+\frac{1}{a}\big)=a^{2}+b^{2}+c^{2}$

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

user71352
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