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

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.)
