I have been trying to find my error in the following question for a while, but am yet to succeed:
Find all triples $(a,b,c)$ of real numbers that satisfy the system of equations:
$$\begin{align} a+b+c&=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\[6pt] a^2+b^2+c^2&=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \end{align}$$
I attempted to do it in the following way:
$(a+b+c)^2=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}$
$a^2+b^2+c^2+2ab+2bc+2ac=\frac{2}{ab}+\frac{2}{ac}+\frac{2}{bc}+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}$
$ab+bc+ac=\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}$
From which we have a solution $(|a|, |b|, |c|)=(1,1,1)$
$(a-\frac{1}{a})^2=(\frac{1}{b}-b+\frac{1}{c}-c)^2$
$a^2-\frac{1}{a}^2=\frac{1}{b^2}+b^2+\frac{1}{c^2}+\frac{2}{bc}-\frac{2c}{b}-\frac{2b}{c}+2bc-2$
So we have: $\frac{1}{b^2}+b^2+\frac{1}{c^2}+c^2+\frac{2}{bc}-\frac{2c}{b}-\frac{2b}{c}+2bc-2=\frac{1}{b^2}-b^2+\frac{1}{c^2}-c^2$(combining the second equation in the statement with the one above)
$b^2+c^2+\frac{1}{bc}+bc-\frac{c}{b}-\frac{b}{c}=2$
How can I use the greatest portion possible of what I have done thus far to reach solve the question. My intuition tells me that we will have to use inequalities, however I can't fathom to see how.