Finding all real $(a,b,c)$ satisfying $a+b+c=\frac1{a}+\frac1{b}+\frac1{c}$ and $a^2+b^2+c^2=\frac1{a^2}+\frac1{b^2}+\frac1{c^2}$ 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.
 A: Let $a+b+c=3u$, $ab+ac+bc=3v^2$, where $v^2$ can be negative,  and $abc=w^3$.
Thus, our conditions give: $$uw^3=v^2$$ and $$(9u^2-6v^2)w^6=9v^4-6uw^3.$$
Now, if $v^2=0$ so $u=0,$ which gives $$ab-(a+b)^2=0$$ or $$a^2-ab+b^2=0,$$ which gives $a=b=0,$ which is impossible.
Thus, $u\neq0$, $v^2\neq0$ and $w^3\neq0$, which gives
$$(3u^2-2v^2)\frac{v^4}{u^2}=3v^4-2uw^3$$ or
$$v^6=u^3w^3$$ or
$$v^2=uw,$$ which with $v^2=uw^3$ gives $w^2=1$ and from here $w^3=1$ or $w^3=-1.$
For $w^3=1$ we obtain that $a$, $b$ and $c$ are roots of the equation:
$$x^3-3ux^2+3ux-1=0$$ or
$$(x-1)(x^2+x+1-3ux)=0,$$ which has three real roots for any real $u$ such that $$(1-3u)^2-4\geq0$$ or $$u\in\left(-\infty,-\frac{1}{3}\right]\cup[1,+\infty).$$
Id est, in this case we obtain the following solution: $\left(1,t,\frac{1}{t}\right)$ and any symmetric permutations of this.
The case $w^3=-1$ is a similar.
Also, by your work: $$(ab+ac+bc)abc=a+b+c$$ and by the given $$(a+b+c)abc=ab+ac+bc$$ and since easy to show that $a+b+c=ab+ac+bc=0$ is impossible (see my post),
we obtain $a^2b^2c^2=1.$
After this see my solution again.
