(Find the triplets $(a, b, c)$ of non-zero real numbers such that

$$
\left\{\begin{array}{l}
a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \\
a^{2}+b^{2}+c^{2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
\end{array}\right.
$$
Let $(a, b, c)$ be a solution, for example $(1,1,1)$. We first observe that

$$
(a+b+c)^{2}=\left(a^{2}+b^{2}+c^{2}\right)+2(a b+b c+c a)
$$

and that

$$
\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^{2}=\left(\frac{1}{a}^{2}+\frac{1}{b}^{2}+\frac{1}{c}^{2}\right)+2\left(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\right)
$$

We deduce that $a b+b c+c a=\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}$.

Now,

$$
a b+b c+c a=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) a b c=(a+b+c) a b c
$$

and

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

Therefore, $a+b+c=0$ or $a b c=\frac{1}{a b c}$, i.e., $a b c= \pm 1$.

If $a+b+c=0$, we have seen that $a b+b c+c a=0$, so $a^{2}+b^{2}+c^{2}=0$, which contradicts the non-zero nature of the real numbers $a, b$, and $c$. Thus, there exists a number $\varepsilon= \pm 1$ such that $a b c=\varepsilon$, in which case

$$
\begin{aligned}
(t-a)(t-b)(t-c) & =t^{3}-(a+b+c) t^{2}+(a b+b c+c a) t-a b c \\
& =t^{2}-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) t^{2}+\left(\frac{1}{a b}+\frac{1}{b c}+\frac{1}{c a}\right) t-\frac{1}{a b c} \\
& =\left(t-\frac{1}{a}\right)\left(t-\frac{1}{b}\right)\left(t-\frac{1}{c}\right)
\end{aligned}
$$

Therefore, the triplets $(a, b, c)$ and $(1 / a, 1 / b, 1 / c)$ are identical up to order, in which case the equalities in the statement are indeed satisfied; in other words, the function $x \mapsto 1 / x$ maps each of the numbers $a, b$, and $c$ to one of the numbers $a, b$, and $c$.

Either each is mapped to itself, which means that $a, b, c= \pm 1$. Or, by permuting $a, b$, and $c$, the number $a$ is mapped to $b$, in which case $b$ is mapped to $a$ and $c$ to itself.

In conclusion, the solutions are the triplets $\left( \pm 1, x, \frac{1}{x}\right)$ and their permutations.