Determine the triplets of real numbers $(a, b, c)$ such that

$$
a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \text { and } a^{2}+b^{2}+c^{2}=\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}
$$
By multiplying the first equation by $s_{3}$ and the second by $s_{3}^{2}$, we get $s_{1} s_{3}=s_{2}$ and $s_{3}^{2}\left(s_{1}^{2}-2 s_{2}\right)=s_{2}^{2}-2 s_{3} s_{1}=s_{2}^{2}-2 s_{2}$, which simplifies to $s_{2}^{2}-2 s_{2} s_{3}^{2}=s_{2}^{2}-2 s_{2}$, and $s_{2}\left(s_{3}^{2}-1\right)=0$. If $s_{2}=0, s_{3} s_{1}=0$. If $s_{3}=0$, one of the numbers is zero, and if $s_{1}=0, a, b, c$ are roots of the polynomial $X^{3}-s_{3}$ which has only one real root (even with multiplicities). Thus, $s_{3}^{2}=1$.

- If $s_{3}=1$, then $s_{1}=s_{2} . a, b, c$ are roots of the polynomial $X^{3}-k X^{2}+k X-1=(X-1)\left(X^{2}-(k-1) X+1\right)$. The solutions in this case are of the form $\left(1, \frac{k-1+\sqrt{(k-1)^{2}-4}}{2}, \frac{k-1-\sqrt{(k-1)^{2}-4}}{2}\right),|k-1| \geqslant 2$, as well as their permutations.
- If $s_{3}=-1$, we reason similarly and obtain the solutions $\left(-1, \frac{k+1+\sqrt{(k+1)^{2}-4}}{2}, \frac{k+1-\sqrt{(k+1)^{2}-4}}{2}\right),|k+1| \geqslant 2$ up to permutations.