To find all triples $(a, b, c)$ of real numbers that satisfy the given system of equations, we start by analyzing the conditions:

1. The first equation is:
$$a + b + c = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}.$$
This can be rewritten as:
$$a + b + c = \frac{ab + ac + bc}{abc}.$$
Multiplying both sides by $abc$, we obtain:
$$abc(a + b + c) = ab + ac + bc.$$

2. The second equation is:
$$a^2 + b^2 + c^2 = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$
Similarly, this can be rewritten as:
$$a^2 + b^2 + c^2 = \frac{b^2c^2 + a^2c^2 + a^2b^2}{a^2b^2c^2}.$$
Multiplying both sides by $a^2b^2c^2$, we have:
$$a^2b^2c^2(a^2 + b^2 + c^2) = b^2c^2 + a^2c^2 + a^2b^2.$$

### Solution Strategy

To find solutions, consider particular simple configurations of the variables that are symmetric or involve known functional forms:

#### Case 1: One of the variables is $\pm 1$

Assume $a = 1$. Then the system of equations simplifies significantly:

- Substituting $a = 1$ in the first equation:
$$1 + b + c = 1 + \frac{1}{b} + \frac{1}{c} \implies b + c = \frac{1}{b} + \frac{1}{c}.$$
Multiplying through by $bc$:
$$bc(b + c) = b + c.$$
This implies if $b + c \neq 0$, then $bc = 1$.

- Substituting $a = 1$ in the second equation:
$$1^2 + b^2 + c^2 = 1 + \frac{1}{b^2} + \frac{1}{c^2} \implies b^2 + c^2 = \frac{1}{b^2} + \frac{1}{c^2}.$$
Again, multiplying through by $b^2c^2$:
$$b^2c^2(b^2 + c^2) = b^2 + c^2,$$
implying $b^2c^2 = 1$, maintaining $bc = 1$.

Given $bc = 1$, this leads to solutions for $(b, c)$ of the form $(x, \frac{1}{x})$.

The symmetric nature implies the solutions are:
- $(1, x, \frac{1}{x})$, with permutations.
- $(-1, x, \frac{1}{x})$, with permutations by a similar argument when $a = -1$.

Thus, the final set of solutions is:
$$\boxed{(\pm 1, x, \frac{1}{x})} \text{ and permutations.}$$