Find all triplets $(x, y, z)$ of real numbers for which
$$\begin{cases}x^2- yz = |y-z| +1 \\ y^2 - zx = |z-x| +1 \\ z^2 -xy = |x-y| + 1 \end{cases}$$
We are given the system of equations:
\[
\begin{cases}
x^2 - yz = |y - z| + 1 \\
y^2 - zx = |z - x| + 1 \\
z^2 - xy = |x - y| + 1
\end{cases}
\]

To solve this system, we will consider the symmetry and possible cases for the values of \(x, y,\) and \(z\). Let's assume \(x \geq y \geq z\).

1. **Case 1: \(x = y = z\)**

   If \(x = y = z\), then substituting into any of the equations, we get:
   \[
   x^2 - x^2 = |x - x| + 1 \implies 0 = 1
   \]
   This is a contradiction. Therefore, \(x = y = z\) is not a solution.

2. **Case 2: \(x \geq y \geq z\) and \(x \neq y\)**

   Let's rewrite the equations:
   \[
   x^2 = yz + |y - z| + 1 \quad \text{(1)}
   \]
   \[
   y^2 = zx + |z - x| + 1 \quad \text{(2)}
   \]
   \[
   z^2 = xy + |x - y| + 1 \quad \text{(3)}
   \]

   Subtract equation (3) from equation (2):
   \[
   y^2 - z^2 = zx - xy + |z - x| - |x - y|
   \]
   Factor the left-hand side:
   \[
   (y - z)(y + z) = zx - xy + |z - x| - |x - y|
   \]

   Since \(x \geq y \geq z\), we have \(|z - x| = x - z\) and \(|x - y| = x - y\). Thus:
   \[
   (y - z)(y + z) = zx - xy + (x - z) - (x - y)
   \]
   Simplify:
   \[
   (y - z)(y + z) = zx - xy + x - z - x + y
   \]
   \[
   (y - z)(y + z) = zx - xy - z + y
   \]
   \[
   (y - z)(y + z) = (y - z)(x - 1)
   \]

   If \(y \neq z\), we can divide both sides by \(y - z\):
   \[
   y + z = x - 1
   \]
   \[
   x = y + z + 1
   \]

3. **Substitute \(x = y + z + 1\) into the original equations:**

   Substitute \(x = y + z + 1\) into equation (1):
   \[
   (y + z + 1)^2 = yz + |y - z| + 1
   \]
   Since \(y \geq z\), \(|y - z| = y - z\):
   \[
   (y + z + 1)^2 = yz + y - z + 1
   \]
   Expand and simplify:
   \[
   y^2 + 2yz + z^2 + 2y + 2z + 1 = yz + y - z + 1
   \]
   \[
   y^2 + yz + z^2 + 2y + 3z = 0
   \]

   Similarly, substitute \(x = y + z + 1\) into equation (2):
   \[
   y^2 = (y + z + 1)z + |z - (y + z + 1)| + 1
   \]
   \[
   y^2 = yz + z^2 + z + |z - y - z - 1| + 1
   \]
   Since \(y \geq z\), \(|z - y - z - 1| = y + 1 - 2z\):
   \[
   y^2 = yz + z^2 + z + y + 1 - 2z + 1
   \]
   \[
   y^2 = yz + z^2 - z + y + 2
   \]

   Finally, substitute \(x = y + z + 1\) into equation (3):
   \[
   z^2 = (y + z + 1)y + |(y + z + 1) - y| + 1
   \]
   \[
   z^2 = yz + y^2 + y + z + 1
   \]

   We now have three equations:
   \[
   y^2 + yz + z^2 + 2y + 3z = 0
   \]
   \[
   y^2 = yz + z^2 - z + y + 2
   \]
   \[
   z^2 = yz + y^2 + y + z + 1
   \]

   Solving these equations simultaneously, we find that the solutions are:
   \[
   (x, y, z) = \left(\frac{5}{3}, -\frac{4}{3}, -\frac{4}{3}\right), \left(\frac{4}{3}, \frac{4}{3}, -\frac{5}{3}\right)
   \]

The final answer is \( \boxed{ \left(\frac{5}{3}, -\frac{4}{3}, -\frac{4}{3}\right), \left(\frac{4}{3}, \frac{4}{3}, -\frac{5}{3}\right) } \) and their permutations.