Find all integers $a,b,c,x,y,z$ such that \[a+b+c=xyz, \; x+y+z=abc, \; a \ge b \ge c \ge 1, \; x \ge y \ge z \ge 1.\]
Given the problem, we need to find all integers \(a, b, c, x, y, z\) such that:
\[
a + b + c = xyz, \quad x + y + z = abc, \quad a \ge b \ge c \ge 1, \quad x \ge y \ge z \ge 1.
\]

Without loss of generality, assume \(a + b + c \ge x + y + z\) (otherwise, we can switch \((a, b, c)\) with \((x, y, z)\)). Then, we have:
\[
a + b + c \ge abc.
\]

We will consider two cases:

### Case 1: \(c = 1\)
If \(c = 1\), then:
\[
a + b + c \ge abc \implies a + b + 1 \ge ab.
\]
Rewriting this, we get:
\[
a + b + 1 \ge ab \implies 1 + a + b \ge ab \implies 2 \ge ab - a - b + 1 \implies 2 \ge (a - 1)(b - 1).
\]
Since \(a \ge b\), we have the following possibilities:
- \(b = 1\)
- \(b = 2\) and \(a = 2\) or \(a = 3\)

Thus, the only triplets \((a, b, c)\) satisfying \(c = 1\) and \(a + b + c \ge abc\) are:
\[
(a, 1, 1), (2, 2, 1), (3, 2, 1).
\]

### Case 2: \(c \ge 2\)
If \(c \ge 2\), then:
\[
a + b + c \ge abc \implies a + b + c \ge 2ab.
\]
Rewriting this, we get:
\[
1 \ge ab - a - b + 1 + ab - c \implies 1 \ge (a - 1)(b - 1) + (ab - c).
\]
However, since \(a \ge b \ge c \ge 2\), we have \((a - 1)(b - 1) \ge 1\) and \(ab - c > 1\). Therefore, the inequality \(1 \ge (a - 1)(b - 1) + (ab - c)\) is false, so the case when \(c \ge 2\) fails.

Now, we need to check the possible values for \((a, b, c)\) and find corresponding \((x, y, z)\):

1. **For \((a, b, c) = (2, 2, 1)\):**
   \[
   x + y + z = 4, \quad xyz = 5.
   \]
   This implies one of \(x, y, z\) must be 5, but since \(x, y, z \ge 1\), we have \(x + y + z > 4\). Thus, \((a, b, c) = (2, 2, 1)\) is not possible.

2. **For \((a, b, c) = (3, 2, 1)\):**
   \[
   x + y + z = 6, \quad xyz = 6.
   \]
   The possible values for \(x, y, z\) are \(\{1, 2, 3, 6\}\). Since none of \(x, y, z\) can be 6 (as \(x + y + z > 6\)), one of \(x, y, z\) must be 3. Thus, \(x = 3\), and we have:
   \[
   y + z = 3, \quad yz = 2.
   \]
   Solving this, we get \(y = 2\) and \(z = 1\). Hence, \((x, y, z) = (3, 2, 1)\).

3. **For \((a, b, c) = (a, 1, 1)\):**
   \[
   x + y + z = a, \quad xyz = a + 2.
   \]
   Since \(x \ge y \ge z\) and \(x + y + z = a\), we must have \(a \ge \frac{a}{3}\). If \(yz \ge 5\), then \(x \le \frac{a + 2}{5}\), so \(\frac{a}{3} \le \frac{a + 2}{5}\), which implies \(a \le 3\). This is impossible as \(a + 2 = xyz \ge 5x > 5 \implies a > 3\). Therefore, \(yz \le 4\).

   - If \(yz = 4\), we can have \(y = z = 2\), giving:
     \[
     x + 4 = a, \quad 4x = a + 2 \implies x = 2, \quad a = 6.
     \]
     Thus, \((a, b, c) = (6, 1, 1)\) and \((x, y, z) = (2, 2, 2)\).

   - If \(yz = 3\), we can have \(y = 3, z = 1\), giving:
     \[
     x + 4 = a, \quad 3x = a + 2 \implies x = 3, \quad a = 7.
     \]
     Thus, \((a, b, c) = (7, 1, 1)\) and \((x, y, z) = (3, 3, 1)\).

   - If \(yz = 2\), we can have \(y = 2, z = 1\), giving:
     \[
     x + 3 = a, \quad 2x = a + 2 \implies x = 5, \quad a = 8.
     \]
     Thus, \((a, b, c) = (8, 1, 1)\) and \((x, y, z) = (5, 2, 1)\).

   - If \(yz = 1\), we can have \(y = z = 1\), giving:
     \[
     x + 2 = a, \quad x = a + 2,
     \]
     which is impossible.

Therefore, if \(a + b + c \ge x + y + z\), we get the following six-tuplets \((a, b, c, x, y, z)\):
\[
(3, 2, 1, 3, 2, 1), (6, 1, 1, 2, 2, 2), (7, 1, 1, 3, 3, 1), (8, 1, 1, 5, 2, 1).
\]

By symmetry of the system and the conditions on \(a, b, c, x, y, z\), if \(x + y + z \ge a + b + c\), we just switch \((a, b, c)\) with \((x, y, z)\). Therefore, all the six-tuplets satisfying the conditions of the problem are:
\[
(3, 2, 1, 3, 2, 1), (6, 1, 1, 2, 2, 2), (7, 1, 1, 3, 3, 1), (8, 1, 1, 5, 2, 1), (2, 2, 2, 6, 1, 1), (3, 3, 1, 7, 1, 1), (5, 2, 1, 8, 1, 1).
\]

The final answer is \( \boxed{ (3, 2, 1, 3, 2, 1), (6, 1, 1, 2, 2, 2), (7, 1, 1, 3, 3, 1), (8, 1, 1, 5, 2, 1), (2, 2, 2, 6, 1, 1), (3, 3, 1, 7, 1, 1), (5, 2, 1, 8, 1, 1) } \).