$a,b,c$ are positive numbers such that $ a^2 + b^2 + c^2 = 2abc + 1 $. Find the maximum value of
\[ (a-2bc)(b-2ca)(c-2ab) \]
1. Given the equation \(a^2 + b^2 + c^2 = 2abc + 1\), we need to find the maximum value of \((a-2bc)(b-2ca)(c-2ab)\).

2. We start by noting that since \(a, b, c\) are positive numbers, the expression \((a-2bc)(b-2ca)(c-2ab)\) must be positive for the maximum value to be positive. Without loss of generality, assume \(a \ge b \ge c\).

3. First, consider the case where \(a \le 2bc\). In this scenario, we have:
   \[
   c \le b \le a \le 2bc \le 2ca \le 2ab
   \]
   This implies:
   \[
   (a - 2bc)(b - 2ca)(c - 2ab) \le 0
   \]
   Therefore, we can assume \(a > 2bc\).

4. Next, we analyze the condition given by the problem:
   \[
   (c - ab)^2 = (a^2 - 1)(b^2 - 1) \ge 0
   \]
   This implies that \((a-1)\) and \((b-1)\) have the same sign, and so does \((c-1)\). We need to consider two cases: \(a, b, c > 1\) and \(a, b, c \le 1\).

5. If \(a, b, c > 1\), then:
   \[
   a^2 + b^2 + c^2 - 2abc - 1 \ge a^2 + 2bc - 2abc - 1 = (a - 1)(a + 1 - 2bc) > 0
   \]
   This leads to a contradiction. Therefore, we must have \(a, b, c \le 1\).

6. Now, consider the expression:
   \[
   (b - 2ca)(c - 2ab) = bc + 4a^2bc - 2a(b^2 + c^2)
   \]
   Using the given condition \(a^2 + b^2 + c^2 = 2abc + 1\), we can rewrite it as:
   \[
   bc + 4a^2bc - 2a(2abc + 1 - a^2) = bc + 2a^3 - 2a
   \]

7. Applying the Arithmetic Mean-Geometric Mean (AM-GM) inequality, we get:
   \[
   \frac{1}{2}(a - 2bc) \cdot 2(b - 2ca)(c - 2ab) = \frac{1}{2}(a - 2bc)(2bc + 4a^3 - 4a)
   \]
   \[
   \le \frac{1}{2} \left( \frac{a - 2bc + 2bc + 4a^3 - 4a}{2} \right)^2 = \frac{1}{8}(4a^3 - 3a)^2
   \]

8. Since \(a \le 1\), it follows that:
   \[
   |4a^3 - 3a| \le 1
   \]
   Therefore:
   \[
   (a - 2bc)(b - 2ca)(c - 2ab) \le \frac{1}{8}(4a^3 - 3a)^2 \le \frac{1}{8}
   \]

9. When \(a = 1\), \(b = c = \frac{1}{2}\), the expression equals \(\frac{1}{8}\). Thus, the maximum value is \(\frac{1}{8}\).

The final answer is \(\boxed{\frac{1}{8}}\).