We define the function $f(x,y)=x^3+(y-4)x^2+(y^2-4y+4)x+(y^3-4y^2+4y)$. Then choose any distinct $a, b, c \in \mathbb{R}$ such that the following holds: $f(a,b)=f(b,c)=f(c,a)$. Over all such choices of $a, b, c$, what is the maximum value achieved by
\[\min(a^4 - 4a^3 + 4a^2, b^4 - 4b^3 + 4b^2, c^4 - 4c^3 + 4c^2)?\]
1. **Symmetry of \( f(x, y) \)**:
   We start by noting that the function \( f(x, y) = x^3 + (y-4)x^2 + (y^2-4y+4)x + (y^3-4y^2+4y) \) is symmetric in \( x \) and \( y \). This means \( f(x, y) = f(y, x) \).

2. **Given Condition**:
   We are given that \( f(a, b) = f(b, c) = f(c, a) \). This implies that the polynomial \( f(x, y) \) evaluated at these points must yield the same value.

3. **Factorization**:
   To find the relationship between \( a, b, \) and \( c \), we consider the polynomial \( f(x, y) - f(x, z) \). Since \( f(a, b) = f(a, c) \), we have:
   \[
   f(a, b) - f(a, c) = 0
   \]
   This implies that \( b - c \) is a factor of \( f(a, b) - f(a, c) \). Dividing out this factor, we get:
   \[
   a^2 + b^2 + c^2 + ab + bc + ca - 4a - 4b - 4c + 4 = 0
   \]
   This equation is symmetric in \( a, b, \) and \( c \).

4. **Substitution**:
   To simplify the expression, we make the substitution \( a = p + 1 \), \( b = q + 1 \), and \( c = r + 1 \). Substituting these into the symmetric equation, we get:
   \[
   (p+1)^2 + (q+1)^2 + (r+1)^2 + (p+1)(q+1) + (q+1)(r+1) + (r+1)(p+1) - 4(p+1) - 4(q+1) - 4(r+1) + 4 = 0
   \]
   Simplifying this, we obtain:
   \[
   p^2 + q^2 + r^2 + pq + qr + rp - 2 = 0
   \]
   This can be rewritten as:
   \[
   (p+q)^2 + (q+r)^2 + (r+p)^2 = 4
   \]

5. **Objective**:
   We need to find the maximum value of:
   \[
   \min(a^4 - 4a^3 + 4a^2, b^4 - 4b^3 + 4b^2, c^4 - 4c^3 + 4c^2)
   \]
   Using the substitution \( a = p + 1 \), \( b = q + 1 \), \( c = r + 1 \), we need to maximize:
   \[
   \min((p^2-1)^2, (q^2-1)^2, (r^2-1)^2)
   \]

6. **Testing Values**:
   - If \( p = q = r \), then \( p^2 = q^2 = r^2 = \frac{1}{3} \), and:
     \[
     (p^2-1)^2 = \left(\frac{1}{3} - 1\right)^2 = \left(-\frac{2}{3}\right)^2 = \frac{4}{9}
     \]
   - If \( p = q = 0 \) and \( r = \sqrt{2} \), then:
     \[
     (p^2-1)^2 = (0^2-1)^2 = 1
     \]
     \[
     (r^2-1)^2 = (\sqrt{2}^2-1)^2 = (2-1)^2 = 1
     \]
   - Thus, the minimum value in this case is \( 1 \).

7. **Conclusion**:
   We have shown that the maximum value of the minimum expression is \( 1 \). Any attempt to achieve a value greater than \( 1 \) leads to contradictions.

\[
\boxed{1}
\]