Find the minimum of the expression  $\frac{a + b + c}{2} -\frac{[a, b] + [b, c] + [c, a]}{a + b + c}$ where the variables $a, b, c$ are any integers greater than $1$ and $[x, y]$ denotes the least common multiple of numbers $x, y$.
1. Define the function \( f(a, b, c) = \frac{a + b + c}{2} - \frac{[a, b] + [b, c] + [c, a]}{a + b + c} \).

2. Let \( k = a + b + c \). We need to find the minimum value of \( f(a, b, c) \).

3. Note that \( [a, b] + [b, c] + [c, a] \leq ab + bc + ca \). This is because the least common multiple (LCM) of two numbers is always less than or equal to their product.

4. Using the identity \( (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \), we can derive:
   \[
   (a + b + c)^2 - 3(ab + bc + ca) = a^2 + b^2 + c^2 - ab - bc - ca
   \]
   Since \( a^2 + b^2 + c^2 \geq ab + bc + ca \) by the rearrangement inequality, we have:
   \[
   (a + b + c)^2 \geq 3(ab + bc + ca)
   \]
   This implies:
   \[
   [a, b] + [b, c] + [c, a] \leq \frac{(a + b + c)^2}{3}
   \]

5. Therefore, we have:
   \[
   f(a, b, c) \geq \frac{k}{2} - \frac{k^2 / 3}{k} = \frac{k}{2} - \frac{k}{3} = \frac{k}{6}
   \]

6. To find the minimum value, we test specific values of \( a, b, \) and \( c \). For example, let \( a = 3 \), \( b = 2 \), and \( c = 2 \):
   \[
   f(3, 2, 2) = \frac{3 + 2 + 2}{2} - \frac{[3, 2] + [2, 2] + [2, 3]}{3 + 2 + 2} = \frac{7}{2} - \frac{6 + 2 + 6}{7} = \frac{7}{2} - \frac{14}{7} = \frac{7}{2} - 2 = \frac{3}{2}
   \]

7. We can assume \( k \leq 9 \) based on the above calculations. If \( \gcd(a, b, c) > 1 \), then \( f(a/d, b/d, c/d) < f(a, b, c) \). So we can assume \( \gcd(a, b, c) = 1 \) and \( a \geq b \geq c \).

8. Testing other values:
   \[
   f(5, 2, 2) = \frac{5 + 2 + 2}{2} - \frac{[5, 2] + [2, 2] + [2, 5]}{5 + 2 + 2} = \frac{9}{2} - \frac{10 + 2 + 10}{9} = \frac{9}{2} - \frac{22}{9} = \frac{81}{18} - \frac{22}{9} = \frac{81}{18} - \frac{44}{18} = \frac{37}{18} > \frac{3}{2}
   \]
   \[
   f(4, 3, 2) = \frac{4 + 3 + 2}{2} - \frac{[4, 3] + [3, 2] + [2, 4]}{4 + 3 + 2} = \frac{9}{2} - \frac{12 + 6 + 4}{9} = \frac{9}{2} - \frac{22}{9} = \frac{81}{18} - \frac{44}{18} = \frac{37}{18} > \frac{3}{2}
   \]
   \[
   f(3, 3, 2) = \frac{3 + 3 + 2}{2} - \frac{[3, 3] + [3, 2] + [2, 3]}{3 + 3 + 2} = \frac{8}{2} - \frac{3 + 6 + 6}{8} = 4 - \frac{15}{8} = 4 - \frac{15}{8} = \frac{17}{8} > \frac{3}{2}
   \]

9. Hence, the minimum value of \( f(a, b, c) \) is \( \frac{3}{2} \).

The final answer is \( \boxed{ \frac{3}{2} } \)