Consider the real function defined by $f(x) =\frac{1}{|x + 3| + |x + 1| + |x - 2| + |x -5|}$ for all $x \in R$.
a) Determine its maximum.
b) Graphic representation.
1. **Identify the critical points**:
   The function \( f(x) = \frac{1}{|x + 3| + |x + 1| + |x - 2| + |x - 5|} \) involves absolute values, which change their behavior at the points where the expressions inside the absolute values are zero. These points are \( x = -3, -1, 2, 5 \).

2. **Analyze the function in different intervals**:
   We need to consider the behavior of the function in the intervals determined by these critical points: \( (-\infty, -3) \), \( (-3, -1) \), \( (-1, 2) \), \( (2, 5) \), and \( (5, \infty) \).

3. **Calculate the sum of absolute values in each interval**:
   - For \( x \in (-\infty, -3) \):
     \[
     |x + 3| = -(x + 3) = -x - 3, \quad |x + 1| = -(x + 1) = -x - 1, \quad |x - 2| = -(x - 2) = -x + 2, \quad |x - 5| = -(x - 5) = -x + 5
     \]
     \[
     |x + 3| + |x + 1| + |x - 2| + |x - 5| = (-x - 3) + (-x - 1) + (-x + 2) + (-x + 5) = -4x + 3
     \]

   - For \( x \in (-3, -1) \):
     \[
     |x + 3| = x + 3, \quad |x + 1| = -(x + 1) = -x - 1, \quad |x - 2| = -(x - 2) = -x + 2, \quad |x - 5| = -(x - 5) = -x + 5
     \]
     \[
     |x + 3| + |x + 1| + |x - 2| + |x - 5| = (x + 3) + (-x - 1) + (-x + 2) + (-x + 5) = -2x + 9
     \]

   - For \( x \in (-1, 2) \):
     \[
     |x + 3| = x + 3, \quad |x + 1| = x + 1, \quad |x - 2| = -(x - 2) = -x + 2, \quad |x - 5| = -(x - 5) = -x + 5
     \]
     \[
     |x + 3| + |x + 1| + |x - 2| + |x - 5| = (x + 3) + (x + 1) + (-x + 2) + (-x + 5) = 2x + 11 - 2x = 11
     \]

   - For \( x \in (2, 5) \):
     \[
     |x + 3| = x + 3, \quad |x + 1| = x + 1, \quad |x - 2| = x - 2, \quad |x - 5| = -(x - 5) = -x + 5
     \]
     \[
     |x + 3| + |x + 1| + |x - 2| + |x - 5| = (x + 3) + (x + 1) + (x - 2) + (-x + 5) = 2x + 7
     \]

   - For \( x \in (5, \infty) \):
     \[
     |x + 3| = x + 3, \quad |x + 1| = x + 1, \quad |x - 2| = x - 2, \quad |x - 5| = x - 5
     \]
     \[
     |x + 3| + |x + 1| + |x - 2| + |x - 5| = (x + 3) + (x + 1) + (x - 2) + (x - 5) = 4x - 3
     \]

4. **Determine the minimum value of the sum of absolute values**:
   From the calculations above, we see that the minimum value of \( |x + 3| + |x + 1| + |x - 2| + |x - 5| \) is 11, which occurs in the interval \( (-1, 2) \).

5. **Find the maximum value of \( f(x) \)**:
   Since \( f(x) = \frac{1}{|x + 3| + |x + 1| + |x - 2| + |x - 5|} \), the maximum value of \( f(x) \) occurs when the denominator is minimized. The minimum value of the denominator is 11.
   \[
   f(x) = \frac{1}{11}
   \]

Conclusion:
The maximum value of the function \( f(x) \) is \( \frac{1}{11} \).

The final answer is \( \boxed{\frac{1}{11}} \)