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The given expression asks for the minimum value of the function:

\[
f(a, b) = a + \frac{1}{b(a-b)}
\]

where \(a > b\) and both \(a\) and \(b\) are positive real numbers.

### Step 1: Rewriting the function in terms of a new variable

Let us define a new variable to simplify the expression. Set \(x = a - b\), so that \(a = b + x\) with \(x > 0\). The function becomes:

\[
f(b, x) = (b + x) + \frac{1}{b \cdot x}
\]

Now the goal is to minimize:

\[
f(b, x) = b + x + \frac{1}{bx}
\]

### Step 2: Consider symmetry

Notice that the minimum of the expression might occur when \(b\) and \(x\) are balanced in some way. One way to check this is to attempt partial differentiation with respect to \(b\) and \(x\), or guess that there might be an equality condition that minimizes the function.

### Step 3: Check for critical points

Let’s differentiate the function with respect to both \(b\) and \(x\), solve for the critical points, and verify that they give a minimum value.

I'll calculate that now to find the minimum value.The critical points for the function are:

- \( b = \frac{1}{\sqrt{x}} \)
- \( x = \frac{1}{\sqrt{b}} \)

These suggest a relationship between \(b\) and \(x\), but it seems the solution simplifies if we equate these expressions, leading to:

\[
b = x = 1
\]

### Step 4: Evaluate the function at the critical point

Substitute \(b = 1\) and \(x = 1\) into the original function \(f(b, x)\):

\[
f(1, 1) = 1 + 1 + \frac{1}{1 \times 1} = 3
\]

Thus, the minimum value of the function is \(3\).

Would you like more details on any part of the solution?

### Related Questions:
1. How can we verify that the critical point corresponds to a minimum?
2. Can this problem be solved using the Cauchy-Schwarz inequality?
3. What would happen if \(a = b\)? Is this allowed?
4. How can this method be generalized for more complex rational functions?
5. How does this approach compare to using the method of Lagrange multipliers?

**Tip:** When solving optimization problems, introducing substitutions or constraints like \(x = a - b\) can often simplify the calculations significantly.

Let a > b be positive real numbers. What is the minimum value of a + 1 / b(a-b)?

This problem involves finding the minimum value of the expression a + 1 / b(a-b) for positive real numbers a > b. Using techniques from calculus, including critical points and optimization, the solution shows the minimum value is 3. Suitable for advanced high school or university students.

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