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.