Let \(\alpha\) and \(\beta\) be the roots of the quadratic equation \(x^2 + bx + c = 0\).
Given:
Step 1: Determine \(\alpha\beta\)
Using the identity \( \left( \frac{1}{\alpha} - \frac{1}{\beta} \right)^2 = \frac{1}{\alpha^2} + \frac{1}{\beta^2} - \frac{2}{\alpha\beta} \), substitute the given values: \( \frac{1}{9} = \frac{5}{9} - \frac{2}{\alpha\beta} \). Rearranging yields \( \frac{2}{\alpha\beta} = \frac{4}{9} \), which simplifies to \( \alpha\beta = \frac{9}{2} \).
Step 2: Determine \(\alpha^2 + \beta^2\)
Using the identity \( \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{\alpha^2 \beta^2} \), and substituting the given value and the derived \(\alpha\beta\): \( \frac{5}{9} = \frac{\alpha^2 + \beta^2}{(\frac{9}{2})^2} \). This gives \( \alpha^2 + \beta^2 = \frac{5}{9} \cdot \frac{81}{4} = \frac{405}{36} = \frac{45}{4} \).
Step 3: Find \((\alpha + \beta)^2\)
Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \), substitute known values: \( \frac{45}{4} = (\alpha + \beta)^2 - 2 \cdot \frac{9}{2} \). This simplifies to \( \frac{45}{4} = (\alpha + \beta)^2 - 9 \), so \( (\alpha + \beta)^2 = \frac{45}{4} + 9 = \frac{81}{4} \). Therefore, \( \alpha + \beta = \pm \frac{9}{2} \).
Step 4: Relate to coefficients \(b\) and \(c\)
From Vieta's formulas, \( \alpha + \beta = -b \) and \( \alpha\beta = c \). Thus, \( b = -(\alpha + \beta) = \mp \frac{9}{2} \) and \( c = \frac{9}{2} \).
Step 5: Calculate the maximum value of \(b + c\)
The expression to maximize is \( b + c = -(\alpha + \beta) + \alpha\beta \). To maximize this, we select the value of \( \alpha + \beta \) that yields the largest sum. If \( \alpha + \beta = -\frac{9}{2} \), then \( b = \frac{9}{2} \). In this case, \( b + c = \frac{9}{2} + \frac{9}{2} = \boxed{9} \).
Final Answer: Option (C): 9