To solve the problem of finding the minimum value of (\frac{1}{a} + \frac{4}{b}) given (a > 0), (b > 0), and (a + b = 1), we can use the AM-GM inequality as follows:

Step 1: Multiply by (a + b = 1)

Since (a + b = 1), we rewrite the expression by multiplying it with (a + b):
[ \left(\frac{1}{a} + \frac{4}{b}\right)(a + b) = 1 + \frac{b}{a} + \frac{4a}{b} + 4 = 5 + \frac{b}{a} + \frac{4a}{b} ]

Step 2: Apply AM-GM inequality

For positive numbers (\frac{b}{a}) and (\frac{4a}{b}):
[ \frac{b}{a} + \frac{4a}{b} \geq 2\sqrt{\frac{b}{a} \cdot \frac{4a}{b}} = 2 \times 2 = 4 ]
Equality holds when (\frac{b}{a} = \frac{4a}{b}), i.e., (b = 2a).

Step 3: Calculate the minimum value

Substitute back into the expression:
[ 5 + \frac{b}{a} + \frac{4a}{b} \geq 5 + 4 = 9 ]

When equality holds: (a + b = 1) and (b = 2a), so (a = \frac{1}{3}), (b = \frac{2}{3}), and (\frac{1}{a} + \frac{4}{b} = 3 + 6 = 9).

Answer: (\boxed{9})

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