To solve the problem of finding the minimum value of (\frac{1}{a} + \frac{4}) given (a > 0), (b > 0), and (a + b = 1), we can use the AM-GM inequality as follows:
Since (a + b = 1), we rewrite the expression by multiplying it with (a + b):
[
\left(\frac{1}{a} + \frac{4}\right)(a + b) = 1 + \frac{a} + \frac{4a} + 4 = 5 + \frac{a} + \frac{4a}
]
For positive numbers (\frac{a}) and (\frac{4a}):
[
\frac{a} + \frac{4a} \geq 2\sqrt{\frac{a} \cdot \frac{4a}} = 2 \times 2 = 4
]
Equality holds when (\frac{a} = \frac{4a}), i.e., (b = 2a).
Substitute back into the expression:
[
5 + \frac{a} + \frac{4a} \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} = 3 + 6 = 9).
Answer: (\boxed{9})
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