Given that $a$ and $b$ are positive real numbers with $a + b = 4$ , the minimum value of $(1 + \frac{1}{a})$ $(1 + \frac{1}{b})$ , is:

2

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8 / 3

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9 / 4

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3

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Solution

The correct option is C $2$
$a = 1, 3, 2$
$b = 1, 3, 2$
therefore minimum value of $(1 + \frac{1}{a}) (1 + \frac{1}{b})$ is 2

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