If $a + b + c = 1$ , where $a, b, c$ are positive real numbers, then the minimum value of $\frac{1}{a b} + \frac{1}{b c} + \frac{1}{c a}$ is

13

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24

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27

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32

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Solution

The correct option is C $27$
$\frac{1}{a b} + \frac{1}{b c} + \frac{1}{a c} = \frac{a + b + c}{a b c} = \frac{1}{a b c}$
We know that,
$A . M . \geq G . M \Rightarrow \frac{a + b + c}{3} \geq (a b c)^{1 / 3} \Rightarrow {(\frac{1}{3})}^{3} \geq a b c \Rightarrow \frac{1}{a b c} \geq 27$

Therefore,
$\frac{1}{a b} + \frac{1}{b c} + \frac{1}{a c} \geq 27$

Hence, the minimum value of the given expression is $27$ .

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