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Q.

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

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a

3

b

9

c

27

d

36

answer is C.

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Detailed Solution

$\begin{array}{l} \frac{1}{a b} + \frac{1}{b c} + \frac{1}{c a} = \frac{1}{a b c} (∵ a + b + c = 1) \\ w e h a v e \frac{a + b + c}{3} \geq (a b c)^{\frac{1}{3}} \\ \Rightarrow (a b c)^{\frac{1}{3}} \leq \frac{1}{3} \\ \Rightarrow a b c \leq \frac{1}{27} \end{array}$

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