If a, b, c are any three positive numbers, then the least value of $(a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$ is

3

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6

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9

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None of these

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Solution

The correct option is C $9$
H.M of $a, b, c$ is $= \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$

And A.M is $= \frac{a + b + c}{3}$

Now we know for any positive numbers, $A . M \geq H . M$

$\Rightarrow \frac{a + b + c}{3} \geq \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}}$

$∴ (a + b + c) (\frac{1}{a} + \frac{1}{b} + \frac{1}{c}) \geq 9$

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