If $a_{1}, a_{2}$ and $a_{3}$ be any positive real numbers, then which of the following statement is true?

3 a_{1} a_{2} a_{3} \leq a_{1}^{3} + a_{2}^{3} + a_{3}^{3}

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\frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \frac{a_{3}}{a_{1}} \geq 3

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(a_{1} + a_{2} + a_{3}) (\frac{1}{a_{2}} + \frac{1}{a_{2}} + \frac{1}{a_{3}}) \geq 9

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(a_{1} . a_{2} . a_{3}) {(\frac{1}{a_{2}} + \frac{1}{a_{2}} + \frac{1}{a_{3}})}^{3} \geq 27

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Solution

The correct option is D $(a_{1} . a_{2} . a_{3}) {(\frac{1}{a_{2}} + \frac{1}{a_{2}} + \frac{1}{a_{3}})}^{3} \geq 27$
We know that, $G M \geq H M$

$\Rightarrow (a_{1} . a_{2} . a_{3})^{1 / 3} \geq \frac{3}{(\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}})}$

$\Rightarrow (a_{1} . a_{2} . a_{3}) \geq \frac{27}{{(\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}})}^{3}}$

$\Rightarrow (a_{1} . a_{2} . a_{3}) {(\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}})}^{3} \geq 27$

Hence, option D is correct.

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Similar questions

a_{1}, a_{2}, a_{3} \in + R

3 a_{1} a_{2} a_{3} \leq a_{1}^{3} + a_{2}^{3} + a_{3}^{3}

a_{1}, a_{2}, a_{3}, . . . . . . . . ., a_{n}

(a_{1} + a_{2} + a_{3} + . . . . . . . + a_{n}) (\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} . . . . . . . . . . . . + \frac{1}{a_{n}}) \geq n^{2}

\frac{a_{1}}{a_{1} + 1 -} \frac{a_{2}}{a_{2} + 1 -} \frac{a_{3}}{a_{3} + 1 -} \dots, a_{1}, a_{2}, a_{3}, \dots

\frac{a_{1}}{a_{1}} + \frac{a_{2}}{a_{2}} + \frac{a_{3}}{a_{3}} + \dots \frac{a_{n}}{a_{n}} = \frac{1}{1} + \frac{1}{a_{1}} + \frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \dots \frac{a_{n - 2}}{a_{n - 1}}

a_{1}, a_{2}, a_{3} . . . . a_{n} \in R^{+}

a_{1} . a_{2} . a_{3} . . . . a_{n} = 1

(1 + a_{1} + a_{1}^{2}) (1 + a_{2} + a_{2}^{2}) (1 + a_{3} + a_{3}^{2}) . . . . (1 + a_{n} + a_{n}^{2})

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