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Inequality

If a1,a2,a3...

Question

If $a_{1}, a_{2}, a_{3} . . . . a_{15} \in R^{+}$ and $a_{1}, a_{2}, a_{3} . . . a_{n} = 1$ , then minimum value of $(1 + a_{1} + a_{1}^{2}) (1 + a_{3} + a_{3}^{2}) . . . . (1 + a_{a} + a_{a}^{2})$ is equal to:_

A

3^{a + 1}

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B

3^{n}

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C

3^{n - 1}

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D

n o n e o f t h e s e

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Solution

The correct option is B $3^{n}$
$a_{1}, a_{2}, a_{3} - - - -, a_{n} \in R^{+}$
and $a_{1}, a_{2}, a_{3} - - - -, a_{n} = 1$ -----eq.1
Therefore the given expression will be minimum when, $a_{1} = a_{2} = a_{3} = - - - - = a_{n}$
$∴$ From eq.1 we get,
$a_{i}^{n} = 1$
Since $a_{i} \in R^{+} ∴ a_{I} = 1 [i = 1, 2, . . . . . . . . . n]$
Hence, $(1 + a_{1} + a_{1}^{2}) (1 + a_{2} + a_{2}^{2}) . . . . . . . . . . (1 + a_{n} + a_{n}^{2}) = (1 + a_{i} + a_{i}^{2})^{n} = 3^{n}$

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Similar questions

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

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

, then minimum value of

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

a_{r} > 0, r \in N

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

are in A.P. then

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}

\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_{r} > 0, r ϵ N

a_{1}

a_{2}

a_{3}

a_{2 n}

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}

a_{1}, a_{2}, a_{3}, a_{4}

are in AP, then

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{1}{\sqrt{a_{3}} + \sqrt{a_{4}}}

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