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If $a_1, a_2, \ldots . a_n$ are positive real numbers whose product is a fixed number $c$, then the minimum value of $a_1+a_2+\ldots+a_{n-1}+2 a_n$ is
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The correct answer is:
$n(2 c)^{1 / n}$
Given that $\mathrm{a}_{1,} \mathrm{a}_2 \ldots, \mathrm{a}_n$ be positive real numbers such that $\mathrm{a}_1 \cdot \mathbf{a}_2 \cdot \ldots \mathrm{a}_{\mathrm{n}}=\mathrm{c} .$. (1)
We know that, A.M z G.M
$\Rightarrow \frac{a_1+a_2+\ldots+a_{n-1}+2 a_n}{n} \geq\left(a_1 a_2 \ldots a_{n-1}\left(2 a_n\right)\right)^{\frac{1}{n}}$
$\Rightarrow a_1+a_2+\ldots+a_{n-1}+2 a_n \geq n(2 c)^{\frac{1}{n}} \ldots($ from(1) $)$
So, the minimum value of ${ }_{a_1+a_2+\ldots+a_{n-1}+2 a_n}$ is
$\mathrm{n}(2 \mathrm{c})^{\frac{1}{\mathrm{n}}}$.
We know that, A.M z G.M
$\Rightarrow \frac{a_1+a_2+\ldots+a_{n-1}+2 a_n}{n} \geq\left(a_1 a_2 \ldots a_{n-1}\left(2 a_n\right)\right)^{\frac{1}{n}}$
$\Rightarrow a_1+a_2+\ldots+a_{n-1}+2 a_n \geq n(2 c)^{\frac{1}{n}} \ldots($ from(1) $)$
So, the minimum value of ${ }_{a_1+a_2+\ldots+a_{n-1}+2 a_n}$ is
$\mathrm{n}(2 \mathrm{c})^{\frac{1}{\mathrm{n}}}$.
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