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If $a_1,a_2,a_3,\cdots,a_n$ are positive real numbers whose product is a fixed number $c$, then the minimum value of $a_1+a_2+a_3+\cdots+a_{n-1}+2a_n$ is
(A) $n(2c)^{1/n}$
(B) $(n+1)c^{1/n}$
(C) $2nc^{1/n}$
(D) $(n+1)(2c)^{1/n}$

My Approch:
$GM=c^{1/n}$
$AM \geq GM $
For minimum value,
$\frac{a_1+a_2+a_3+\cdots+a_{n-1}+a_n}{n}=c^{1/n}$
$a_1+a_2+a_3+\cdots+a_{n-1}+a_n=nc^{1/n}$

But the question asks for $a_1+a_2+a_3+\cdots+a_{n-1}+2a_n$.
The closest approximation would be to consider $a_n=GM$. But that wouldn't give you the accurate answer. How do you solve this question then?

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Put in $b_n=2a_n$. Then you are tp minimize $a_1+a_2+...+a_{n-1}+b_n$ given the product of those terms $a_1·a_2·...·a_{n-1}·b_n$ being equal to $2c$, and no coefficients in the terms other than $1$. You apparently know how to solve this, and the answer follows.

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  • $\begingroup$ Why would the product of those terms equal 2c? $\endgroup$ – Osheen Sachdev Nov 25 '16 at 3:11
  • $\begingroup$ What is the product with the last factor mase equal to $b_n =2a_n$, if it's $c$ with the last factor being $a_n$? $\endgroup$ – Oscar Lanzi Nov 25 '16 at 3:14
  • $\begingroup$ Oh right! I was thinking that c stands for the product of double of all the terms :P $\endgroup$ – Osheen Sachdev Nov 25 '16 at 3:16

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