Question

If $a_1{a}_{}......a_n$ are positive real numbers whose product is a fixed number $c$, then the minimum value of $a_1+a_2+.......+a_{(n-1)}+2a_n$ is

• A. $n{\left(2c\right)}^{1/n}$
• B. $(n+1)c^{1/n}$
• C. $\left(2n\right)c^{1/n}$
• D. $(n+1){\left(2c\right)}^{1/n}$

Answer 1

The correct option is A

$n{\left(2c\right)}^{1/n}$

Calculate the minimum value of ${\mathit{a}}_{\mathbf{1}}\mathbf{+}\mathbf{}{\mathit{a}}_{\mathbf{2}}\mathbf{+}\mathbf{}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{}\mathbf{+}\mathbf{}{\mathit{a}}_{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{1}\mathbf{)}}\mathbf{+}\mathbf{}\mathbf{2}{\mathit{a}}_{\mathbf{n}}$ :

Given product, $a_1\times a_2\times \dots ....\times a_n=c$

Multiplying both sides by $2$ we get,

$a_1{a}_2\dots ......\left(a_{n-1}\right)\left(2a_n\right)=2c.........\left(1\right)$

We know $AM\ge GM$

$$\begin{gathered} \frac{(a_1+a_2+a_3+\dots .+2a_n)}{n}\ge {\left[a_1{a}_2.....\left(a_{n-1}\right)\left(2a_n\right)\right]}^{1/n} \\ \Rightarrow \frac{(a_1+a_2+a_3+\dots .+2a_n)}{n}\ge {\left(2c\right)}^{1/n}\left(fromequation\left(1\right)\right) \\ \Rightarrow (a_1+a_2+a_3+\dots .+2a_n)\ge {n\left(2c\right)}^{1/n} \end{gathered}$$

So the minimum value of $a_1+a_2+\dots \dots .+a_{n-1}+2a_n=n{\left(2c\right)}^{1/n}$

Hence, the correct option is( A).