Question

. The minimum value of the sum of real numbers $a^{-5},a^{-4},3a^{-3},1,a^8,a^{10}$ , where $a>0$ is

• A. $5$
• B. $6$
• C. $8$
• D. $7$

Answer 1

The correct option is C

$8$

Explanation for the correct answer:

The given numbers are $a^{-5},a^{-4},a^{-3},a^{-3},a^{-3},1,a^8,a^{10}$ (we split the $3a^{-3}$ into three separate $a^{-3}$)

Applying the A.M-G,M inequality we get

$\frac{a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^8+a^{10}}{8}\ge {\left(a^{-5}\times a^{-4}\times a^{-3}\times a^{-3}\times a^{-3}\times 1\times a^8\times a^{10}\right)}^{\frac{1}{8}}$

$\Rightarrow$ $\frac{a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^8+a^{10}}{8}\ge {\left(a^{-18}.a^{18}\right)}^{\frac{1}{8}}$

$\Rightarrow$ $a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^8+a^{10}\ge 8{\left(1\right)}^{\frac{1}{8}}$

$\Rightarrow$ $a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^8+a^{10}\ge 8$

Hence, the minimum value of the sum of the given numbers is $8$.

Hence, option $\left(C\right)$ is the correct answer.