Question

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

• A. $9$
• B. $8$
• C. $2$
• D. $1$

Answer 1

The correct option is B

$8$

Find the minimum value of the sum of given numbers

Given : $a^{-5},a^{-4},3a^{-3},1,a^8$and $a^{10}$

All the numbers are positive as $a>0$.

Since we know that A.M.$\ge$G.M.,

$$\begin{gathered} \Rightarrow \frac{1}{8}\left[a^{-5}+a^{-4}+3a^{-3}+1+a^8+a^{10}\right]\ge {\left[a^{-5}.a^{-4}.3a^{-3}.1.a^8.a^{10}\right]}^{\frac{1}{8}} \\ \Rightarrow \frac{1}{8}\left[a^{-5}+a^{-4}+3a^{-3}+1+a^8+a^{10}\right]\ge 1^{\frac{1}{8}} \\ \Rightarrow \left[a^{-5}+a^{-4}+3a^{-3}+1+a^8+a^{10}\right]\ge 8 \end{gathered}$$

Therefore, the minimum value of the sum is $8$.

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