Q42 / 69 · JEE Advanced 2011 (Paper 1) · Mathematics
The minimum value of the sum of real numbers $a^{-5}, a^{-4}, 3 a^{-3}, 1, a^8$ and $a^{10}$ with $a>0$ is
Correct answer: 8
Solution
Using $A M \geq G M$,
$$
\begin{aligned}
& \frac{a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^8+a^{10}}{8} \\
& \geq\left(a^{-5} \cdot a^{-4} \cdot a^{-3} \cdot a^{-3} \cdot a^{-3} \cdot 1 \cdot a^8 \cdot a^{10}\right)^{\frac{1}{8}} \\
& \Rightarrow a^{-5}+a^{-4}+3 a^{-3}+1+a^8+a^{10} \geq 8 \cdot 1
\end{aligned}
$$
Hence, minimum value is 8 .
Hence, the correct answer is 8.