Discussion Forum

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 

Answer:

Explanation:

Using AM $\geq$ GM

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

 $\geq (a^{-5}.a^{-4}.a^{-3}.a^{-3}.a^{-3}.1.a^{8}.a^{10})^{1/8}$

$\Rightarrow$  $ a^{-5}+a^{-4}+3a^{-3}+1+a^{8}+a^{10}\geq 8.1$

 Hence, minimum value is 8