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

These real numbers can be taken as 8 quantities,a^-5, a^-4, a^-3, a^-3, a^-3, 1, a^8, a^10 .All are positive as a >0 Therefore AM> GM (as all quantities are not distinct) Therefore (1 / 8)[a^-5+a^-4+a^-3+a^-3+a^-3+1+a^8+a^10] >= (a^-5 * a^-4 * a^-3 * a^-3 * a^-3 * 1 * a^8 * a^10)^1 / 8 => (1 / 8)[a^-5+a^-4+a^-3+a^-3+a^-3+1+a^8+a^10] >= 1^1 / 8 => (a^-5+a^-4+a^-3+a^-3+a^-3+1+a^8+a^10) >= 8 ∴ Minimum value of the sum is mathbf8$.

The minimum value of the sum of real numbers a^{- 5},... | Sequence & Series (Mathematics)

These real numbers can be taken as 8 quantities,

$a^{-5}, a^{-4}, a^{-3}, a^{-3}, a^{-3}, 1, a^8, a^{10} .$

All are positive as a $>0\\$

$\therefore \mathrm{AM}>\mathrm{GM}$ (as all quantities are not distinct)$

$\therefore \frac{1}{8}\left[a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^8+a^{10}\right] \\$

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

$\Rightarrow \frac{1}{8}\left[a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^8+a^{10}\right] \geq 1^{1 / 8} \\$

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

∴ Minimum value of the sum is $\mathbf{8}$ .

Explanation

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