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(D) Given the expression $S = a^{-5} + a^{-4} + 3a^{-3} + 1 + a^8 + a^{10}$ for $a > 0$.Since all terms are positive,we apply the Arithmetic Mean-Geom

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$8$

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(D) Given the expression $S = a^{-5} + a^{-4} + 3a^{-3} + 1 + a^8 + a^{10}$ for $a > 0$.Since all terms are positive,we apply the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality.We have $7$ terms: $a^{-5}, a^{-4}, a^{-3}, a^{-3}, a^{-3}, a^8, a^{10}$.$\frac{a^{-5} + a^{-4} + a^{-3} + a^{-3} + a^{-3} + a^8 + a^{10}}{7} \geq (a^{-5} \cdot a^{-4} \cdot a^{-3} \cdot a^{-3} \cdot a^{-3} \cdot a^8 \cdot a^{10})^{1/7}$$\frac{a^{-5} + a^{-4} + 3a^{-3} + a^8 + a^{10}}{7} \geq (a^{-5-4-3-3-3+8+10})^{1/7}$$\frac{a^{-5} + a^{-4} + 3a^{-3} + a^8 + a^{10}}{7} \geq (a^0)^{1/7} = 1$$a^{-5} + a^{-4} + 3a^{-3} + a^8 + a^{10} \geq 7$Adding the constant $1$ from the original expression:$S = (a^{-5} + a^{-4} + 3a^{-3} + a^8 + a^{10}) + 1 \geq 7 + 1 = 8$The minimum value is $8$.

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

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