[JEE Advanced 2011] The minimum value of $a^{-5}+a^{-4}+3a^{-3}+1+a^{8}+a^{10}$ for $a>0$ is ____
The minimum value of $a^{-5}+a^{-4}+3a^{-3}+1+a^{8}+a^{10}$ for $a>0$ is ____
1 Answer
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✓ Vidaara Team
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· 1mo ago
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Answer: $8$.
These are eight terms (with $3a^{-3}$ as three copies) whose exponents sum to $0$, so their product is $1$; by AM–GM the sum $\ge8$.
JEE Advanced 2011 · Quadratic Equations and Inequations — verified solution by the Vidaara Team.
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