IIT-JEE 2011 · Paper 1 · Mathematics Q68

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MathematicsInteger+4Question 68 of 69

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

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Answer and worked solution

Official answer: 8

AM-GM on eight terms: $a^{-5},a^{-4},a^{-3},a^{-3},a^{-3},1,a^8,a^{10}$ (splitting $3a^{-3}$ ). Their product is $a^{-5-4-9+8+10}=1$ , so the sum is at least 8, with equality at $a=1$ .

Answer: 8

Topics:AM-GM-HM inequalities

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IIT-JEE 2011 · Paper 1, Mathematics question 68