JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Sequences and Series

We need to minimize $S=a^{-5}+a^{-4}+3a^{-3}+1+a^8+a^{10}$ for $a>0$.

By the AM-GM inequality applied to the terms $a^{-5},a^{-4},a^{-3},a^{-3},a^{-3},1,a^8,a^{10}$ (note that $3a^{-3}$ is split into three terms $a^{-3}$):

$\frac{a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^8+a^{10}}{8}\ge \sqrt[8]{a^{-5}\cdot a^{-4}\cdot a^{-3}\cdot a^{-3}\cdot a^{-3}\cdot 1\cdot a^8\cdot a^{10}}$.

The product is $a^{-5-4-3-3-3+8+10}=a^0=1$.

Therefore $\frac{S}{8}\ge 1$, giving $S\ge 8$.

Equality holds when all terms are equal, which occurs at $a=1$. At $a=1$, $S=1+1+3+1+1+1=8$.