Problem:
Given $n \ge 3$, $a_i \ge 1$ for $i \in \{1, 2, \dots, n\}$. Prove the following inequality:
$$(a_1+a_2+\dots +a_n) (\frac{1}{a_1}+\frac{1}{a_2} + \dots +\frac{1}{a_n}) \le n^2+\sum_{1\le i<j\le n}|a_i-a_j|.$$
This inequality seems like a reversed form of AM-HM inequality (which states that $\operatorname{LHS}\ge n^2$). I know that Kantorovich inequality is of the same direction, but that is not helpful. I tried to apply Cauchy–Schwarz inequality, but failed.
Thanks for any help.
Best Answer
We need to prove that $$\sum_{1\leq i<j\leq n}\frac{(a_i-a_j)^2}{a_ia_j}\leq\sum_{1\leq i<j\leq n}|a_i-a_j|,$$ for which it's enough to prove that $$a_ia_j\geq|a_i-a_j|,$$ which is true because if $a_i\geq a_j$ we obtain: $$a_ia_j-|a_i-a_j|=a_ia_j-(a_i-a_j)=a_i(a_j-1)+a_j\geq0.$$ The case $a_i\leq a_j$ is the similar.