Proving a reversed AM-HM inequality

Question

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.

Answer 1

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.

Answer 2

I will prove this via induction. Without loss of generality, assume $a_1 \le \cdots \le a_n$. It turns out this inequality remains true for $n = 1, 2$ as well, reducing to $1 \le 1$ in the first case. We actually need to prove the $n = 2$ case in order to elicit intuition for the inductive step.

Base Case $n = 2$

If $a_1 = a_2$, the inequality is trivially true, so assume $a_1 < a_2$. Then we seek to prove \begin{align*} (a_1 + a_2)(a_1^{-1} + a_2^{-1}) \le 2^2 + |a_2 - a_1| &\iff \frac{a_1}{a_2} + \frac{a_2}{a_1} \le 2 + (a_2 - a_1) \tag{*}\\ &\iff 0 \le \frac{a_1^2}{a_2 - a_1} + a_1 - 1 \end{align*} which is true from the conditions $a_i \ge 1$.

Inductive Step

Define \begin{align*} S^A_n = \sum_{i=1}^{n}a_i \qquad \text{and} \qquad S^H_n = \sum_{i=1}^{n}a_i^{-1} \end{align*}

We will use the relation \begin{align*} \sum_{1 \le i < j \le n}|a_i - a_j| = 2\sum_{i=1}^{n}ia_i - (n+1)S^A_n \end{align*} So we have \begin{align*} S^A_{n+1}S^H_{n+1} &= S^A_n S^H_n + a_{n+1}S^H_n + a^{-1}_{n+1}S^A_n + 1 \\ &\le n^2 + \left[2\sum_{i=1}^{n}ia_i - (n+1)S^A_n\right] + a_{n+1}S^H_n + a^{-1}_{n+1}S^A_n + 1 & \text{(Inductive Hypothesis)} \\ &\overset{\text{def}}{=} M \end{align*} To conclude, we want to show that $M$ does not exceed \begin{align*} N &\overset{\text{def}}{=} (n+1)^2 + \left[2\sum_{i=1}^{n+1}ia_i - (n+2)S^A_{n+1}\right] \end{align*} Computing the difference, \begin{align*} N - M &= 2n + 1 + 2(n+1)a_{n+1} - S^A_n - (n+2)a_{n+1} - a_{n+1}S^H_n - a^{-1}_{n+1}S^A_n -1\\ &= 2n + na_{n+1} - S^A_n - a_{n+1}S^H_n - a^{-1}_{n+1}S^A_n \end{align*} Finally, we have \begin{align*} na_{n+1} - S^A_n &= \sum_{i=1}^{n}(a_{n+1} - a_i) \\ &\ge\sum_{i=1}^{n}\left(\frac{a_i}{a_{n+1}}+\frac{a_{n+1}}{a_{i}}-2\right) & \text{From (*) in base step}\\ &=a_{n+1}S^H_n + a^{-1}_{n+1}S^A_n - 2n \end{align*}