# How prove Reversing the Arithmetic mean – Geometric mean inequality？

Let $x_1,x_2,\cdots,x_n$ $(n\geq2)$ be a non-decreasing monotonous sequence of positive numbers such that $x_1,\frac{x_2}{2},\cdots,\frac{x_n}{n}$ be a non-increasing monotonous sequence .Prove that $$\dfrac{\displaystyle\sum_{i=1}^{n} x_i }{n\left (\displaystyle\prod_{i=1}^{n}x_i \right )^{\frac{1}{n}}}\le \dfrac{n+1}{2\sqrt[n]{n!}}$$

It is well know $$\sum_{i=1}^{n}x_{i}\ge n\left(\prod_{i=1}^{n}x_{n}\right)^{1/n}$$

But I don't How to prove this Reversing the AM-GM inequality

WLOG we can set $\sum x_k=\frac12n(n+1)$, as the inequality and conditions are all homogeneous. Now the inequality is equivalent to: $$\prod_k x_k \ge n! \iff \sum_k \log x_k \ge \sum_k \log k$$

which would follow from the concavity of $\log$ and Karamata's inequality applied to the sequences $(k)$ and $(x_k)$, provided we can show $(k) \succ (x_k)$.

Clearly $\sum k = \sum x_k = \frac12n(n+1)$ and $(k), (x_k)$ are both non-decreasing. It is also easy to show $x_1 \ge 1$ as otherwise $x_k < k$ for all $k$ which would contradict the sum condition.

Suppose $\sum_{k=1}^{m-1} x_k \ge \sum_{k=1}^{m-1} k$ for some $m \in \{2, \dots, n\}$. Then if possible let $\sum_{k=1}^m x_k < \sum_{k=1}^m k$. This would mean $x_m < m \implies x_{m+j} < m+j$ for all allowable positive $j$, so $\sum_{k=1}^n x_k = \sum_{k=1}^m x_k+\sum_{k=m+1}^n x_k < \sum_{k=1}^m k + \sum_{k=m+1}^n k = \frac12n(n+1)$, which is a contradiction.

Hence $(k) \succ (x_k)$, and we are done.

The inequality can be written as: $$\sum_{i=1}^{n} x_i \leq \frac{n(n+1)}{2}\left(\prod_{i=1}^{n}\frac{x_i}{i}\right)^{1/n}=\left(\sum_{i=1}^{n}i\right)\left(\prod_{i=1}^{n}\frac{x_i}{i}\right)^{1/n}\tag{1}$$ or as: $$A_n=\frac{1}{\sum_{i=1}^{n}i}\sum_{i=1}^{n}i\cdot\frac{x_i}{i}\leq \left(\prod_{i=1}^{n}\frac{x_i}{i}\right)^{1/n}=G_n\tag{2}$$ And we can assume $x_1=1$ WLOG.

Since $\{x_n\}$ is non-decreasing while $\{x_i/i\}$ is non-increasing, we have $1\leq \frac{x_{n+1}}{x_n}\leq \frac{n+1}{n}$.

Moreover, $$A_{n+1}=\frac{1}{\binom{n+2}{2}}\left(\binom{n+1}{2}A_n+x_{n+1}\right)=\frac{n}{n+2}A_n+\frac{x_{n+1}}{\binom{n+2}{2}}\leq\frac{nA_n+2}{n+2}\tag{3}$$ hence we just need to prove: $$\frac{n}{n+2}\cdot G_n + \frac{2}{n+2}\cdot 1 \le G_{n+1} \tag{4}$$ in order that $(2)$ is granted by induction.

• Nice,can you post $(4)$ solution?
– user223800
Commented Mar 26, 2015 at 11:16