Problem 1739

Solution. By using Karamata's inequality for the concave function $f(x)=\ln x$, we only need to prove that the number sequence $\left(x^{*}\right)$ majorizes the number sequence $\left(y^{*}\right)$, where $(x)=\left(x_{1}, x_{2}, \ldots, x_{n}\right)$ and $(y)=\left(y_{1}, y_{2}, \ldots, y_{n}\right)$, and for each $i \in\{1,2, \ldots, n\}$,
$$x_{i}=\frac{a_{i}+a_{i+1}}{2}, y_{i}=\frac{a_{i}+a_{i+1}+a_{i+2}}{3}$$
(with the common notation $a_{n+1}=a_{1}$ and $a_{n+2}=a_{2}$). According to the Symmetric Majorization Criterion, it suffices to prove the following inequality
$$3\left(\sum_{i=1}^{n}\left|z_{i}+z_{i+1}\right|\right) \geq 2\left(\sum_{i=1}^{n}\left|z_{i}+z_{i+1}+z_{i+2}\right|\right)$$
for all real numbers $z_{1} \geq z_{2} \geq \ldots \geq z_{n}$ and $z_{n+1}, z_{n+2}$ stand for $z_{1}, z_{2}$ respectively.
Notice that (*) is obviously true if $z_{i} \geq 0$ for all $i=1,2, \ldots, n$. Otherwise, assume that $z_{1} \geq z_{2} \geq \ldots \geq z_{k} \geq 0>z_{k+1} \geq \ldots \geq z_{n}$. We realize first that it's enough to consider ( $\star$ ) for 8 numbers (instead of $n$ numbers). Now consider it for 8 numbers $z_{1}, z_{2}, \ldots, z_{8}$. For each number $i \in\{1,2, \ldots, 8\}$, we denote $c_{i}=\left|z_{i}\right|$, then $c_{i} \geq 0$. To prove this problem, we will prove first the most difficult case $z_{1} \geq z_{2} \geq z_{3} \geq z_{4} \geq$ $0 \geq z_{5} \geq z_{6} \geq z_{7} \geq z_{8}$. Giving up the absolute value signs, the problem becomes
$$\begin{array}{c} 3\left(c_{1}+2 c_{2}+2 c_{3}+c_{4}+c_{5}+2 c_{6}+2 c_{7}+c_{8}+\left|c_{4}-c_{5}\right|+\left|c_{8}-c_{1}\right|\right) \\ \geq 2\left(c_{1}+2 c_{2}+2 c_{3}+c_{4}+\left|c_{3}+c_{4}-c_{5}\right|+\left|c_{4}-c_{5}-c_{6}\right|+c_{5}+2 c_{6}+2 c_{7}+c_{8}+\left|c_{7}+c_{8}-c_{1}\right|+\left|c_{8}-c_{1}-c_{2}\right|\right) \\ \Leftrightarrow c_{1}+2 c_{2}+2 c_{3}+c_{4}+c_{5}+2 c_{6}+2 c_{7}+c_{8}+3\left|c_{4}-c_{5}\right|+3\left|c_{8}-c_{1}\right| \\ \geq 2\left|c_{3}+c_{4}-c_{5}\right|+2\left|c_{4}-c_{5}-c_{6}\right|+2\left|c_{7}+c_{8}-c_{1}\right|+2\left|c_{8}-c_{1}-c_{2}\right| \end{array}$$

Clearly, this inequality is obtained by adding the following results
$$\begin{aligned} 2\left|c_{4}-c_{5}\right|+2 c_{3} & \geq 2\left|c_{3}+c_{4}+c_{5}\right| \\ 2\left|c_{8}-c_{1}\right|+2 c_{7} & \geq 2\left|c_{7}+c_{8}-c_{1}\right| \\ \left|c_{4}-c_{5}\right|+c_{4}+c_{5}+2 c_{6} & \geq 2\left|c_{4}-c_{5}-c_{6}\right| \\ \left|c_{8}-c_{1}\right|+c_{8}+c_{1}+2 c_{2} & \geq 2\left|c_{8}-c_{1}-c_{2}\right| \end{aligned}$$

For other cases when there exist exactly three (or five); two (or six); only one (or seven) non-negative numbers in $\left\{z_{1}, z_{2}, \ldots, z_{8}\right\}$, the problem is proved completely similarly (indeed, notice that, for example, if $z_{1} \geq z_{2} \geq z_{3} \geq 0 \geq z_{4} \geq z_{5} \geq z_{6} \geq$ $z_{7} \geq z_{8}$ then we only need to consider the similar but simpler inequality of seven numbers after eliminating $z_{6}$). Therefore $(\star)$ is proved and the conclusion follows immediately.