since I have post $n=5$ case show this inequaliy $\sum_{cyc}\frac{a}{|b-c|}\ge 3$ and somedays ago,I have prove $n=3$ it is easyto prove it, I have conjecture :

let $x_{1},x_{2},\cdots,x_{2n-1}\ge 0$,and $x_{i}\neq x_{j}\forall i\neq j=1,2,\cdots,2n-1$

show this inequality $$\dfrac{x_{1}}{|x_{2}-x_{3}|}+\dfrac{x_{2}}{|x_{3}-x_{4}|}+\cdots+\dfrac{x_{2n-1}}{|x_{1}-x_{2}|}\ge n$$

$n=3$ it easy to prove because $$\dfrac{a}{|b-c|}+\dfrac{b}{|c-a|}+\dfrac{c}{|a-b|}=\dfrac{a}{b-c}+\dfrac{b}{a-c}+\dfrac{c}{|a-b|}\ge\dfrac{a}{b}+\dfrac{b}{a}\ge 2$$ where $c=\min(a,b,c)$


1 Answer 1


The question is to prove $$ \sum_{i = 1}^{2n-1}\frac{x_{i}}{|x_{i+1}-x_{i+2}|}\ge n $$ The problem is cyclic. Here and in the following the indices are understood modulo $2n-1$.

The idea of the proof can be taken from the proofs in the cases $2n-1 = 3$ and $2n-1 = 5$ which have been done individually. We generalize these ideas to show that, for $n \ge 3$,

$$ \sum_{i = 1}^{2n-1}\frac{x_{i}}{|x_{i+1}-x_{i+2}|}\\ \ge \sum_{i = 1}^{2n-1}\frac{x_{i}}{\max \{x_{i+1}, x_{i+2}\}}\\ \ge \frac{x_{j_1}}{x_{j_2}} + \frac{x_{j_2}}{x_{j_3}} + \cdots + \frac{x_{j_{k-1}}}{x_{j_k}} +\frac{x_{j_{k}}}{x_{j_1}} \ge k \ge n $$ In here, the first step is evident, and the third step $\frac{x_{j_1}}{x_{j_2}} + \frac{x_{j_2}}{x_{j_3}} + \cdots + \frac{x_{j_{k-1}}}{x_{j_k}} +\frac{x_{j_{k}}}{x_{j_1}} \ge k$ follows from the AM-GM inequality. It remains to be shown that a cyclic series of length $k\ge n$ of values $x_{j_1}, \cdots, x_{j_k}$ always exists which confirms with the terms in the sum $\sum_{i = 1}^{2n-1}\frac{x_{i}}{\max \{x_{i+1}, x_{i+2}\}}$, possibly dropping terms in this sum (which are positive, hence the inequality is preserved).

To illustrate, consider the case $x_1 > x_2 > \cdots > x_{2n-1}$, then we have $$ \sum_{i = 1}^{2n-1}\frac{x_{i}}{\max \{x_{i+1}, x_{i+2}\}}\\ = \sum_{i = 1}^{2n-3}\frac{x_{i}}{x_{i+1}} + \frac{x_{2n-2}}{x_{1}}+ \frac{x_{2n-1}}{x_{1}} \ge \sum_{i = 1}^{2n-3}\frac{x_{i}}{x_{i+1}} + \frac{x_{2n-2}}{x_{1}} $$ which is of the form $\frac{x_{j_1}}{x_{j_2}} + \frac{x_{j_2}}{x_{j_3}} + \cdots + \frac{x_{j_{k-1}}}{x_{j_k}} +\frac{x_{j_{k}}}{x_{j_1}}$ with $k = 2n-2$ terms, and obviously $2n-2 \ge n$ holds for $n \ge 3$. One term in the sum was dropped.

Now consider the general case. W.l.o.g., let the highest value be $x_1$, i.e. $x_1 > x_i$ for all $i \ne 1$. For the series of values $x_{j_1}, \cdots, x_{j_k}$, we therefore start with $x_{j_1} = x_1$.

Note that, by the maximality of $x_1$, we have $$ \sum_{i = 1}^{2n-1}\frac{x_{i}}{\max \{x_{i+1}, x_{i+2}\}} = \sum_{i = 1}^{2n-{\color{red} 3}}\frac{x_{i}}{\max \{x_{i+1}, x_{i+2}\}} + \frac{x_{2n-2}}{x_{1}}+ \frac{x_{2n-1}}{x_{1}} $$ so, after dropping terms, start and end confirm to the cyclic structure $\frac{x_{j_1}}{x_{j_2}} + \frac{x_{j_2}}{x_{j_3}} + \cdots + \frac{x_{j_{k-1}}}{x_{j_k}} +\frac{x_{j_{k}}}{x_{j_1}}$ with $x_{j_1} = x_1$ and for the last element, either $x_{j_k} = x_{2n-2}$ or $x_{j_k} = x_{2n-1}$, dropping the other "last" term.

Now it remains to be shown that such a series can always be constructed, and what is the minimal length $k$ of this series. Since the terms in the sum are $\frac{x_{i}}{\max \{x_{i+1}, x_{i+2}\}}$, each succeeding term can at most be two indices ahead of the previous one. Dropping the terms in between, to achieve the cyclic form, we can have a situation with the minimal length of $k$ as follows: $$ \sum_{i = 1}^{2n-1}\frac{x_{i}}{\max \{x_{i+1}, x_{i+2}\}}\\ = \sum_{i = 1}^{2n-3}\frac{x_{i}}{\max \{x_{i+1}, x_{i+2}\}} + \frac{x_{2n-2}}{x_{1}}+ \frac{x_{2n-1}}{x_{1}}\\ \ge \frac{x_1}{x_3} + \frac{x_3}{x_5} + \cdots + \frac{x_{2n-3}}{x_{2n-1}} + \frac{x_{2n-1}}{x_{1}} $$ Since for the last element, either $x_{j_k} = x_{2n-2}$ or $x_{j_k} = x_{2n-1}$, the last term will always have the denominator $x_1$, i.e. we could also have $$ \sum_{i = 1}^{2n-1}\frac{x_{i}}{\max \{x_{i+1}, x_{i+2}\}}\\ = \sum_{i = 1}^{2n-3}\frac{x_{i}}{\max \{x_{i+1}, x_{i+2}\}} + \frac{x_{2n-2}}{x_{1}}+ \frac{x_{2n-1}}{x_{1}}\\ \ge \frac{x_1}{x_3} + \frac{x_3}{x_5} + \cdots + \frac{x_{{\color{red} q}}}{x_{2n-2}} + \frac{x_{2n-2}}{x_{1}} $$ with $q = 2n-3$ or $q = 2n-4$. This establishes cyclicity.

Now it is easy to count the minimal length $k_{\min}$. It occurs when every second term is dropped. With the two conclusion options just demonstrated for the last term, we have indeed $k \ge k_{\min} = n$.

For illustration again, consider the series of values $(x_1, x_2, \cdots, x_{2n-1}) = (2n-1, 2n-3, 2n-2, 2n-5, 2n-4, \cdots, 3,4,1,2)$. Then we have $$ \sum_{i = 1}^{2n-1}\frac{x_{i}}{\max \{x_{i+1}, x_{i+2}\}}\\ = \frac{2n-1}{2n-2} + \frac{2n-3}{2n-2} + \frac{2n-2}{2n-4} + \frac{2n-5}{2n-4} + \cdots +\frac{4}{2}+\frac{1}{2n-1} +\frac{2}{2n-1}\\ \ge \frac{2n-1}{2n-2} + \frac{2n-2}{2n-4} + \frac{2n-4}{2n-6} + \cdots +\frac{4}{2} +\frac{2}{2n-1} $$ and the last sum is of the cyclic structure $\frac{x_{j_1}}{x_{j_2}} + \frac{x_{j_2}}{x_{j_3}} + \cdots + \frac{x_{j_{k-1}}}{x_{j_k}} +\frac{x_{j_{k}}}{x_{j_1}}$ and has $k=n$ terms.

This concludes the proof. $\qquad \Box$


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