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if $n$ be odd postive integers,let $x_{i}\in R$,and such $$x_{1}+x_{2}+\cdots+x_{n}=0$$

show that :$$\left(\sum_{i=1}^{n}|x_{i+1}-x_{i}|\right)^2\le 4(n-1)\sum_{i=1}^{n}x^2_{i}\tag{1}$$ where $x_{n+1}=x_{1}$

This problem is from this problem :find the minimum of the positive real value $c$ such $x_{1}+x_{2}+\cdots+x_{n}=1$

I'm more interested in this question with this inequality (1). because I think it might have a very nice answer.such as Cauchy-Schwarz inequality and also $$(\sum_{i=1}^{n}|x_{i+1}-x_{i}|)^2\le (\sum_{i=1}^{n}(|x_{i}|+|x_{i+1}|))^2=4(\sum_{i=1}^{n}|x_{i}|)^2\le 4n\sum_{i=1}^{n}x^2_{i}$$ But I can't use Cauchy-Schwarz inequality to prove $4n-4$

math110
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    Let $S\in\mathbb R^{n\times n}$ be the circular shift matrix. Then the claim is equivalent to $|(S-I)x|_1\le 2\sqrt{n-1}|x|_2$. We always have$$|(S-I)x|_1\le\sqrt n|(S-I)x|_2\le\sqrt n|S-I||x|_2\le 2\sqrt n|x|,$$which is close to the inequality. Well, the inequality is only supposed to hold for $x$ in the $1$-codimensional subspace $M = {x : x_1+\ldots+x_n=0}$. I guess this allows to reduce the dimension by one. Note that $S-I$ leaves the subspace $M$ invariant. – amsmath Sep 01 '19 at 01:27
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    For $n=3$ I was able to prove the following for $x\in M$: (1) $|x|_1^2\le \frac{8}3|x|_2^2$ and (2) $|(S-I)x|_2^2 = 3|x|_2^2$. Then for $x\in M$ we have$$|(S-I)x|_1^2\le \frac{8}3|(S-I)x|_2^2 = 4(n-1)|x|_2^2,$$which is to be proved. I also observed that for general $n$ we have $\operatorname{im}(S-I) = \operatorname{im}(S-I)^T = M$. – amsmath Sep 01 '19 at 03:34
  • @amsmath,How to prove it.Thanks – math110 Sep 01 '19 at 04:55
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    An orthogonal system for $M$ is given by ${u=(1,0,-1)^T,,v=(1,-2,1)^T}$. You'll easily see that these two vectors are eigenvectors of $K := (S-I)^T(S-I)$ corresponding to the eigenvalue $3$. Therefore $K$ acts on $M$ as $3I$. Thus, $|(S-I)x|_2^2 = \langle Kx,x\rangle = 3|x|_2^2$ for $x\in M$. For (1) let $x\in M$. Then $x=su+tv$ with $s,t\in\mathbb R$. We want to prove that $|x|_1^2\le\frac 83|x|_2^2$. We have$$|x|_2^2 = |su+tv|_2^2 = s^2|u|_2^2+t^2|v|_2^2 = 2s^2+6t^2.$$ Also$$|x|_1 = |s+t|+2|t|+|t-s| = 2\max{|s|,|t|}+2|t|.$$ – amsmath Sep 01 '19 at 05:22
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    It thus remains to prove that $4(\max{a,b}+b)^2\le\frac 83(2a^2+6b^2)$ for $a,b\ge 0$, which is easy to do. – amsmath Sep 01 '19 at 05:22
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    BTW, I just saw that Alex Ravsky proved this inequality in an answer to your question that you have linked above. So, why do you ask it again instead of dealing with the proof given there? – amsmath Sep 01 '19 at 06:05

2 Answers2

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Let $x_{n+2} = x_2$, and note that $$ \prod_{i = 1}^n \left[(x_{i + 2} - x_{i + 1})(x_{i+1} - x_{i})\right] = \left[\prod_{i = 1}^n (x_{i+1} - x_{i})\right]^2 \geq 0. $$ Since $n$ is odd, there exsits $j \in \{1, \dots, n\}$, such that $(x_{j + 2} - x_{j + 1})(x_{j+1} - x_{j}) \geq 0$. Without loss of generality we assume that $j = n - 1$, i.e., $(x_{n+1} - x_{n})(x_{n} - x_{n-1}) \geq 0$. Therefore, we have $$ |x_{n} - x_{n-1}| + |x_{n+1} - x_n| = |x_{n} - x_{n-1} + x_{n+1} - x_n| = |x_{n +1} - x_{n - 1}| = |x_1 - x_{n - 1}|, $$ and thus $$ \sum_{i = 1}^n |x_{i + 1} - x_i| = \sum_{i = 1}^{n-2} |x_{i + 1} - x_i| + |x_1 - x_{n - 1}| = \sum_{i = 1}^{n-1} |y_{i + 1} - y_i|, $$ where we have defined $y_i = x_i$ for $i = 1, \dots, n - 1$ and $y_{n} = x_1$.

Finally, following either the argument in the problem or the first comment, we have $$ \left(\sum_{i = 1}^{n-1} |y_{i + 1} - y_i|\right)^2 \leq 4(n-1)\sum_{i = 1}^{n - 1} y_i^2 = 4(n-1)\sum_{i = 1}^{n - 1} x_i^2 \leq 4(n-1)\sum_{i = 1}^{n} x_i^2. $$

Xiangxiang Xu
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Any cycle $C=(x_1,x_2,...,x_n)$ has a subcycle $(u_1,d_1,...,u_m,d_m)$, with $u_{m+1}=u_1$, such that $C$ is non-increasing between each $u_i$ and $d_i$ and non-decreasing between each $d_i$ and $u_{i+1}$. Then $$\sum_{i=1}^n|x_{i+1}-x_i|=2\sum_{i=1}^mu_i-2\sum_{i=1}^md_i\le 2\sum_{i=1}^m(|u_i|+|d_i|).$$ By Cauchy-Schwarz $$\left(\sum_{i=1}^{n}|x_{i+1}-x_i|\right)^2\le 8m\sum_{i=1}^m(|u_i|^2+|d_i|^2)\le 4(n-1)\sum_{i=1}^n x_i^2.$$

Note

The condition $x_{1}+x_{2}+\cdots+x_{n}=0$ is irrelevant.