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Given the odd positive integer $n>1$, find the minimum positive real value of $c$ such that for all $x_{i}\in \Bbb R$ with $$x_{1}+x_{2}+\cdots+x_{n}=1,$$ it holds that

$$c\left(\sum_{i=1}^{n}x^2_{i}\right)^3\ge \left(\sum_{i=1}^{n}|x_{i+1}-x_{i}|\right)^2\left(\sum_{1\le i<j\le n}(x_{i}-x_{j})^2\right)$$ where $x_{n+1}=x_{1}.$

My attempt: since $$\sum_{1\le i<j\le n}(x_{i}-x_{j})^2=n\sum_{i=1}^{n}x^2_{i}-(\sum_{i=1}^{n}x_{i})^2=n\sum_{i=1}^{n}x^2_{i}-n$$ and $$(\sum_{i=1}^{n}|x_{i+1}-x_{i}|)^2=\sum_{i=1}^{n}(x_{i+1}-x_{i})^2+2\sum_{1\le i<j\le n}|x_{i+1}-x_{i}||x_{j+1}-x_{j}|=2\sum_{i=1}^{n}x^2_{i}-2\sum_{i=1}^{n}x_{i+1}x_{i}+2\sum_{1\le i<j\le n}|x_{i+1}-x_{i}||x_{j+1}-x_{j}|$$ Then I can't ,maybe this problem is from a Integral discrete it?

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  • $\begingroup$ I fixed the order of logical quantifiers to actually make sense. $\endgroup$ – mathworker21 Aug 27 at 13:59
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    $\begingroup$ If $\sum x_i=1$ then $n\sum_{i=1}^{n}x^2_{i}-(\sum_{i=1}^{n}x_{i})^2=n\sum_{i=1}^{n}x^2_{i}-1$. $\endgroup$ – Alex Ravsky Aug 28 at 15:26
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We can show that River Li’s bound $c_n\ge \frac{16}{27}(n-1)n^2=b_n$ is tight as follows.

Given any real $x_1,\dots x_n$ with $\sum x_i=1$ for each $i$ put $y_i=x_i-\tfrac 1n$. Then $x_i-x_j=y_i-y_j$ for each $i$ and $j$, $\sum y_i=0$, and

$$\sum x_i^2=\sum \left(y_i+\frac 1n\right)^2=\sum y_i^2+\frac 2n y_i+\frac 1{n^2}=$$ $$ \sum y_i^2+ \frac 2n\left(\sum y_i\right)+n\frac 1{n^2}=\sum y_i^2+\frac 1{n}.$$

Then

$$\sum_{1\le i<j\le n}(x_{i}-x_{j})^2=n\sum_{i=1}^{n}x^2_{i}-1=n\sum y_i^2=nY\ge 0.$$

Thus we have to show that

$$b_n(Y+\tfrac 1n)^3\ge n\left(\sum_{i=1}^{n}|y_{i+1}-y_{i}|\right)^2Y.$$

For this we need the following auxiliary result.

For each natural $k$ and each point $z=(z_1,\dots, z_k)\in \Bbb R^k$ let $\|z\|=\sqrt{\sum_{i=1}^k z_i^2}$, $B_k=\{z\in\Bbb R^k: \|z\|\le 1\}$ be the unit ball in the space $\Bbb R^k$, and $f:B_k\to\Bbb R$ be a function such that $$f(z)=\sum_{i=1}^{k}|z_{i+1}-z_{i}|$$ for each $z\in B_k$, where $z_{k+1}=z_1$. Since the set $B_k$ is compact and the function $f_k$ is continuous, it attains its maximum $M_k$ at some point $t=(t_1,\dots,t_k)\in B_k$. Clearly, $\sum_{i=1}^{n}|y_{i+1}-y_{i}|\le \sqrt{Y}M_k$.

By the inequality between arithmetic and quadratic means, we have $$M_k=\sum_{i=1}^{k}|t_{i+1}-t_{i}|\le \sum_{i=1}^{k}|t_{i+1}|+|t_{i}|=$$ $$2\sum_{i=1}^{k}|t_i|\le 2\sqrt{k}\sqrt{\sum_{i=1}^{k}t_i^2}=2\sqrt{k}\|t\|\le 2\sqrt{k}.$$ It is easy to check that if $k$ is even then the equality above is attained when $t_i=\tfrac {(-1)^i}{\sqrt{k}}$ for each $i$.

Now assume that $k>1$ is odd. Suppose to the contrary that all $t_i$ are non-zero. We claim that there exists a natural $i$ such that both $t_i$ and $t_{i+1}$ (in the circular order) have the same sign. Indeed, otherwise the sign of each $t_i$ is $(-1)^{i-1}\operatorname{sign} t_1$. Since $k$ is odd, $t_1$ and $t_k$ have the same sign, a contradiction. Let $t_j,\dots, t_l$ be a longest circular subsequence of consecutive $t_i$’s with the same sign. Let $t_m$ be a number in this subsequence with the smallest absolute value. Let a point $t’\in\Bbb R^k$ has the same coordinates as $t$, but only its $m$-th coordinate is zero, and $$\lambda=\frac {\|t\|}{\|t’\|}>1.$$ Then $\|\lambda t’\|=\|t\|\le 1$, so $\lambda t’\in B_k$. On the other hand, it is easy to check that $f(t’)\ge f(t)$, so $f(\lambda t’)=\lambda f(t’)>f(t)$, a contradiction with the maximality of $f(t)$. Thus there exists $m$ such that $t_m=0$. Then similarly to the above case of even $k$ we can show that $$f(t)\le 2\sum_{i=1}^{k}|t_i|\le 2\sqrt{k-1}\sqrt{\sum_{i=1}^{k}t_i^2}=2\sqrt{k-1}\|t\|\le 2\sqrt{k-1}.$$ It is easy to check that if the equality above is attained when $t_i=\tfrac {(-1)^i}{\sqrt{k-1}}$ for each $i<k$ and $t_k=0$. Thus $M_k=2\sqrt{k-1}$.

Since $n$ is odd, $$\left(\sum_{i=1}^{n}|y_{i+1}-y_{i}|\right)^2\le YM_n^2=4Y(n-1).$$

So it remains to show that

$b_n(Y+\tfrac 1n)^3\ge 4(n-1)nY^2$

$\frac{16}{27}(n-1)n^2(Y+\frac 1n)^3\ge 4(n-1)nY^2$

$4n(Y+\frac 1n)^3\ge 27Y^2$

$4(nY+1)^3\ge 27n^2Y^2$

Put $y=nY\ge 0$. We have to show that

$4(y+1)^3\ge 27y^2$

$4y^3-15y^2+12y+4\ge 0$

$(y-2)^2(4y+1)\ge 0$.

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  • $\begingroup$ But I think we can find the minumim of the $c(n)$,I think your step maybe is not usefull.But Thanks $\endgroup$ – inequality Aug 29 at 15:38
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    $\begingroup$ @inequality My long mathematical experience taught me that when we start to solve an unfamiliar problem, simple results, even weak, often help to understand the problem, develop initial intuition for it, find its crucial issues. In the considered case, my simple upper bound for $c(n)$ complements computationally suggested River Li’s lower bound and differs from it only by a constant factor. $\endgroup$ – Alex Ravsky Aug 30 at 1:52
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    $\begingroup$ (+1) Impressive! $\endgroup$ – River Li Aug 30 at 15:09
  • $\begingroup$ @RiverLi Thank you for your kind words. But, saying the truth, you did the key part of the work, so I propose to give the bounty to you. Going to the details, after my initial attack yielding loose bounds I saw no ways to continue, so I gave up. I also had an idea to give to $x_i$’s two interchanging values providing the best lower bound, but, being a professional mathematician, was to busy and too lazy to develop it. $\endgroup$ – Alex Ravsky Aug 30 at 17:36
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    $\begingroup$ Actually, what I did is easy since optimization algorithms can reveal the rule sometimes (everyone can do). You did the hard part. By the way, there are some interesting but hard problems in MSE etc. People, from students to mathematicians, enjoy the problems and nice solutions. $\endgroup$ – River Li Aug 31 at 1:06
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With numerical searching, I guess that the best constant $c_n$ is $\frac{16}{27}(n-1)n^2$. Some values: $c_3 = 32/3, \ c_5 = 1600/27, \ c_7 = 1568/9, \ c_9 = 384, \ c_{11} = 19360/27, \cdots$
Are there any counterexamples (by computer)?

Details: We have $$c_n = \sup_{x_1 + x_2 + \cdots + x_n = 1} \frac{\left(\sum_{i=1}^{n}|x_{i+1}-x_{i}|\right)^2\left(n\sum_{i=1}^{n}x^2_{i} -1\right)}{\left(\sum_{i=1}^{n}x^2_{i}\right)^3}.$$

Let $(x_1, x_2, \cdots, x_n) = (a, b, a, b, \cdots, a, b, \frac{1}{n})$ where $$a = \frac{1}{n} + \frac{1}{\sqrt{\frac{n(n-1)}{2}}}, \quad b = \frac{1}{n} - \frac{1}{\sqrt{\frac{n(n-1)}{2}}}.$$ We have $$\frac{\left(\sum_{i=1}^{n}|x_{i+1}-x_{i}|\right)^2\left(n\sum_{i=1}^{n}x^2_{i} -1\right)}{\left(\sum_{i=1}^{n}x^2_{i}\right)^3} = \frac{16}{27}(n-1)n^2.$$

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As seen in Alex Ravsky's answer, the desired minimum is the same as the maximum of the expression $$ (nY+1)^{-3}n^4Yf^2(y) $$ subject to $y_1+\cdots+y_n = 0$. For this, let us define the function $g(x) := \frac{x^2}{(x+1)^3}$. It has a global maximum at $x=2$ with $g(2) = \frac 4{27}$. As shown by Alex, we have $$ (nY+1)^{-3}n^4Yf^2(y)\le 4(n-1)(nY+1)^{-3}n^4Y^2 = 4(n-1)n^2g(nY)\le\frac{16}{27}(n-1)n^2. $$ On the other hand, set $\tau =\sqrt{2/n}$ and choose $y\in\mathbb R^n$ such that $y_k = \frac {(-1)^k\tau}{\sqrt{n-1}}$ for $k=1,\ldots,n-1$ and $y_n = 0$. Then $Y = \tau^2 = 2/n$ and $y_1+\ldots+y_n = 0$. Plugging $y$ into our expression gives $$ \frac{2n^3}{27}f(y)^2 = \frac{2n^3}{27}\cdot 4(n-1)Y = \frac{16}{27}(n-1)n^2. $$

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  • $\begingroup$ You are right about $f$ and $c$. I edited. $\endgroup$ – amsmath Sep 3 at 14:07
  • $\begingroup$ Arrgh, yes. Thanks. $\endgroup$ – amsmath Sep 3 at 17:40

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