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Given a finite increasing sequence of real numbers $\{x_i\}_{i=1}^n$ consisting of at least two elements, how can we show that $$ \frac{1}{n}\sum_{i=1}^{n}x_{i}x_{n-i+1} < \left(\frac{1}{n}\sum_{i=1}^{n}x_{i}\right)^2 $$ without expanding. I attempted applying Holder's inequality but failed. Maybe I missed something.

venrey
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4 Answers4

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Hints:

1) the sum in LHS is by Rearrangement Inequality the lowest possible of form $x_ix_{\pi(i)}$ where $\pi$ is a permutation.

2) unless the $x_i$ are strictly increasing somewhere, equality is possible.

P.S. For some more detail, note that $n\sum x_ix_{n−i+1}<\left(\sum x_i\right)^2$ follows from $n$ rearrangements summed together. Each such rearrangement has an RHS of form $\sum x_i x_{i+j}$ where $j=0,1,...n−1$, and the sum in LHS is lower than each of them.

Macavity
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This should be wrong. For example for $x_i = 1$ this is an equality isn't it? $$ \frac{1}{n} \sum \limits_{i=1}^{n} 1^2 = 1 = 1^2 = (\frac{1}{n} \ n)^2 = (\frac{1}{n} \sum \limits_{i=1}^{n} 1)^2 $$

justabit
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Let $$ \bar{x}=\frac1n\sum_{i=1}^nx_i\tag1 $$ If $x_i\le x_{i+1}$ and $x_1\lt x_n$, we can find a $k$ so that $x_k\le\bar{x}\lt x_{k+1}$. Then $$ \bar{x}-x_{n-i+1}\ge0\iff i\ge n-k+1\tag2 $$ Therefore, $$ \begin{align} \left(\frac1n\sum_{i=1}^nx_i\right)^2-\frac1n\sum_{i=1}^nx_ix_{n-i+1} &=\frac1n\sum_{i=1}^nx_i\bar{x}-\frac1n\sum_{i=1}^nx_ix_{n-i+1}\tag3\\ &=\frac1n\sum_{i=1}^nx_i\left(\bar{x}-x_{n-i+1}\right)\tag4\\ &=\frac1n\sum_{i=1}^n\left(x_i-x_{n-k+1}\right)\left(\bar{x}-x_{n-i+1}\right)\tag5 \end{align} $$ Explanation:
$(3)$: apply $(1)$
$(4)$: distributive property
$(5)$: subtract $\frac1nx_{n-k-1}\sum\limits_{i=1}^n\left(\bar{x}-x_{n-i+1}\right)=0$

Applying $(2)$, $$ \begin{align} i\ge n-k+1&\implies\bar{x}-x_{n-i+1}\ge0\quad\text{and}\quad x_i-x_{n-k+1}\ge0\tag6\\ i\lt n-k+1&\implies\bar{x}-x_{n-i+1}\lt0\quad\text{and}\quad x_i-x_{n-k+1}\le0\tag7 \end{align} $$ Thus, the terms in $(5)$ are non-negative. Furthermore, either the $i=1$ term or the $i=n$ term of $(5)$ is positive; that is, $\bar{x}$ cannot equal either $x_1$ or $x_n$, and $x_{n-k+1}$ cannot equal both $x_1$ and $x_n$.

Summarizing, if $x_i\le x_{i+1}$ and $x_1\lt x_n$, $$ \frac1n\sum_{i=1}^nx_ix_{n-i+1}\lt\left(\frac1n\sum_{i=1}^nx_i\right)^2\tag8 $$

robjohn
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  • This seems quite a neat argument @robjohn. Question: How did you come up with the last equality? In other words, I can't seem to get how did you utilized the existence of $k$. – venrey Oct 22 '17 at 03:13
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    @venrey: I have added some explanation. If that does not answer your question, please let me know. – robjohn Oct 22 '17 at 04:13
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$$\frac{1}{n}\sum_{i=1}^{n}x_{i}x_{n-i+1} < \left(\frac{1}{n}\sum_{i=1}^{n}x_{i}\right)^2\iff\sum_{i=1}^{n}x_{i}x_{n-i+1} < \frac{1}{n}\left(\sum_{i=1}^{n}x_{i}\right)^2$$

When $n$ is odd one has $LHS=2(x_1x_n+\cdots+x_{\frac{n-1}{2}}x_{\frac{n+3}{2}})+x^2_{\frac{n+1}{2}}$

When $n$ is even one has $LHS=2(x_1x_n+\cdots+x_{\frac{n}{2}}x_{\frac{n+2}{2}})$

In both cases we have trivially that a sum of squares distinct of zero is greater than $0$.

$$0\lt n((x_n-x_1)^2+\cdots +(x_{\frac{n+3}{2}}-x_{\frac{n-1}{2}})^2)\text{ when n is odd }\\0\lt n((x_n-x_1)^2+\cdots +(x_{\frac{n+2}{2}}-x_{\frac{n}{2}})^2)\text{ when n is even }$$

Piquito
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  • Where do you use the monotonicity of $x_i$? Without that, the inequality is not necessarily true. – robjohn Oct 22 '17 at 01:51
  • All the $(x_i-x_j)^2$ should be positive for this because the square of both $a_i-a_j$ and $a_j-a_i$ are equal and greater than zero. – Piquito Oct 23 '17 at 02:55
  • Why the sign of edit? I don't have edit anything. Who put this "edit"? – Piquito Oct 23 '17 at 02:57