1 Introduction
We first recall two definitions.
Definition 1.1
([1])
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For fixed \(n\geq 2\), let \(x=(x_{1},x_{2},\ldots ,x_{n})\) and \(y=(y_{1},y_{2},\ldots ,y_{n})\) be two n-tuples of real numbers.
(i)
x is said to be majorized by y (in symbols, \(x \prec y\)) if
where \(x_{[1]}\ge \cdots \ge x_{[n]}\) and \(y_{[1]}\ge \cdots \ge y _{[n]}\) are rearrangements of x and y in descending order.
$$ \sum_{i = 1}^{k} x_{[i]} \le \sum _{i = 1}^{k} y_{[i]}\quad\mbox{for }k = 1,2,\ldots ,n -1,\quad \mbox{and} \quad \sum_{i = 1}^{n} x_{i} = \sum_{i = 1}^{n} y_{i}, $$
(ii)
Let \(\Omega \subset \mathbb{R}^{n}\). A function \(\varphi :\Omega \rightarrow \mathbb{R}\) is said to be a Schur-convex function (shortly, an S-convex function) if
$$ x \prec y \Rightarrow \varphi (x) \leq \varphi (y). $$
For example:
$$\begin{aligned}& a^{(2)}= \biggl( \frac{a_{1}+a_{2}}{2},\frac{a_{2}+a_{3}}{2},\ldots , \frac{a _{n}+a_{1}}{2} \biggr) , \\& a^{(3)}= \biggl( \frac{a_{1}+a_{2}+a_{3}}{3},\frac{a_{2}+a_{3}+a_{4}}{3}, \ldots , \frac{a_{n}+a_{1}+a_{2}}{3} \biggr) . \end{aligned}$$
In 2006, I. Olkin, one of the authors of the book [1], wrote a letter to K. Z. Guan, referring to the following interesting question: is it true that
However, a proof for \(a^{(k+1)}\prec a^{(k)}\) remains elusive (see [1], p. 63).
$$ a^{(k+1)}\prec a^{(k)},\quad 1\leq k\leq n-1? $$
(1)
In 2010, Shi [2] proved that (1) holds when \(n= 4\), \(k= 2 \) and \(n=5\), \(k=3 \). In this paper, we prove that (1) holds for any \(n\geq 2\) and \(1\leq k\leq n-1\).
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For any \(1\leq k\leq n\), let
be the ordered component of the sequence \(a^{(k)}_{1},a^{(k)}_{2}, \ldots ,a^{(k)}_{n}\). We denote
$$ a^{(k)}_{[1]}\geq a^{(k)}_{[2]}\geq \cdots \geq a^{(k)}_{[n]} $$
$$ S_{h}^{(k)}=\sum _{i=1}^{h}a^{(k)}_{[i]}, \quad 1\leq h\leq n. $$
2 Lemmas and corollaries
For proving our main results, we need the following lemmas.
Lemma 2.1
Let
\(n\geq 2\)
and
\(1\le k\le n-1\). Then
$$ \textstyle\begin{cases} a^{(k)}_{1}\geq a^{(k)}_{2}\geq \cdots \geq a^{(k)}_{n-k+1}, \\ a^{(k)}_{n}\geq a^{(k)}_{n-1}\geq \cdots \geq a^{(k)}_{n-k+1}, \\ a^{(k)}_{1}\geq a^{(k)}_{n}. \end{cases} $$
(2)
Proof
For any \(1\le k\le n-1\) and \(1\le i\le n\), we have
Note that \(a_{1}\geq a_{2}\geq \cdots \geq a_{n}\), so we can induce that □
$$ k\bigl(a_{i}^{(k)}-a_{i+1}^{(k)}\bigr)= ( a_{i}+a_{i+1}+\cdots +a_{i+k-1} ) - ( a_{i+1}+a_{i+2}+\cdots +a_{i+k} ) =a_{i}-a_{i+k}. $$
(1)
if \(i+k\leq n\), that is, \(1\leq i\leq n-k\), then \(a_{i}\geq a_{i+k}\). It follows that \(a_{i}^{(k)}\geq a_{i+1}^{(k)}\).
(2)
if \(i+k> n\), that is, \(n-k< i\leq n\), then \(a_{i+k}=a_{n+(i+k-n)}=a _{i+k-n}\). Since \(1\le k\le n-1\), we have \(n\geq i> i+k-n\). So \(a_{i}\geq a_{i+k-n}=a_{i+k}\). It follows that \(a_{i}^{(k)}\geq a_{i+1} ^{(k)}\).
(3)
\(k(a_{1}^{(k)}-a_{n}^{(k)})= ( a_{1}+a_{2}+\cdots +a_{k} ) - ( a_{n}+a_{1}+\cdots +a_{k-1} ) =a_{k}-a_{n}\geq 0\). Therefore (2) holds.
From the proof of Lemma 2.1 it is easy to deduce the following:
Corollary 2.2
Let
\(n\geq 2\), \(1\leq k\leq n-1\), and let
\(\sum_{i=0}^{-1}a^{(k)}_{n-i}=0\). For any
\(1\leq h\leq n-1\), there exist
\(1\leq h_{1}\leq n-k\)
and
\(-1\leq h_{2}\leq k-2\)
such that
\(h=h_{1}+h _{2}+1\)
and
$$ S_{h}^{(k)}= \sum_{i=1}^{h_{1}}a^{(k)}_{i}+ \sum_{i=0}^{h_{2}}a^{(k)} _{n-i}. $$
Lemma 2.3
Let
\(n\geq 4\), \(2\leq k\leq n-2\), and
\(0\leq r\leq k-2\).
(i)
If
then
$$ a^{(k)}_{2}\leq a^{(k)}_{n-r} \leq a^{(k)}_{1}, $$
(3)
$$ a^{(k+1)}_{2}\leq a^{(k+1)}_{n-r} \leq a^{(k+1)}_{1}. $$
(ii)
If
then
$$ a^{(k)}_{n-k+1}\leq a^{(k)}_{n-r} \leq a^{(k)}_{n-k}, $$
(4)
$$ a^{(k+1)}_{n-k}\leq a^{(k+1)}_{n-r-1} \leq a^{(k+1)}_{n-k-1}. $$
(iii)
For
\(2\leq m\leq n-k-1\), if
then
$$ a^{(k)}_{m+1}\leq a^{(k)}_{n-r} \leq a^{(k)}_{m}, $$
(5)
$$ a^{(k+1)}_{m+1}\leq a^{(k+1)}_{n-r}, \qquad a^{(k+1)}_{n-r-1}\leq a^{(k+1)}_{m-1}. $$
(6)
(iv)
If
then
$$ a^{(k)}_{n}\leq a^{(k)}_{n-k} , $$
(7)
$$ a^{(k+1)}_{n-1}\leq a^{(k+1)}_{n-k-1} . $$
(v)
For
\(2\leq m\leq n-k-1\), if
then
$$ a^{(k)}_{n}\leq a^{(k)}_{m} , $$
(8)
$$ a^{(k+1)}_{n-1}\leq a^{(k+1)}_{m} . $$
(vi)
For
\(2\leq m\leq n-k\), if
then
$$ a^{(k)}_{n-r-1}\leq a^{(k)}_{m} \leq a^{(k)}_{n-r}, $$
(9)
$$ a^{(k+1)}_{n-r-2}\leq a^{(k+1)}_{m-1}, \qquad a^{(k+1)}_{m}\leq a^{(k+1)}_{n-r}. $$
Proof
(i)
By Lemma 2.1 we have
It follows that
Thus we have
By the left inequality of (3) we have
Note that \(a_{k+2}\leq a_{k-r}=a_{n+k-r}\), so we have
Therefore
$$ \textstyle\begin{cases} a^{(k+1)}_{1}\geq a^{(k+1)}_{2}\geq \cdots \geq a^{(k+1)}_{n-k}, \\ a^{(k+1)}_{n}\geq a^{(k+1)}_{n-1}\geq \cdots \geq a^{(k+1)}_{n-k}, \\ a^{(k+1)}_{1}\geq a^{(k+1)}_{n}. \end{cases} $$
$$ a^{(k+1)}_{1}=\max \bigl\{ a^{(k+1)}_{1}, a^{(k+1)}_{2},\ldots , a^{(k+1)} _{n}\bigr\} . $$
$$ a^{(k+1)}_{n-r}\leq a^{(k+1)}_{1}. $$
$$ a_{2}+ \cdots + a_{k+1}\leq a_{n-r}+ \cdots + a_{n-r+k-1}. $$
$$ a_{2}+ \cdots + a_{k+1}+ a_{k+2}\leq a_{n-r}+\cdots + a_{n-r+k-1}+ a _{n+k-r}. $$
$$ a^{(k+1)}_{2}\leq a^{(k+1)}_{n-r}. $$
(ii)
By Lemma 2.1 we have
By the right inequality of (4) we get
Note that \(a_{n-r-1}\leq a_{n-k-1}\), so we have
This means that
$$ a^{(k+1)}_{n-k}\leq a^{(k+1)}_{n-r-1}. $$
$$ a_{n-r} +\cdots + a_{n-r+k-1}\leq a_{n-k} +\cdots +a_{n-1}. $$
$$ a_{n-r} +\cdots + a_{n-r+k-1} + a_{n-r-1}\leq a_{n-k} +\cdots +a_{n-1}+ a_{n-k-1}. $$
$$ a^{(k+1)}_{n-r-1}\leq a^{(k+1)}_{n-k-1}. $$
(iii)
By (5) we get
Note that \(n\geq m+k+1\geq k-r \geq 1\) and \(n\geq n-r-1\geq m-1 \geq 1\), so we have
It follows that
and
Therefore (6) holds.
$$ a_{m+1}+ \cdots + a_{m+k}\leq a_{n-r} + \cdots + a_{n-r+k-1} \leq a _{m}+ \cdots +a_{m+k-1}. $$
$$ a_{m+k+1}\leq a_{n-r+k}=a_{k-r},\qquad a_{n-r-1} \leq a_{m-1}. $$
$$ a_{m+1}+ \cdots + a_{m+k} + a_{m+k+1}\leq a_{n-r} + \cdots + a_{n-r+k-1} + a_{n-r+k} $$
$$ a_{n-r-1} + a_{n-r} + \cdots + a_{n-r+k-1} \leq a_{m-1} +a_{m}+ \cdots +a_{m+k-1}. $$
(iv)
By (7) we get
Since \(a_{n-1}\leq a_{n-k-1}\), we have
It follows that
$$ a_{n} + \cdots + a_{n+k-1}\leq a_{n-k} + \cdots + a_{n-1}. $$
$$ a_{n} + \cdots + a_{n+k-1} + a_{n-1}\leq a_{n-k} + \cdots + a_{n-1} + a_{n-k-1}. $$
$$ a^{(k+1)}_{n-1}\leq a^{(k+1)}_{n-k-1}. $$
(v)
By (8) we have
Note that \(n-1\geq m+k\) and \(a_{n-1}\leq a_{m+k}\), so we have
This means that
$$ a_{n} + \cdots + a_{n+k-1}\leq a_{m} +\cdots + a_{m+k-1}. $$
$$ a_{n} + \cdots + a_{n+k-1}+ a_{n-1}\leq a_{m} +\cdots + a_{m+k-1}+ a _{m+k}. $$
$$ a^{(k+1)}_{n-1}\leq a^{(k+1)}_{m}. $$
(vi)
By (9) we get
Note that \(0\leq r\leq k-2\) and \(2\leq m\leq n-k\), so we have
It follows that
So we get
Therefore
Note that \(m\leq n-k\), so we have
It follows that
Therefore
$$ a_{n-r-1}+\cdots + a_{n-r+k-2}\leq a_{m} +\cdots + a_{m+k-1}\leq a _{n-r}+\cdots + a_{n-r+k-1}. $$
$$ n\geq n-r-2\geq n-k\geq m\geq m-1\geq 1. $$
$$ a_{n-r-2}\leq a_{m-1}. $$
$$ a_{n-r-2} + a_{n-r-1}+\cdots + a_{n-r+k-2}\leq a_{m-1} + a_{m} + \cdots + a_{m+k-1}. $$
$$ a^{(k+1)}_{n-r-2}\leq a^{(k+1)}_{m-1}. $$
$$ k-r\leq m+k\leq n,\qquad a_{m+k}\leq a_{k-r}= a_{n+k-r}. $$
$$ a_{m} +\cdots + a_{m+k-1} + a_{m+k}\leq a_{n-r}+\cdots + a_{n-r+k-1} + a_{n+k-r}. $$
$$ a^{(k+1)}_{m}\leq a^{(k+1)}_{n-r}. $$
Lemma 2.4
Let
\(n\geq 4\), \(2\leq k\leq n-1\), \(1\leq m\leq n-k\), and
\(0\leq r \leq k-2\).
(i)
If
\(a^{(k)}_{m+1}\leq a^{(k)}_{n-r}\leq a^{(k)}_{m}\), then
$$ S_{m+r+1}^{(k)}= \sum _{i=1}^{m}a^{(k)}_{i}+\sum _{i=0}^{r}a^{(k)}_{n-i}. $$
(10)
(ii)
If
\(a^{(k)}_{n}\leq a^{(k)}_{m} \), then
$$ S_{m}^{(k)}= \sum_{i=1}^{m}a^{(k)}_{i}. $$
(iii)
For
\(2\leq m\leq n-k\), if
\(a^{(k)}_{n-r-1}\leq a^{(k)} _{m}\leq a^{(k)}_{n-r}\), then
$$ S_{m+r+1}^{(k)}= \sum_{i=1}^{m}a^{(k)}_{i}+ \sum_{i=0}^{r}a^{(k)}_{n-i}. $$
Proof
We only prove (i). Using a similar method, we can obtain (ii) and (iii).
By Lemma 2.1 we have
By Corollary 2.2 we let
where \(1\leq h_{1}\leq n-k+1\), \(-1\leq h_{2}\leq k-2\), and \(\sum_{i=0} ^{-1}a^{(k)}_{n-i}=0\). It is clear that
$$ \textstyle\begin{cases} a_{1}^{(k)}\geq a_{2}^{(k)}\geq \cdots \geq a^{(k)}_{m} \geq a^{(k)} _{n-r}\geq a^{(k)}_{m+1}, \\ a^{(k)}_{n}\geq a^{(k)}_{n-1}\geq \cdots \geq a^{(k)}_{n-r+1}\geq a ^{(k)}_{n-r}\geq a^{(k)}_{m+1}. \end{cases} $$
(11)
$$ S_{m+r+1}^{(k)}=\sum_{i=1}^{h_{1}}a^{(k)}_{i}+ \sum_{i=0}^{h_{2}}a^{(k)} _{n-i}, $$
(12)
$$ m+r+1= h_{1}+h_{2}+1. $$
(13)
Next, we prove that \(h_{1}=m\) and \(h_{2}=r\).
(1)
If \(h_{1}\geq m+1\), which means that the right-hand side of (12) includes \(a^{(k)}_{m+1}\), then by (11) the right-hand side of (12) should include \(a^{(k)}_{1},a^{(k)} _{2},\ldots ,a^{(k)}_{m+1}\) and \(a^{(k)}_{n-r}, a^{(k)}_{n-r+1}, \ldots , a^{(k)}_{n}\), so we have
This is a contradiction with (13).
$$ h_{1}+h_{2}+1\geq m+r+2>m+r+1. $$
(2)
If \(h_{1}\leq m-1\), then by (13) we have \(k-2\geq h_{2} \geq r+1\). Together with (11), we get
So the right-hand side of (12) must include \(a_{m}^{(k)}\), which means that \(h_{1}\geq m\). This is a contradiction with \(h_{1}\leq m-1\).
$$ a^{(k)}_{n-h_{2}}\leq a^{(k)}_{n-r-1}\leq a^{(k)}_{n-r}\leq a^{(k)} _{m}. $$
Corollary 2.5
Let
\(n\geq 4\), \(2\leq k\leq n-2\), \(1\leq m\leq n-k\), and
\(0\leq r \leq k-2\), and let
$$ a^{(k)}_{m+1}\leq a^{(k)}_{n-r} \leq a^{(k)}_{m}. $$
(14)
(i)
For
\(m=1\), we have
$$ S_{r+2}^{(k+1)}=a^{(k+1)}_{1}+ \sum_{i=0}^{r}a^{(k+1)}_{n-i}. $$
(15)
(ii)
For
\(m=n-k\), we have
$$ S_{n-k+r+1}^{(k+1)}= \sum _{i=1}^{n-k-1}a^{(k+1)}_{i}+ \sum _{i=0}^{r+1}a^{(k+1)}_{n-i}. $$
(16)
(iii)
For
\(2\leq k\leq n-3\)
and
\(2\leq m\leq n-k-1\), \(S_{m+r+1}^{(k+1)}\)
must be one of the following two cases:
or
$$ S_{m+r+1}^{(k+1)}=\sum_{i=1}^{m}a^{(k+1)}_{i}+ \sum_{i=0}^{r}a^{(k+1)} _{n-i}, $$
$$ S_{m+r+1}^{(k+1)}=\sum_{i=1}^{m-1}a^{(k+1)}_{i}+ \sum_{i=0}^{r+1}a^{(k+1)} _{n-i}. $$
Proof
(i)
(ii)
(iii)
By Corollary 2.2 we let
where \(1\leq p\leq n-k\), \(-1\leq q\leq k-2\), and \(\sum_{i=0}^{-1}a^{(k)} _{n-i}=0\). Then we have
$$ S_{m+r+1}^{(k+1)}=\sum_{i=1}^{p}a^{(k+1)}_{i}+ \sum_{i=0}^{q}a^{(k+1)} _{n-i}, $$
(17)
$$ p+q+1= m+r+1. $$
(18)
Next, we prove that \(p=m\) or \(p=m-1\).
(1)
If \(p\geq m+1\), then by Lemma (2.3)(iii) we have \(a^{(k+1)}_{m+1}\leq a^{(k+1)}_{n-r}\). So we get
Thus the right-hand side of (17) includes \(a^{(k+1)}_{n-r}\), which means that \(q \geq r\). Therefore
This is a contradiction with (18).
$$ a^{(k+1)}_{p}\leq a^{(k+1)}_{m+1}\leq a^{(k+1)}_{n-r}. $$
$$ p+q+1\geq m+r+2> m+r+1. $$
(2)
If \(1\leq p\leq m-2\), then by (18) we have \(n-q\leq n-r-2\). By Lemma 2.3(iii) we get \(a^{(k+1)}_{n-r-1}\leq a^{(k+1)} _{m-1}\). It follows that
So the right-hand side of (17) must include \(a^{(k+1)}_{m-1}\). Therefore
This is a contradiction with \(1\leq p\leq m-2\).
$$ a^{(k+1)}_{n-q}\leq a^{(k+1)}_{n-r-2}\leq a^{(k+1)}_{n-r-1}\leq a^{(k+1)} _{m-1}. $$
$$ p \geq m-1. $$
Thus \(p=m\) or \(p=m-1\). □
In a similar way as in Corollary 2.5, we can prove the following corollaries.
Corollary 2.6
Let
\(n\geq 4\), \(2\leq k\leq n-2\), \(2\leq m\leq n-k\), and
\(0\leq r \leq k-2\).
(i)
If
\(a^{(k)}_{n}\leq a^{(k)}_{m} \), then
\(S_{m}^{(k+1)}\)
must be one of the following two cases:
or
$$ S_{m}^{(k+1)}= \sum_{i=1}^{m}a^{(k+1)}_{i} $$
$$ S_{m}^{(k+1)}=\sum_{i=1}^{m-1}a^{(k+1)}_{i}+a^{(k+1)}_{n}. $$
(ii)
If
\(a^{(k)}_{n-r-1}\leq a^{(k)}_{m}\leq a^{(k)}_{n-r}\), then
\(S_{m+r+1}^{(k+1)}\)
must be one of the following two cases:
or
$$ S_{m+r+1}^{(k+1)}=\sum_{i=1}^{m}a^{(k+1)}_{i}+ \sum_{i=0}^{r}a^{(k+1)} _{n-i} $$
$$ S_{m+r+1}^{(k+1)}=\sum_{i=1}^{m-1}a^{(k+1)}_{i}+ \sum_{i=0}^{r+1}a^{(k+1)} _{n-i}. $$
Corollary 2.7
Let
\(n\geq 4\), \(2\leq k\leq n-2\), \(1\leq m\leq n-k\), and
\(-1\leq r \leq k-2\), and let
\(\sum_{i=0}^{-1}a^{(k)}_{n-i}=0\). If
then we have:
$$ S_{m+r+1}^{(k)}= \sum _{i=1}^{m}a^{(k)}_{i}+\sum _{i=0}^{r}a^{(k)}_{n-i}, $$
(i)
if
\(m=1\), then
$$ S_{r+2}^{(k+1)}=a^{(k+1)}_{1}+ \sum_{i=0}^{r}a^{(k+1)}_{n-i}; $$
(ii)
if
\(2\leq m\leq n-k\), then
\(S_{m+r+1}^{(k+1)}\)
must be one of the following two cases:
or
$$ S_{m+r+1}^{(k+1)}=\sum_{i=1}^{m}a^{(k+1)}_{i}+ \sum_{i=0}^{r}a^{(k+1)} _{n-i} $$
$$ S_{m+r+1}^{(k+1)}=\sum_{i=1}^{m-1}a^{(k+1)}_{i}+ \sum_{i=0}^{r+1}a^{(k+1)} _{n-i}. $$
Lemma 2.8
Let
\(n\geq 4\), \(2\leq k\leq n-2\), \(1\leq m\leq n-k\), and
\(-1\leq r \leq k-2\), and let
\(\sum_{i=0}^{-1}a^{(k)}_{n-i}=0\). If
then
$$ \textstyle\begin{cases} S_{m+r+1}^{(k)}= \sum_{i=1}^{m}a^{(k)}_{i}+\sum_{i=0}^{r}a^{(k)}_{n-i}, \\ S_{m+r+1}^{(k+1)}= \sum_{i=1}^{m}a^{(k+1)}_{i}+\sum_{i=0}^{r}a^{(k+1)}_{n-i}, \end{cases} $$
(19)
$$ S_{m+r+1}^{(k)}\geq S_{m+r+1}^{(k+1)}. $$
(20)
Proof
By a simple calculation we obtain
By (19) we have
It follows that
So we have
Note that
Thus we can induce that
This means that (20) holds. □
$$ S_{m+r+1}^{(k)}-S_{m+r+1}^{(k+1)}= \frac{1}{k+1} \Biggl( S_{m+r+1}^{(k)}- \sum _{i=k-r}^{k+m}a_{i} \Biggr) . $$
$$ a^{(k)}_{m+1}\leq a^{(k)}_{n-r}. $$
$$ a_{m+1}+a_{m+2}+\cdots +a_{k+m}\leq a_{n-r}+a_{n-r+1}+\cdots +a_{n-r+k-1}. $$
$$ a_{k-r}+a_{k-r+1}+\cdots +a_{k+m}\leq a_{n-r}+a_{n-r+1}+\cdots +a_{n+m}. $$
$$ \textstyle\begin{cases} a_{n-r}+a_{n-r+1}+\cdots +a_{n+m}\leq a_{n-r+j}+a_{n-r+1+j}+\cdots +a _{n+m+j},& 0\leq j\leq r, \\ a_{k-r}+a_{k-r+1}+\cdots +a_{k+m}\leq a_{n-r+j}+a_{n-r+1+j}+\cdots +a _{n+m+j}, &r+1\leq j\leq k-1. \end{cases} $$
$$ S_{m+r+1}^{(k)}= \sum _{i=1}^{m}a^{(k)}_{i}+\sum _{i=0}^{r}a^{(k)}_{n-i}= \frac{1}{k}\sum_{i=n-r}^{n+m}\sum _{j=0}^{k-1} a_{j+i} = \frac{1}{k} \sum_{j=0}^{k-1}\sum _{i=n-r}^{n+m}a_{j+i} \geq \sum _{i=k-r}^{k+m}a _{i}. $$
Lemma 2.9
Let
\(n\geq 4\), \(2\leq k\leq n-2\), \(2\leq m\leq n-k\), and
\(-1\leq r \leq k-2\), and let
\(\sum_{i=0}^{-1}a^{(k)}_{n-i}=0\). If
then
$$ \textstyle\begin{cases} S_{m+r+1}^{(k)}= \sum_{i=1}^{m}a^{(k)}_{i}+\sum_{i=0}^{r}a^{(k)}_{n-i}, \\ S_{m+r+1}^{(k+1)}= \sum_{i=1}^{m-1}a^{(k+1)}_{i}+\sum_{i=0}^{r+1}a^{(k+1)}_{n-i}, \end{cases} $$
(21)
$$ S_{m+r+1}^{(k)}\geq S_{m+r+1}^{(k+1)}. $$
(22)
Proof
Note that
By (21) we have
It follows that
So we can induce that
Since
we have
This means that (22) holds. □
$$\begin{aligned} &S_{m+r+1}^{(k)}-S_{m+r+1}^{(k+1)} \\ &\quad = \frac{1}{k+1} \Biggl( \sum _{i=1}^{m-1}a^{(k)}_{i}+ \sum _{i=0}^{r}a^{(k)}_{n-i} \Biggr) +a^{(k)}_{m}-a^{(k+1)}_{n-r-1}- \frac{1}{k+1} \sum _{i=k-r}^{k+m-1}a_{i} \\ &\quad = \frac{1}{k+1} \Biggl( \sum _{i=1}^{m}a^{(k)}_{i}+ \sum _{i=0}^{r}a^{(k)}_{n-i}- \sum _{i=n-r-1}^{n+m-1}a_{i} \Biggr) \\ &\quad =\frac{1}{k+1} \Biggl( S_{m+r+1}^{(k)}-\sum _{i=n-r-1}^{n+m-1}a _{i} \Biggr) . \end{aligned}$$
$$ a^{(k)}_{n-r-1}\leq a^{(k)}_{m}. $$
$$ a_{n-r-1}+a_{n-r}+\cdots +a_{n-r+k-2}\leq a_{m}+a_{m+1}+\cdots +a_{m+k-1}. $$
$$ a_{n-r-1}+a_{n-r}+\cdots +a_{n+m-1}\leq a_{k-r-1}+a_{k-r}+\cdots +a _{k+m-1}. $$
$$ \textstyle\begin{cases} a_{n-r-1}+a_{n-r}+\cdots +a_{n+m-1}\leq a_{n-r+j}+a_{n-r+1+j}+\cdots +a_{n+m+j}, &0\leq j\leq r, \\ a_{k-r-1}+a_{k-r}+\cdots +a_{k+m-1}\leq a_{n-r+j}+a_{n-r+1+j}+\cdots +a_{n+m+j}, &r+1\leq j\leq k-1, \end{cases} $$
$$ S_{m+r+1}^{(k)} =\frac{1}{k} \sum _{j=0}^{k-1}\sum_{i=n-r}^{n+m}a_{j+i} \geq \sum_{i=n-r-1}^{n+m-1}a_{i}. $$
3 Main results
We are now in a position to prove our main results (1) in two cases: \(k=1\) and \(2 \leq k\leq n-1\).
Theorem 3.1
For any
\(n\geq 2\), we have
$$ a^{(2)}\prec a^{(1)}=a. $$
(23)
Proof
It is clear that (23) holds if \(n=2\). Next, let \(n\geq 3\). Then we have
$$ \frac{a_{1}+a_{2}}{2}= S_{1}^{(2)}\leq S_{1}^{(1)}=a_{1}, \qquad S_{n}^{(2)}=S _{n}^{(1)}. $$
For \(2\leq m\leq n-1\), we prove that \(S_{m}^{(2)}\leq S_{m}^{(1)}\) in the following two cases: □
(i)
If \(S_{m}^{(2)}= \sum_{i=1}^{m}a^{(2)}_{i}\), then
$$ S_{m}^{(2)}-S_{m}^{(1)}= \frac{a_{m+1}-a_{1}}{2}\leq 0. $$
(ii)
If \(S_{m}^{(2)}= a^{(2)}_{n}+\sum_{i=1}^{m-1}a^{(2)}_{i}\), then
So (23) holds.
$$ S_{m}^{(2)}-S_{m}^{(1)}= \frac{a_{n}-a_{m}}{2}\leq 0. $$
Theorem 3.2
For any
\(n\geq 3\)
and
\(2\leq k\leq n-1\), we have
$$ a^{(k+1)}\prec a^{(k)}. $$
(24)
Proof
Next, let \(n\geq 4\) and \(2\leq k\leq n-2\).
For any \(1\leq m\leq n-k\) and \(-1\leq r\leq k-2\), let \(\sum_{i=0}^{-1}a ^{(k)}_{n-i}=0\), and let
Next, we prove that
in the following two cases: □
$$ S_{m+r+1}^{(k)}= \sum_{i=1}^{m}a^{(k)}_{i}+ \sum_{i=0}^{r}a^{(k)}_{n-i}. $$
$$ S_{m+r+1}^{(k)}\geq S_{m+r+1}^{(k+1)} $$
(i)
4 Discussion
In the theory of majorizations, there are two key concepts, majorizing relations and Schur-convex functions. Majorizing relations are weaker ordered relations among vectors, and Shur-convex functions are an extension of classical convex functions. Combining these two objects is an effective method of constructing inequalities.
In the theory of majorization, there are two important and fundamental objects, establishing majorizing relations among vectors and finding various Schur-convex functions. Majorizing relations deeply characterize intrinsic connections among vectors, and combining a new majorizing relation with suitable Schur-convex functions can lead to various interesting inequalities; see [3‐13].
Competing interests
The authors declare that they have no competing interests.
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