# On strengthened form of Copson’s inequality

## Abstract

In this paper, the famous Copson inequality has been improved. We obtain some new results by a different method.

MSC:26D15, 25D05.

## 1 Introduction

Suppose that ${a}_{k}>0$, $p>1$, then we obtain the following Hardy inequality:

${\left(\frac{p}{p-1}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{p}>\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{p}.$
(1.1)

Hardy’s inequality plays an important role in the field of analysis; see [14]. In recent decades, some generalizations and strengthening of Hardy’s inequality have been obtained in [15]. We list some previous results as follows.

Copson’s inequality [4, 5]

Suppose that ${a}_{n}>0$ ($n=1,2,3\dots$). If $p>1$, then

$\sum _{n=1}^{\mathrm{\infty }}{\left(\sum _{k=n}^{\mathrm{\infty }}\frac{{a}_{k}}{k}\right)}^{p}<{p}^{p}\sum _{n=1}^{\mathrm{\infty }}{{a}_{n}}^{p}.$
(1.2)

If $0, then

$\sum _{n=1}^{\mathrm{\infty }}{\left(\sum _{k=n}^{\mathrm{\infty }}\frac{{a}_{k}}{k}\right)}^{p}>{p}^{p}\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{p}.$
(1.3)

And in [4] and [6], the authors pay much attention to the generalization of Copson’s inequality.

In this paper, inequalities (1.2) and (1.3) were strengthened by using a new method.

## 2 Relational lemmas and definitions

In this section, some relational lemmas and definitions will be introduced.

Theorem A [[7], Th. 1.1]

Suppose that $a,b\in \mathbb{R}$, $a, $c\in \left[a,b\right]$, and $f:{\left[a,b\right]}^{n}\to \mathbb{R}$ has continuous partial derivatives, and

${D}_{i}=\left\{\left({x}_{1},{x}_{2},\dots ,{x}_{n-1},c\right)|\underset{1\le k\le n-1}{min}\left\{{x}_{k}\right\}\ge c,{x}_{i}=\underset{1\le k\le n-1}{min}\left\{{x}_{k}\right\}\ne c\right\},\phantom{\rule{1em}{0ex}}i=1,2,\dots ,n-1.$

If $\frac{\partial f\left(x\right)}{\partial {x}_{i}}>0$ holds for all $x\in {D}_{i}$ ($i=1,2,\dots ,n-1$), then

$f\left({y}_{1},{y}_{2},\dots ,{y}_{n-1},c\right)\ge f\left(c,c,\dots ,c,c\right),$

where ${y}_{i}\in \left[c,b\right]$ ($i=1,2,\dots ,n-1$).

Theorem B [[7], Cor. 1.3]

Suppose that $a,b\in \mathbb{R}$, $a, and $f:{\left[a,b\right]}^{n}\to \mathbb{R}$ has continuous partial derivatives and

${D}_{i}=\left\{\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right)|a\le \underset{1\le k\le n}{min}\left\{{x}_{k}\right\}<{x}_{i}=\underset{1\le k\le n}{max}\left\{{x}_{k}\right\}\le b\right\},\phantom{\rule{1em}{0ex}}i=1,2,\dots ,n.$

If $\frac{\partial f\left(x\right)}{\partial {x}_{i}}>0$ holds for all $x\in {D}_{i}$, where $i=1,2,\dots ,n$, then

$f\left({x}_{1},{x}_{2},\dots ,{x}_{n}\right)\ge f\left({x}_{min},{x}_{min},\dots ,{x}_{min}\right),$

where ${x}_{i}\in \left[a,b\right]$ ($i=1,2,\dots ,n$), ${x}_{min}={min}_{1\le k\le n}\left\{{x}_{k}\right\}$.

Definition 1 [1]

Let $G\subseteq {\mathbb{R}}^{n}$ be a convex set, $\varphi :H\to \mathbb{R}$ be a continuous function. If

$\varphi \left(\alpha x+\left(1-\alpha \right)y\right)\le \left(\ge \right)\alpha \varphi \left(x\right)+\left(1-\alpha \right)\varphi \left(y\right)$

holds for all $x,y\in G$, $\alpha \in \left[0,1\right]$, then the function φ is convex (concave).

Let $\varphi :\left[a,b\right]\to \mathbb{R}$ be a convex (concave) function. Then

$\varphi \left(\frac{a+b}{2}\right)\le \left(\ge \right)\frac{1}{b-a}{\int }_{a}^{b}\varphi \left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le \left(\ge \right)\frac{\varphi \left(a\right)+\varphi \left(b\right)}{2},$
(2.1)

and the equality holds if and only if ϕ is linear.

Lemma 2 Suppose that $p>0$.

1. (1)

If $p\ge 1$, or $0, then

$2{p}^{p}\ge {2}^{p};$
(2.2)
2. (2)

If $\frac{1}{2}, then

${2}^{p}>2{p}^{p};$
(2.3)
3. (3)

If $\frac{1}{2}, then

${p}^{p}+\left(p-1\right){2}^{p}>0.$
(2.4)

Proof Set ${f}_{1}:p\in \left(0,+\mathrm{\infty }\right)\to plnp+ln2-pln2$. Then we have

${{f}_{1}}^{\prime }\left(p\right)=lnp+1-ln2=ln\left(\frac{ep}{2}\right).$

Obviously, ${f}_{1}$ is monotone decreasing for $p\in \left(0,\frac{2}{e}\right)$, ${f}_{1}$ is monotone increasing for $p\in \left(\frac{2}{e},+\mathrm{\infty }\right)$, and ${f}_{1}\left(1\right)=0$, ${f}_{1}\left(\frac{1}{2}\right)=-\frac{1}{2}ln2+ln2-\frac{1}{2}ln2=0$, then (2.2) and (2.3) hold. Let

${f}_{2}:p\in \left(0,1\right)\to plnp-pln2-ln\left(1-p\right).$

We get

and

${h}^{″}\left(p\right)=-\frac{1}{p}-\frac{1}{{p}^{2}}<0.$

Then h is concave for $p\in \left(0,1\right)$. Because $h\left(\frac{1}{2}\right)>0$ and ${lim}_{p\to {1}^{-}}h\left(p\right)>0$, then $h\left(p\right)>0$ and ${f}_{2}^{\mathrm{\prime }}\left(p\right)>0$ hold for $p\in \left[\frac{1}{2},1\right)$. From ${f}_{2}\left(\frac{1}{2}\right)=0$, we have ${f}_{2}\left(p\right)>0$ for $p\in \left(\frac{1}{2},1\right)$. Inequality (2.4) is proved. □

Lemma 3

1. (1)

If $p>1$, then the equation

${p}^{p}\left(1-x\right){\left(\frac{1}{2}-x\right)}^{p-1}=1$
(2.5)

has only a positive root for $x\in \left(0,\frac{1}{2}\right)$.

1. (2)

If $\frac{1}{2}, then the equation

${p}^{p}\left(1+x\right)={\left(\frac{1}{2}+x\right)}^{1-p}$
(2.6)

has only a positive root for $x\in \left(0,\frac{1}{2}\right)$.

Proof

1. (1)

Let ${g}_{1}:x\in \left[0,\frac{1}{2}\right]\to {p}^{p}\left(1-x\right){\left(\frac{1}{2}-x\right)}^{p-1}-1$. Then ${g}_{1}$ is monotone decreasing. According to inequality (2.2), we have

${g}_{1}\left(0\right)={p}^{p}{\left(\frac{1}{2}\right)}^{p-1}-1={\left(\frac{1}{2}\right)}^{p-1}\left[{p}^{p}-{2}^{p-1}\right]>0$

and ${g}_{1}\left(\frac{1}{2}\right)=-1$. So, equation (2.5) has only a positive root for $x\in \left(0,\frac{1}{2}\right)$.

1. (2)

Let ${g}_{2}:x\in \left[0,\frac{1}{2}\right]\to {p}^{p}\left(1+x\right)-{\left(\frac{1}{2}+x\right)}^{1-p}$. Thus,

${g}_{2}^{\prime }\left(x\right)={p}^{p}-\left(1-p\right){\left(\frac{1}{2}+x\right)}^{-p}>{\left(\frac{1}{2}+x\right)}^{-p}\left[\frac{1}{{2}^{p}}\cdot {p}^{p}-\left(1-p\right)\right].$

By inequality (2.4), ${g}_{2}$ is monotone increasing. According to inequality (2.3), we get

$\underset{x\to {0}^{+}}{lim}{g}_{2}\left(x\right)={p}^{p}-\frac{1}{{2}^{1-p}}<0$

and

$\underset{x\to {\left(1/2\right)}^{+}}{lim}{g}_{2}\left(x\right)=\frac{3}{2}{p}^{p}-1\ge \frac{3}{2}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}-1>0.$

Therefore, equation (2.6) has only a positive root for $x\in \left(0,\frac{1}{2}\right)$.

□

Lemma 4 If $p>1$, $m>0$, $m\in \mathrm{N}$ and $c\in \left(0,\frac{1}{2}\right)$ is the only one positive root of equation (2.5), then

$\sum _{n=1}^{m}{\left[\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{\left(k-c\right)}^{1+\frac{1}{p}}}\right]}^{p-1}<{p}^{p}{\left(m-c\right)}^{\frac{1}{p}}$
(2.7)

and

$\sum _{n=1}^{m}{\left[\sum _{k=n}^{m}\frac{1}{{\left(k-c\right)}^{1+1/p}}\right]}^{p}<{p}^{p}\sum _{n=1}^{m}\frac{{n}^{p}}{{\left(n-c\right)}^{p+1}}.$
(2.8)

Proof (1) If $m=1$, by Lemma 1, we get

$\begin{array}{rcl}\sum _{n=1}^{m}{\left[\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{\left(k-c\right)}^{1+1/p}}\right]}^{p-1}& =& {\left[\sum _{k=1}^{\mathrm{\infty }}\frac{1}{{\left(k-c\right)}^{1+1/p}}\right]}^{p-1}\\ <& {\left[{\int }_{\frac{1}{2}}^{\mathrm{\infty }}\frac{1}{{\left(x-c\right)}^{1+1/p}}\phantom{\rule{0.2em}{0ex}}dx\right]}^{p-1}={p}^{p-1}{\left(\frac{1}{2}-c\right)}^{-\left(p-1\right)/p}.\end{array}$

If $m\ge 2$, by Lemma 1, we get

$\begin{array}{rcl}\sum _{n=1}^{m}{\left[\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{\left(k-c\right)}^{1+1/p}}\right]}^{p-1}& <& \sum _{n=1}^{m}{\left[{\int }_{n-\frac{1}{2}}^{\mathrm{\infty }}\frac{1}{{\left(x-c\right)}^{1+1/p}}\right]}^{p-1}\\ =& {p}^{p-1}\sum _{n=1}^{m}{\left(n-\frac{1}{2}-c\right)}^{-\left(p-1\right)/p}\\ =& {p}^{p-1}\left[{\left(\frac{1}{2}-c\right)}^{-\left(p-1\right)/p}+\sum _{n=2}^{m}{\left(n-\frac{1}{2}-c\right)}^{-\left(p-1\right)/p}\right]\\ <& {p}^{p-1}\left[{\left(\frac{1}{2}-c\right)}^{-\left(p-1\right)/p}+{\int }_{\frac{3}{2}}^{m+\frac{1}{2}}{\left(x-\frac{1}{2}-c\right)}^{-\left(p-1\right)/p}\phantom{\rule{0.2em}{0ex}}dx\right]\\ =& {p}^{p-1}\left[{\left(\frac{1}{2}-c\right)}^{-\left(p-1\right)/p}+p{\left(m-c\right)}^{1/p}-p{\left(1-c\right)}^{1/p}\right].\end{array}$

So,

$\sum _{n=1}^{m}{\left[\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{\left(k-c\right)}^{1+1/p}}\right]}^{p-1}<{p}^{p-1}\left[{\left(\frac{1}{2}-c\right)}^{-\left(p-1\right)/p}+p{\left(m-c\right)}^{1/p}-p{\left(1-c\right)}^{1/p}\right]$
(2.9)

holds for every $m>0$ and $m\in \mathrm{N}$. Since inequalities (2.9), (2.10) and

${\left(\frac{1}{2}-c\right)}^{-\left(p-1\right)/p}=p{\left(1-c\right)}^{1/p},$
(2.10)

inequality (2.7) holds.

1. (2)

Let $q>1$ and $1/p+1/q=1$. Using Hölder’s inequality, we have

Since

${\left[\sum _{n=1}^{m}{\left(\sum _{k=n}^{m}\frac{1}{{\left(k-c\right)}^{1+1/p}}\right)}^{p}\right]}^{1/p}

inequality (2.8) holds. □

Lemma 5 If $\frac{1}{2}, $m>0$, $m\in \mathrm{N}$ and $d\in \left(0,\frac{1}{2}\right)$ is the only one positive root of equation (2.6), then

$\sum _{n=1}^{m}{\left[\sum _{k=n}^{m}\frac{1}{{\left(k+d\right)}^{1+1/p}}\right]}^{p}>{p}^{p}\sum _{n=1}^{m}\frac{{n}^{p}}{{\left(n+d\right)}^{p+1}}.$
(2.11)

Proof

(2.12)

By Hölder’s inequality, we have

And by using inequality (2.12), we obtain

(2.13)

From inequality (2.12) and inequality (2.13), we get

$\sum _{n=1}^{m}{\left(\sum _{k=n}^{m}\frac{1}{{\left(k+d\right)}^{1+1/p}}\right)}^{p}>p\frac{{\left[{\sum }_{n=1}^{m}\frac{{n}^{p}}{{\left(n+d\right)}^{p+1}}\right]}^{1/p}}{{\left\{{\sum }_{n=1}^{m}{\left[{\sum }_{k=n}^{m}\frac{1}{{\left(k+d\right)}^{1+1/p}}\right]}^{p}\right\}}^{\frac{1-p}{p}}}$

and

${\left\{\sum _{n=1}^{m}{\left[\sum _{k=n}^{m}\frac{1}{{\left(k+d\right)}^{1+1/p}}\right]}^{p}\right\}}^{1/p}>p{\left[\sum _{n=1}^{m}\frac{{n}^{p}}{{\left(n+d\right)}^{p+1}}\right]}^{1/p}.$

Then inequality (2.11) holds. □

## 3 Strengthened Copson’s inequality ($p>1$)

Theorem 1 Assume that $p>1$, $m>0$, $m\in \mathrm{N}$, ${a}_{n}>0$ ($n=1,2,\dots ,m$), $c\in \left(0,\frac{1}{2}\right)$ is the only one positive root of equation (2.5) and ${B}_{m}={min}_{1\le n\le m}\left\{{\left(n-c\right)}^{1/p}{a}_{n}\right\}$. Then

(3.1)

Proof Set ${b}_{n}={\left(n-c\right)}^{1/p}{a}_{n}$ ($n=1,2,\dots ,m$). Then inequality (3.1) is equivalent to

(3.2)

where ${B}_{m}={min}_{1\le n\le N}\left\{{b}_{n}\right\}$. Let

$f:b=\left({b}_{1},{b}_{2},\dots ,{b}_{m}\right)\in {\left[0,+\mathrm{\infty }\right)}^{m}\to {p}^{p}\sum _{n=1}^{m}\frac{{b}_{n}^{p}}{n-c}-\sum _{n=1}^{m}{\left(\sum _{k=n}^{m}\frac{{b}_{k}}{{\left(k-c\right)}^{1+1/p}}\right)}^{p}$

and

${D}_{i}=\left\{\left({b}_{1},{b}_{2},\dots ,{b}_{n}\right)|0\le \underset{1\le n\le m}{min}\left\{{b}_{n}\right\}<{b}_{i}=\underset{1\le n\le m}{max}\left\{{b}_{n}\right\}\right\}.$

If $\left({b}_{1},{b}_{2},\dots ,{b}_{n}\right)\in {D}_{i}$, then

$\begin{array}{rcl}\frac{\partial f}{\partial {b}_{i}}& =& {p}^{p}\frac{p{b}_{i}^{p-1}}{i-c}-\frac{p}{{\left(i-c\right)}^{1+1/p}}\sum _{n=1}^{i}{\left(\sum _{k=n}^{m}\frac{{b}_{k}}{{\left(k-c\right)}^{1+1/p}}\right)}^{p-1}\\ >& \frac{p{b}_{i}^{p-1}}{{\left(i-c\right)}^{1+\frac{1}{p}}}\left[{p}^{p}{\left(i-c\right)}^{1/p}-\sum _{n=1}^{i}{\left(\sum _{k=n}^{m}\frac{1}{{\left(k-c\right)}^{1+1/p}}\right)}^{p-1}\right]\\ >& \frac{p{b}_{i}^{p-1}}{{\left(i-c\right)}^{1+\frac{1}{p}}}\left[{p}^{p}{\left(i-c\right)}^{1/p}-\sum _{n=1}^{i}{\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{\left(k-c\right)}^{1+1/p}}\right)}^{p-1}\right].\end{array}$

By inequality (2.7), we know $\frac{\partial f}{\partial {b}_{i}}>0$. By Theorem B, inequality (3.2) holds, the proof is completed. □

Corollary 1 If $p>1$, $m>0$, $m\in \mathrm{N}$, ${a}_{n}>0$ ($n=1,2,\dots ,m$), $c\in \left(0,\frac{1}{2}\right)$ is the only one positive root of equation (2.5), and ${B}_{m}={min}_{1\le n\le m}\left\{{\left(n-c\right)}^{1/p}{a}_{n}\right\}$, then

${p}^{p}\sum _{n=1}^{m}{a}_{n}^{p}-\sum _{n=1}^{m}{\left(\sum _{k=n}^{m}\frac{{a}_{k}}{k-c}\right)}^{p}>-{p}^{p}{B}_{m}^{p}\sum _{n=1}^{m}\frac{{n}^{p}-{\left(n-c\right)}^{p}}{{\left(n-c\right)}^{p+1}}.$
(3.3)

Proof By (3.1) and (2.8), we can obtain

$\begin{array}{rcl}{p}^{p}\sum _{n=1}^{m}{a}_{n}^{p}-\sum _{n=1}^{m}{\left(\sum _{k=n}^{m}\frac{{a}_{k}}{k-c}\right)}^{p}& >& {p}^{p}{B}_{m}^{p}\left[\sum _{n=1}^{m}\frac{1}{n-c}-\sum _{n=1}^{m}\frac{{n}^{p}}{{\left(n-c\right)}^{p+1}}\right]\\ =& -{p}^{p}{B}_{m}^{p}\sum _{n=1}^{m}\frac{{n}^{p}-{\left(n-c\right)}^{p}}{{\left(n-c\right)}^{p+1}}.\end{array}$

□

Corollary 2 If $p>1$, ${a}_{n}>0$ ($n=1,2,\dots$), ${\sum }_{n=1}^{\mathrm{\infty }}{a}_{n}^{p}<+\mathrm{\infty }$ and $c\in \left(0,\frac{1}{2}\right)$ is the only one positive root of equation (2.5), then

$\sum _{n=1}^{\mathrm{\infty }}{\left(\sum _{k=n}^{\mathrm{\infty }}\frac{{a}_{k}}{k-c}\right)}^{p}\le {p}^{p}\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{p}.$
(3.4)

Proof Because of ${\sum }_{n=1}^{\mathrm{\infty }}{a}_{n}^{p}<+\mathrm{\infty }$, the infimum of ${\left\{{\left(n-c\right)}^{1/p}{a}_{n}\right\}}_{n=1}^{\mathrm{\infty }}$ is zero. Then there exists a sequence such that ${\left\{{\left({m}_{i}-c\right)}^{1/p}{a}_{{m}_{i}}\right\}}_{i=1}^{\mathrm{\infty }}$ decrease to zero. Since (3.3), we have

${p}^{p}\sum _{n=1}^{{m}_{i}}{a}_{n}^{p}-\sum _{n=1}^{{m}_{i}}{\left(\sum _{k=n}^{{m}_{i}}\frac{{a}_{k}}{k-c}\right)}^{p}>-{p}^{p}{\left[{\left({m}_{i}-c\right)}^{1/p}{a}_{{m}_{i}}\right]}^{p}\sum _{n=1}^{{m}_{i}}\frac{{n}^{p}-{\left(n-c\right)}^{p+1}}{{\left(n-c\right)}^{p+1}}.$
(3.5)

Let $i\to +\mathrm{\infty }$ in inequality (3.5), we have ${m}_{i}\to +\mathrm{\infty }$ and

$\underset{i\to +\mathrm{\infty }}{lim}{\left[{\left({m}_{i}-c\right)}^{1/p}{a}_{{m}_{i}}\right]}^{p}\sum _{n=1}^{{m}_{i}}\frac{{n}^{p}-{\left(n-c\right)}^{p+1}}{{\left(n-c\right)}^{p+1}}=0.$

Then by (3.5), we can obtain

${p}^{p}\sum _{n=1}^{{m}_{i}}{a}_{n}^{p}-\sum _{n=1}^{{m}_{i}}{\left(\sum _{k=n}^{{m}_{i}}\frac{{a}_{k}}{k-c}\right)}^{p}\ge 0.$

Therefore, inequality (3.4) holds. □

Remark Obviously, inequality (3.4) strengthens inequality (1.2).

## 4 Strengthened Copson’s inequality ($1/2)

Theorem 2 If $\frac{1}{2}, $m>0$, $m\in \mathrm{N}$, ${a}_{n}>0$ ($n=1,2,\dots ,m$), $d\in \left(0,\frac{1}{2}\right)$ is the only one positive root of equation (2.6) and ${B}_{m}={min}_{1\le n\le m}\left\{{\left(n+d\right)}^{1/p}{a}_{n}\right\}$. Then

$\sum _{n=1}^{m}{\left(\sum _{k=n}^{m}\frac{{a}_{k}}{k+d}\right)}^{p}-{p}^{p}\sum _{n=1}^{m}{a}_{n}^{p}\ge {B}_{m}^{p}\left[\sum _{n=1}^{m}{\left(\sum _{k=n}^{m}\frac{1}{{\left(k+d\right)}^{1+1/p}}\right)}^{p}-{p}^{p}\sum _{n=1}^{m}\frac{1}{n+d}\right].$
(4.1)

Proof Let ${b}_{n}={\left(n+d\right)}^{1/p}{a}_{n}$ ($n=1,2,\dots ,m$). Then inequality (4.1) is equivalent to

$\sum _{n=1}^{m}{\left(\sum _{k=n}^{m}\frac{{b}_{k}}{{\left(k+d\right)}^{1+1/p}}\right)}^{p}-{p}^{p}\sum _{n=1}^{m}\frac{{b}_{n}^{p}}{n+d}\ge {B}_{m}^{p}\left[{p}^{p}\sum _{n=1}^{m}\frac{1}{n}-\sum _{n=1}^{m}{\left(\sum _{k=n}^{m}\frac{1}{{k}^{1+1/p}}\right)}^{p}\right],$
(4.2)

where ${B}_{m}={min}_{1\le n\le m}\left\{{b}_{n}\right\}$. Set

$f:b\in {\left(0,+\mathrm{\infty }\right)}^{m}\to \sum _{n=1}^{m}{\left(\sum _{k=n}^{m}\frac{{b}_{k}}{{\left(k+d\right)}^{1+1/p}}\right)}^{p}-{p}^{p}\sum _{n=1}^{m}\frac{{b}_{n}^{p}}{n+d}$

and ${D}_{i}=\left\{\left({b}_{1},{b}_{2},\dots ,{b}_{n}\right)|0\le {min}_{1\le n\le m}\left\{{b}_{n}\right\}<{b}_{i}={max}_{1\le n\le m}\left\{{b}_{n}\right\}\right\}$. If $\left({b}_{1},{b}_{2},\dots ,{b}_{n}\right)\in {D}_{i}$, then

$\begin{array}{rcl}\frac{\partial f}{\partial {b}_{i}}& =& \frac{p}{{\left(i+d\right)}^{1+1/p}}\sum _{n=1}^{i}{\left(\sum _{k=n}^{m}\frac{{b}_{k}}{{\left(k+d\right)}^{1+1/p}}\right)}^{p-1}-{p}^{p+1}\frac{{b}_{i}^{p-1}}{i+d}\\ =& \frac{p{b}_{i}^{p-1}}{{\left(i+d\right)}^{1+1/p}}\left[\sum _{n=1}^{i}{\left(\sum _{k=n}^{m}\frac{{b}_{k}}{{\left(k+d\right)}^{1+1/p}{b}_{i}}\right)}^{-\left(1-p\right)}-{p}^{p}{\left(i+d\right)}^{1/p}\right]\\ >& \frac{p{b}_{i}^{p-1}}{{\left(i+d\right)}^{1+1/p}}\left[\sum _{n=1}^{i}{\left(\sum _{k=n}^{m}\frac{1}{{\left(k+d\right)}^{1+1/p}}\right)}^{-\left(1-p\right)}-{p}^{p}{\left(i+d\right)}^{1/p}\right]\\ >& \frac{p{b}_{i}^{p-1}}{{\left(i+d\right)}^{1+1/p}}\left[\sum _{n=1}^{i}{\left(\sum _{k=n}^{\mathrm{\infty }}\frac{1}{{\left(k+d\right)}^{1+1/p}}\right)}^{-\left(1-p\right)}-{p}^{p}{\left(i+d\right)}^{1/p}\right].\end{array}$

By Lemma 1, we have

$\begin{array}{rcl}\frac{\partial f}{\partial {b}_{i}}& >& \frac{p{b}_{i}^{p-1}}{{\left(i+d\right)}^{1+1/p}}\left[\sum _{n=1}^{i}{\left({\int }_{n-\frac{1}{2}}^{\mathrm{\infty }}\frac{1}{{\left(x+d\right)}^{1+1/p}}\phantom{\rule{0.2em}{0ex}}dx\right)}^{-\left(1-p\right)}-{p}^{p}{\left(i+d\right)}^{1/p}\right]\\ =& \frac{p{b}_{i}^{p-1}}{{\left(i+d\right)}^{1+1/p}}\left[{p}^{-\left(1-p\right)}\sum _{n=1}^{i}{\left(n-\frac{1}{2}+d\right)}^{\left(1-p\right)/p}-{p}^{p}{\left(i+d\right)}^{1/p}\right].\end{array}$

As $i=1$, by the definition of d, we have

$\frac{\partial f}{\partial {b}_{1}}>\frac{p{b}_{1}^{p-1}}{{\left(1+d\right)}^{1+1/p}}\left[{p}^{-\left(1-p\right)}{\left(\frac{1}{2}+d\right)}^{\left(1-p\right)/p}-{p}^{p}{\left(1+d\right)}^{1/p}\right]=0.$

As $2\le i\le m$, because $\frac{1}{2}, $0<\frac{p-1}{p}\le 1$ and $g:x\in \left(0,+\mathrm{\infty }\right)\to {x}^{\left(1-p\right)/p}$ is concave, we have

$\begin{array}{rcl}\frac{\partial f}{\partial {b}_{i}}& >& \frac{p{b}_{i}^{p-1}}{{\left(i+d\right)}^{1+1/p}}\left[{p}^{-\left(1-p\right)}\left({\left(\frac{1}{2}+d\right)}^{\left(1-p\right)/p}+\sum _{n=2}^{i}{\left(n-\frac{1}{2}+d\right)}^{\left(1-p\right)/p}\right)-{p}^{p}{\left(i+d\right)}^{1/p}\right]\\ >& \frac{p{b}_{i}^{p-1}}{{\left(i+d\right)}^{1+1/p}}\left[{p}^{-\left(1-p\right)}\left({\left(\frac{1}{2}+d\right)}^{\left(1-p\right)/p}\\ +{\int }_{\frac{3}{2}}^{i+\frac{1}{2}}{\left(x-\frac{1}{2}+d\right)}^{\left(1-p\right)/p}\phantom{\rule{0.2em}{0ex}}dx\right)-{p}^{p}{\left(i+d\right)}^{1/p}\right]\\ =& \frac{p{b}_{i}^{p-1}}{{\left(i+d\right)}^{1+1/p}}\left[{p}^{-\left(1-p\right)}\left({\left(\frac{1}{2}+d\right)}^{\left(1-p\right)/p}+p{\left(i+d\right)}^{1/p}-p{\left(1+d\right)}^{1/p}\right)-{p}^{p}{\left(i+d\right)}^{1/p}\right]\\ =& \frac{p{b}_{i}^{p-1}}{{\left(i+d\right)}^{1+1/p}}\left[{p}^{-\left(1-p\right)}\cdot p{\left(i+d\right)}^{1/p}-{p}^{p}{\left(i+d\right)}^{1/p}\right]=0.\end{array}$

Thus, for every ${D}_{i}$, $\frac{\partial f}{\partial {b}_{i}}>0$. By Theorem B, inequality (4.2) holds. □

Corollary 3 If $\frac{1}{2}, $m>0$, $m\in \mathrm{N}$, ${a}_{n}>0$ ($n=1,2,\dots ,m$), $d\in \left(0,\frac{1}{2}\right)$ is the only one positive root of equation (2.6) and ${B}_{m}={min}_{1\le n\le m}\left\{{\left(n+d\right)}^{1/p}{a}_{n}\right\}$. Then

$\sum _{n=1}^{m}{\left(\sum _{k=n}^{m}\frac{{a}_{k}}{k+d}\right)}^{p}-{p}^{p}\sum _{n=1}^{m}{a}_{n}^{p}\ge {p}^{p}{B}_{m}^{p}\sum _{n=1}^{m}\frac{{n}^{p}-{\left(n+d\right)}^{p}}{{\left(n+d\right)}^{p+1}}.$
(4.3)

Proof From Theorem 2 and Lemma 5, we have

$\sum _{n=1}^{m}{\left(\sum _{k=n}^{m}\frac{{a}_{k}}{k+d}\right)}^{p}-{p}^{p}\sum _{n=1}^{m}{a}_{n}^{p}\ge {B}_{m}^{p}\left[{p}^{p}\sum _{n=1}^{m}\frac{{n}^{p}}{{\left(n+d\right)}^{p+1}}-{p}^{p}\sum _{n=1}^{m}\frac{1}{n+d}\right].$

Then inequality (4.3) holds. □

Corollary 4 If $\frac{1}{2}, ${a}_{n}>0$ ($n=1,2,\dots$), $d\in \left(0,\frac{1}{2}\right)$ is the only one positive root of equation (2.6) and series ${\sum }_{n=1}^{\mathrm{\infty }}{\left({\sum }_{k=n}^{\mathrm{\infty }}\frac{{a}_{k}}{k+d}\right)}^{p}<+\mathrm{\infty }$. Then

$\sum _{n=1}^{\mathrm{\infty }}{\left(\sum _{k=n}^{\mathrm{\infty }}\frac{{a}_{k}}{k+d}\right)}^{p}\ge {p}^{p}\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{p}.$
(4.4)

Proof According to inequality (4.3), we obtain

$\sum _{n=1}^{m}{\left(\sum _{k=n}^{m}\frac{{a}_{k}}{k+d}\right)}^{p}+{p}^{p}{B}_{m}^{p}\sum _{n=1}^{m}\frac{{\left(n+d\right)}^{p}-{n}^{p}}{{\left(n+d\right)}^{p+1}}\ge {p}^{p}\sum _{n=1}^{m}{a}_{n}^{p}.$

The following proof is the same as the relevant proof for Corollary 2, omitted here. □

Remark For $\frac{1}{2}, there is no doubt that inequality (4.4) strengthens inequality (1.3).

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## Acknowledgements

The research is supported by the Nature Science Foundation of China (No. 110771069) and the NS Foundation of the Educational Committee of Zhejiang Province under Grant Y201223283.

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Correspondence to Qian Xu.

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Xu, Q. On strengthened form of Copson’s inequality. J Inequal Appl 2012, 305 (2012). https://doi.org/10.1186/1029-242X-2012-305