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 [1–4]. In recent decades, some generalizations and strengthening of Hardy’s inequality have been obtained in [1–5]. 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 $0<p<1$, then

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

(1.3)

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

$$D_i=\{(x_1,x_2,\dots ,x_{n-1},c)|\underset{1\le k\le n-1}{min}\{x_k\}\ge c,x_i=\underset{1\le k\le n-1}{min}\{x_k\}\ne c\},i=1,2,\dots ,n-1.$$

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

$$f(y_1,y_2,\dots ,y_{n-1},c)\ge f(c,c,\dots ,c,c),$$

where $y_i\in [c,b]$ ($i=1,2,\dots ,n-1$).

Theorem B [[7], Cor. 1.3]

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

$$D_i=\{(x_1,x_2,\dots ,x_n)|a\le \underset{1\le k\le n}{min}\{x_k\}<x_i=\underset{1\le k\le n}{max}\{x_k\}\le b\},i=1,2,\dots ,n.$$

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

$$f(x_1,x_2,\dots ,x_n)\ge f(x_{min},x_{min},\dots ,x_{min}),$$

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

Definition 1 [1]

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

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

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

Lemma 1 (Hermite-Hadamard’s inequality)

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

$$\varphi \left(\frac{a+b}{2}\right)\le (\ge )\frac{1}{b-a}{\int }_a^b\varphi (x)dx\le (\ge )\frac{\varphi (a)+\varphi (b)}{2},$$

(2.1)

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

Lemma 2 Suppose that $p>0$.

• (1) If $p\ge 1$, or $0<p\le \frac{1}{2}$, then

  $$2p^p\ge 2^p;$$

  (2.2)

• (2) If $\frac{1}{2}<p<1$, then

  $$2^p>2p^p;$$

  (2.3)

• (3) If $\frac{1}{2}<p<1$, then

  $$p^p+(p-1)2^p>0.$$

  (2.4)

Proof Set $f_1:p\in (0,+\mathrm{\infty })\to plnp+ln2-pln2$. Then we have

$${f_1}^{\prime }(p)=lnp+1-ln2=ln\left(\frac{ep}{2}\right).$$

Obviously, $f_1$ is monotone decreasing for $p\in (0,\frac{2}{e})$, $f_1$ is monotone increasing for $p\in (\frac{2}{e},+\mathrm{\infty })$, and $f_1(1)=0$, $f_1(\frac{1}{2})=-\frac{1}{2}ln2+ln2-\frac{1}{2}ln2=0$, then (2.2) and (2.3) hold. Let

$$f_2:p\in (0,1)\to plnp-pln2-ln(1-p).$$

We get

and

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

Then h is concave for $p\in (0,1)$. Because $h(\frac{1}{2})>0$ and ${lim}_{p\to 1^{-}}h(p)>0$, then $h(p)>0$ and $f_2^{\mathrm{\prime }}(p)>0$ hold for $p\in [\frac{1}{2},1)$. From $f_2(\frac{1}{2})=0$, we have $f_2(p)>0$ for $p\in (\frac{1}{2},1)$. Inequality (2.4) is proved. □

Lemma 3

• (1) If $p>1$, then the equation

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

  (2.5)

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

• (2) If $\frac{1}{2}<p<1$, then the equation

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

  (2.6)

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

Proof

• (1) Let $g_1:x\in [0,\frac{1}{2}]\to p^p(1-x){(\frac{1}{2}-x)}^{p-1}-1$. Then $g_1$ is monotone decreasing. According to inequality (2.2), we have

  $$g_1(0)=p^p{\left(\frac{1}{2}\right)}^{p-1}-1={\left(\frac{1}{2}\right)}^{p-1}[p^p-2^{p-1}]>0$$

and $g_1(\frac{1}{2})=-1$. So, equation (2.5) has only a positive root for $x\in (0,\frac{1}{2})$.

• (2) Let $g_2:x\in [0,\frac{1}{2}]\to p^p(1+x)-{(\frac{1}{2}+x)}^{1-p}$. Thus,

  $$g_2^{\prime }(x)=p^p-(1-p){(\frac{1}{2}+x)}^{-p}>{(\frac{1}{2}+x)}^{-p}[\frac{1}{2^p}\cdot p^p-(1-p)].$$

By inequality (2.4), $g_2$ is monotone increasing. According to inequality (2.3), we get

$$\underset{x\to 0^{+}}{lim}{g}_2(x)=p^p-\frac{1}{2^{1-p}}<0$$

and

$$\underset{x\to {(1/2)}^{+}}{lim}{g}_2(x)=\frac{3}{2}{p}^p-1\ge \frac{3}{2}\cdot {\left(\frac{1}{2}\right)}^{\frac{1}{2}}-1>0.$$

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

 □

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

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

(2.7)

and

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

(2.8)

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

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

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

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

So,

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

(2.9)

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

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

(2.10)

inequality (2.7) holds.

• (2)
  

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

Since

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

inequality (2.8) holds. □

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

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

(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

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

and

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

Then inequality (2.11) holds. □

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

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

(3.1)

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

(3.2)

where $B_m={min}_{1\le n\le N}\{b_n\}$. Let

$$f:b=(b_1,b_2,\dots ,b_m)\in {[0,+\mathrm{\infty })}^m\to p^p\sum _{n=1}^m\frac{b_n^p}{n-c}-\sum _{n=1}^m{\left(\sum _{k=n}^m\frac{b_k}{{(k-c)}^{1+1/p}}\right)}^p$$

and

$$D_i=\{(b_1,b_2,\dots ,b_n)|0\le \underset{1\le n\le m}{min}\{b_n\}<b_i=\underset{1\le n\le m}{max}\{b_n\}\}.$$

If $(b_1,b_2,\dots ,b_n)\in D_i$, then

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

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

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

(3.3)

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

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

 □

Corollary 2 If $p>1$, $a_n>0$ ($n=1,2,\dots$), ${\sum }_{n=1}^{\mathrm{\infty }}{a}_n^p<+\mathrm{\infty }$ and $c\in (0,\frac{1}{2})$ 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 ${\{{(n-c)}^{1/p}{a}_n\}}_{n=1}^{\mathrm{\infty }}$ is zero. Then there exists a sequence such that ${\{{(m_i-c)}^{1/p}{a}_{m_i}\}}_{i=1}^{\mathrm{\infty }}$ decrease to zero. Since (3.3), we have

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

(3.5)

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

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

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 ($1/2<p<1$)

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

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

(4.1)

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

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

(4.2)

where $B_m={min}_{1\le n\le m}\{b_n\}$. Set

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

and $D_i=\{(b_1,b_2,\dots ,b_n)|0\le {min}_{1\le n\le m}\{b_n\}<b_i={max}_{1\le n\le m}\{b_n\}\}$. If $(b_1,b_2,\dots ,b_n)\in D_i$, then

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

By Lemma 1, we have

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

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

$$\frac{\partial f}{\partial b_1}>\frac{p{b}_1^{p-1}}{{(1+d)}^{1+1/p}}[p^{-(1-p)}{(\frac{1}{2}+d)}^{(1-p)/p}-p^p{(1+d)}^{1/p}]=0.$$

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

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

Thus, for every $D_i$, $\frac{\partial f}{\partial b_i}>0$. By Theorem B, inequality (4.2) holds. □

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

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

(4.3)

Proof From Theorem 2 and Lemma 5, we have

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

Then inequality (4.3) holds. □

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

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

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

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