# A note on Hardy’s inequality

## Abstract

In this paper, we prove that the inequality ${\sum }_{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}{\sum }_{k=1}^{n}{a}_{k}\right)}^{p}\le {\left(\frac{p}{p-1}\right)}^{p}{\sum }_{n=1}^{\mathrm{\infty }}\left(1-\frac{d\left(p\right)}{{\left(n-1/2\right)}^{1-1/p}}\right){a}_{n}^{p}$ holds for $p\le -1$ and $d\left(p\right)=\left(1+\left({2}^{-1/p}-1\right)p\right)/\left[8\left(1+\left({2}^{-1/p}-1\right)p\right)+2\right]$ if ${a}_{n}>0$ ($n=1,2,\dots$), and ${\sum }_{n=1}^{\mathrm{\infty }}{a}_{n}^{p}<+\mathrm{\infty }$.

MSC:26D15.

## 1 Introduction

Let $p>1$ and ${a}_{n}>0$ ($n=1,2,\dots$) with ${\sum }_{n=1}^{\mathrm{\infty }}{{a}_{n}}^{p}<+\mathrm{\infty }$, then Hardy’s well-known inequality [1] is given by

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

Recently, the refinement, improvement, generalization, extension, and application for Hardy’s inequality have attracted the attention of many researchers [210].

Yang and Zhu [11] presented an improvement of Hardy’s inequality (1.1) for $p=2$ as follows:

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{2}<4\sum _{n=1}^{\mathrm{\infty }}\left(1-\frac{1}{3\sqrt{n}+5}\right){a}_{n}^{2}.$

For $7/6\le p\le 2$, Huang [12] proved that

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{p}<{\left(\frac{p}{p-1}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left(1-\frac{15}{196\left({n}^{1-1/p}+3\text{,}436\right)}\right){a}_{n}^{p}.$

In [13], Wen and Zhang proved that the inequality

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{p}<{\left(\frac{p}{p-1}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left(1-\frac{{C}_{p}}{2{n}^{1-1/p}}\right){a}_{n}^{p}$
(1.2)

holds for $p>1$ if ${a}_{n}>0$ ($n=1,2,\dots$), with ${\sum }_{n=1}^{\mathrm{\infty }}{a}_{n}^{p}<+\mathrm{\infty }$, where ${C}_{p}=1-{\left(1-1/p\right)}^{p-1}$ for $p\ge 2$ and ${C}_{p}=1-1/p$ for $1.

Xu et al. [14] gave a further improvement of the inequality (1.2):

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{p}<{\left(\frac{p}{p-1}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left(1-\frac{{Z}_{p}}{2{\left(n-1\right)}^{1-1/p}}\right){a}_{n}^{p},$

where ${Z}_{p}=p-1-\frac{{\left(p-1\right)}^{2}}{p}{2}^{1/p}$ for $1 and ${Z}_{p}=1-{\left(\frac{p-1}{p}\right)}^{p-1}{2}^{\frac{p-1}{p}}$ for $p>2$.

For the special parameter $p=5/4$, Deng et al. [15] established

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{5/4}\le {5}^{5/4}\sum _{n=1}^{\mathrm{\infty }}\left(1-\frac{1}{10\left({n}^{1/5}+{\eta }_{5/4}\right)}\right){a}_{n}^{5/4},$

where ${\eta }_{5/4}={5}^{5/4}/\left[10\left({5}^{5/4}-\left({\sum }_{n=1}^{\mathrm{\infty }}\frac{1}{{n}^{5/4}}{\left({\sum }_{m=1}^{n}{m}^{-4/5}\right)}^{1/4}\right)\right)-1\right]=0.46\cdots$ .

In [16], Long and Linh discussed Hardy’s inequality with the parameter $p<0$, and proved that

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

for $p\le -1$ and

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{p}<\frac{{2}^{1-p}}{1-p}\sum _{n=1}^{\mathrm{\infty }}{a}_{n}^{p}$

for $-1 if ${a}_{n}>0$ ($n=1,2,\dots$) with ${\sum }_{n=1}^{\mathrm{\infty }}{a}_{n}^{p}<+\mathrm{\infty }$.

It is the aim of this paper to present an improvement of inequality (1.3) for the parameter $p\le -1$. Our main result is Theorem 1.1.

Theorem 1.1 Let $p\le -1$, $d\left(p\right)=\left(1+\left({2}^{-1/p}-1\right)p\right)/\left[8\left(1+\left({2}^{-1/p}-1\right)p\right)+2\right]$ and ${a}_{n}>0$ ($n=1,2,\dots$) with ${\sum }_{n=1}^{\mathrm{\infty }}{a}_{n}^{p}<+\mathrm{\infty }$, then

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{n}\sum _{k=1}^{n}{a}_{k}\right)}^{p}\le {\left(\frac{p}{p-1}\right)}^{p}\sum _{n=1}^{\mathrm{\infty }}\left(1-\frac{d\left(p\right)}{{\left(n-1/2\right)}^{1-1/p}}\right){a}_{n}^{p}.$

## 2 Lemmas

In order to establish our main result we need several lemmas, which we present in this section.

Lemma 2.1 (see [[17], Corollary 1.3])

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

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

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

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

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

Lemma 2.2 Let $n\in \mathbb{R}$ be a positive natural number and $r\in \mathbb{R}$ with $r\ge 1$. Then

$\sum _{k=1}^{n}{\left(k-\frac{1}{2}\right)}^{1/r}\ge \frac{r}{r+1}\left({n}^{1+1/r}-1\right)+{2}^{-1/r}.$
(2.1)

Proof We use mathematical induction to prove inequality (2.1). We clearly see that inequality (2.1) becomes equality for $n=1$. We assume that inequality (2.1) holds for $n=i$ ($i\in \mathbb{N}$, $i\ge 1$), namely

$\sum _{k=1}^{i}{\left(k-\frac{1}{2}\right)}^{1/r}\ge \frac{r}{r+1}\left({i}^{1+1/r}-1\right)+{2}^{-1/r}.$

Then for $n=i+1$ we have

$\begin{array}{rl}\sum _{k=1}^{i+1}{\left(k-\frac{1}{2}\right)}^{1/r}& =\sum _{k=1}^{i}{\left(k-\frac{1}{2}\right)}^{1/r}+{\left(i+\frac{1}{2}\right)}^{1/r}\\ \ge \frac{r}{r+1}\left({i}^{1+1/r}-1\right)+{2}^{-1/r}+{\left(i+\frac{1}{2}\right)}^{1/r}\\ =\frac{r}{r+1}\left[{\left(i+1\right)}^{1+1/r}-1\right]+{2}^{-1/r}+{\left(i+\frac{1}{2}\right)}^{1/r}-{\int }_{i}^{i+1}{x}^{1/r}\phantom{\rule{0.2em}{0ex}}dx.\end{array}$
(2.2)

Note that ${x}^{1/r}$ ($r\ge 1$) is concave on $\left(0,+\mathrm{\infty }\right)$, therefore Hermite-Hadamard’s inequality implies that

${\left(i+\frac{1}{2}\right)}^{1/r}\ge {\int }_{i}^{i+1}{x}^{1/r}\phantom{\rule{0.2em}{0ex}}dx.$
(2.3)

From (2.2) and (2.3) we know that inequality (2.1) holds for $n=i+1$. □

Remark 2.1 The inequality

${2}^{-1/r}\ge \frac{r}{r+1}$
(2.4)

holds for all $r\ge 1$ with equality if and only if $r=1$.

Proof We clearly see that inequality (2.4) becomes equality for $r=1$.

If $r>1$, then it is well known that the function ${\left(1+1/r\right)}^{r}$ is strictly increasing on $\left(1,+\mathrm{\infty }\right)$, so we get

${\left(1+\frac{1}{r}\right)}^{r}>2.$
(2.5)

Therefore, inequality (2.4) follows from (2.5). □

Lemma 2.3 The inequality

$\frac{\left(3r+1\right){2}^{-1-1/r}}{r+1}>\left({2}^{-1/r}-\frac{r}{r+1}\right)\frac{3{r}^{2}+1}{r}$
(2.6)

holds for all $r\ge 1$.

Proof Let $r\ge 1$, then we clearly see that

$\left(6log2-3\right){r}^{2}+r+2log2-2\ge 4\left(2log2-1\right)>0.$
(2.7)

${e}^{\left(3{r}^{2}-r+2\right)/\left(6{r}^{2}+2\right)}<2.$
(2.8)

It follows from the well-known inequality ${\left(1+x\right)}^{1/x} ($x>0$) that

$e>{\left(1+\frac{3{r}^{2}-r+2}{6{r}^{3}+2r}\right)}^{\left(6{r}^{3}+2r\right)/\left(3{r}^{2}-r+2\right)}.$
(2.9)

From (2.8) and (2.9) we have

${2}^{-1/r}<\frac{6{r}^{3}+2r}{6{r}^{3}+3{r}^{2}+r+2}.$
(2.10)

Therefore, inequality (2.6) follows easily from (2.10). □

Lemma 2.4 Let $r\ge 1$ and

$f\left(x\right)=\frac{{x}^{r}}{{\left(\frac{r}{r+1}{x}^{1+1/r}+{2}^{-1/r}-\frac{r}{r+1}\right)}^{r+1}}.$
(2.11)

Then f is convex on $\left[1/2,+\mathrm{\infty }\right)$.

Proof From (2.11) we have

$\begin{array}{r}{f}^{\prime }\left(x\right)=\frac{-\frac{2r+1}{r+1}{x}^{r+1/r}+\left({2}^{-1/r}-\frac{r}{r+1}\right)r{x}^{r-1}}{{\left(\frac{r}{r+1}{x}^{1+1/r}+{2}^{-1/r}-\frac{r}{r+1}\right)}^{r+2}},\\ {f}^{″}\left(x\right)=\frac{\frac{\left(3r+1\right)\left(2r+1\right)}{{\left(r+1\right)}^{2}}{x}^{2+2/r}-\left({2}^{-1/r}-\frac{r}{r+1}\right)\frac{\left(3{r}^{2}+1\right)\left(2r+1\right)}{r\left(r+1\right)}{x}^{1+1/r}+r\left(r-1\right){\left({2}^{-1/r}-\frac{r}{r+1}\right)}^{2}}{{\left(\frac{r}{r+1}{x}^{1+1/r}+{2}^{-1/r}-\frac{r}{r+1}\right)}^{r+3}}{x}^{r-2}.\end{array}$
(2.12)

It follows from Lemma 2.3 and (2.12) that

${f}^{″}\left(x\right)\ge \frac{\frac{\left(3r+1\right)\left(2r+1\right)}{{\left(r+1\right)}^{2}}{2}^{-1-1/r}-\left({2}^{-1/r}-\frac{r}{r+1}\right)\frac{\left(3{r}^{2}+1\right)\left(2r+1\right)}{r\left(r+1\right)}}{{\left(\frac{r}{r+1}{x}^{1+1/r}+{2}^{-1/r}-\frac{r}{r+1}\right)}^{r+3}}{x}^{r+1/r-1}>0$
(2.13)

for all $x\in \left[1/2,+\mathrm{\infty }\right)$.

Therefore, Lemma 2.4 follows from inequality (2.13). □

Lemma 2.5 Let $r\ge 1$, $0\le t\le 4$ and $c=\left(r+1-{2}^{1/r}r\right)/\left[8\left(r+1-{2}^{1/r}r\right)+2\right]$, then

$\left(r+1\right)\left({2}^{-1/r}-\frac{r}{r+1}\right)\left(1-ct\right)t\ge 1-{\left[1+\left(\frac{1+r}{{2}^{1/r}r}-1\right)t\right]}^{-r}.$
(2.14)

Proof If $r=1$, then we clearly see that inequality (2.14) becomes equality. Next, we assume that $r>1$. Let

$f\left(t\right)=\left(r+1\right)\left({2}^{-1/r}-\frac{r}{r+1}\right)\left(1-ct\right)t-1+{\left[1+\left(\frac{1+r}{{2}^{1/r}r}-1\right)t\right]}^{-r}.$
(2.15)

$f\left(0\right)=0,$
(2.16)
${f}^{\prime }\left(t\right)=\frac{\left(r+1\right)\left({2}^{-1/r}-\frac{r}{r+1}\right)}{{\left[1+\left(\frac{r+1}{{2}^{1/r}r}-1\right)t\right]}^{r+1}}\left\{\left(1-2ct\right){\left[1+\left(\frac{r+1}{{2}^{1/r}r}-1\right)t\right]}^{r+1}-1\right\}.$
(2.17)

Note that

$1-2ct\ge 1-8c=\frac{1}{4\left(r+1-{2}^{1/r}r\right)+1}>0,$
(2.18)
${\left[1+\left(\frac{r+1}{{2}^{1/r}r}-1\right)t\right]}^{r+1}\ge 1+\left(r+1\right)\left(\frac{r+1}{{2}^{1/r}r}-1\right)t.$
(2.19)

It follows from Remark 2.1 and (2.17)-(2.19) that

${f}^{\prime }\left(t\right)\ge \frac{\left(r+1\right)\left({2}^{-1/r}-\frac{r}{r+1}\right)}{{\left[1+\left(\frac{r+1}{{2}^{1/r}r}-1\right)t\right]}^{r+1}}\left\{\left(1-2ct\right)\left[1+\left(r+1\right)\left(\frac{r+1}{{2}^{1/r}r}-1\right)t\right]-1\right\}.$
(2.20)

Let

$g\left(t\right)=\left(1-2ct\right)\left[1+\left(r+1\right)\left(\frac{r+1}{{2}^{1/r}r}-1\right)t\right]-1.$
(2.21)

Then from $g\left(0\right)=0$ and $g\left(4\right)=4{\left(r+1-{2}^{1/r}r\right)}^{2}/\left[{2}^{1/r}r\left(4\left(r+1-{2}^{1/r}r\right)+1\right)\right]\ge 0$ together with the fact that $g\left(t\right)$ is a concave parabola we know that

$g\left(t\right)\ge 0$
(2.22)

for $t\in \left[0,4\right]$.

Therefore, Lemma 2.5 follows easily from (2.15) and (2.16) together with (2.20)-(2.22). □

Lemma 2.6 Let $r\ge 1$, $c=\left(r+1-{2}^{1/r}r\right)/\left[8\left(r+1-{2}^{1/r}r\right)+2\right]$, N is a positive natural number, ${a}_{k}>0$ ($k=1,2,\dots ,N$) and ${B}_{N}={min}_{1\le k\le N}\left\{{\left(k-1/2\right)}^{1/r}{a}_{k}\right\}$, then

$\begin{array}{r}{\left(\frac{r+1}{r}\right)}^{r}\sum _{n=1}^{N}\left(1-\frac{c}{{\left(n-1/2\right)}^{1+1/r}}\right){a}_{n}^{r}-\sum _{n=1}^{N}{\left(\frac{n}{{\sum }_{k=1}^{n}1/{a}_{k}}\right)}^{r}\\ \phantom{\rule{1em}{0ex}}\ge {B}_{N}^{r}\left[{\left(\frac{r+1}{r}\right)}^{r}\sum _{n=1}^{N}\left(1-\frac{c}{{\left(n-1/2\right)}^{1+1/r}}\right)\frac{1}{n-1/2}\\ \phantom{\rule{2em}{0ex}}-\sum _{n=1}^{N}{\left(\frac{n}{{\sum }_{k=1}^{n}{\left(k-1/2\right)}^{1/r}}\right)}^{r}\right].\end{array}$
(2.23)

Proof Let ${a}_{k}={b}_{k}/{\left(k-1/2\right)}^{1/r}$ ($k=1,2,\dots ,N$), then ${B}_{N}={min}_{1\le k\le N}\left\{{b}_{k}\right\}$ and inequality (2.23) becomes

$\begin{array}{r}{\left(\frac{r+1}{r}\right)}^{r}\sum _{n=1}^{N}\left(1-\frac{c}{{\left(n-1/2\right)}^{1+1/r}}\right)\frac{{b}_{n}^{r}}{n-1/2}-\sum _{n=1}^{N}{\left(\frac{n}{{\sum }_{k=1}^{n}\frac{{\left(k-1/2\right)}^{1/r}}{{b}_{k}}}\right)}^{r}\\ \phantom{\rule{1em}{0ex}}\ge {B}_{N}^{r}\left[{\left(\frac{r+1}{r}\right)}^{r}\sum _{n=1}^{N}\left(1-\frac{c}{{\left(n-1/2\right)}^{1+1/r}}\right)\frac{1}{n-1/2}\\ \phantom{\rule{2em}{0ex}}-\sum _{n=1}^{N}{\left(\frac{n}{{\sum }_{k=1}^{n}{\left(k-1/2\right)}^{1/r}}\right)}^{r}\right].\end{array}$
(2.24)

Let ${D}_{m}=\left\{\mathbf{b}=\left({b}_{1},{b}_{2},\dots ,{b}_{N}\right)|{b}_{m}={max}_{1\le k\le N}\left\{{b}_{k}\right\}>{min}_{1\le k\le N}\left\{{b}_{k}\right\}\right\}$ ($m=1,2,\dots ,N$), and

$\begin{array}{rl}f\left({b}_{1},{b}_{2},\dots ,{b}_{N}\right)=& {\left(\frac{r+1}{r}\right)}^{r}\sum _{n=1}^{N}\left(1-\frac{c}{{\left(n-1/2\right)}^{1+1/r}}\right)\frac{{b}_{n}^{r}}{n-1/2}\\ -\sum _{n=1}^{N}{\left(\frac{n}{{\sum }_{k=1}^{n}\frac{{\left(k-1/2\right)}^{1/r}}{{b}_{k}}}\right)}^{r}.\end{array}$
(2.25)

Then for any $\mathbf{b}\in {D}_{m}$ ($m=1,2,\dots ,N$) we have

$\begin{array}{rl}\frac{\partial f\left(\mathbf{b}\right)}{\partial {b}_{m}}=& \left(1-\frac{c}{{\left(m-1/2\right)}^{1+1/r}}\right)\frac{{\left(r+1\right)}^{r}{b}_{m}^{r-1}}{\left(m-1/2\right){r}^{r-1}}\\ -\frac{r{\left(m-1/2\right)}^{1/r}}{{b}_{m}^{2}}\sum _{n=m}^{N}\frac{{n}^{r}}{{\left({\sum }_{k=1}^{n}\frac{{\left(k-1/2\right)}^{1/r}}{{b}_{k}}\right)}^{r+1}}\\ >& \left(1-\frac{c}{{\left(m-1/2\right)}^{1+1/r}}\right)\frac{{\left(r+1\right)}^{r}{b}_{m}^{r-1}}{\left(m-1/2\right){r}^{r-1}}\\ -r{\left(m-1/2\right)}^{1/r}{b}_{m}^{r-1}\sum _{n=m}^{+\mathrm{\infty }}\frac{{n}^{r}}{{\left({\sum }_{k=1}^{n}{\left(k-1/2\right)}^{1/r}\right)}^{r+1}}.\end{array}$
(2.26)

From Lemma 2.2 and (2.26) one has

$\begin{array}{rl}\frac{1}{r{\left(m-1/2\right)}^{1/r}{b}_{m}^{r-1}}\frac{\partial f\left(\mathbf{b}\right)}{\partial {b}_{m}}>& {\left(\frac{r+1}{r}\right)}^{r}\left(1-\frac{c}{{\left(m-1/2\right)}^{1+1/r}}\right)\frac{1}{{\left(m-1/2\right)}^{1+1/r}}\\ -\sum _{n=m}^{+\mathrm{\infty }}\frac{{n}^{r}}{{\left(\frac{r}{r+1}{n}^{1+1/r}+{2}^{-1/r}-\frac{r}{r+1}\right)}^{r+1}}.\end{array}$
(2.27)

It clearly follows from Lemma 2.4 and the Hermite-Hadamard inequality that

${\int }_{m-1/2}^{m+1/2}\frac{{x}^{r}}{{\left(\frac{r}{r+1}{x}^{1+1/r}+{2}^{-1/r}-\frac{r}{r+1}\right)}^{r+1}}\ge \frac{{m}^{r}}{{\left(\frac{r}{r+1}{m}^{1+1/r}+{2}^{-1/r}-\frac{r}{r+1}\right)}^{r+1}}$

and

${\int }_{m-1/2}^{+\mathrm{\infty }}\frac{{x}^{r}}{{\left(\frac{r}{r+1}{x}^{1+1/r}+{2}^{-1/r}-\frac{r}{r+1}\right)}^{r+1}}\ge \sum _{n=m}^{+\mathrm{\infty }}\frac{{n}^{r}}{{\left(\frac{r}{r+1}{n}^{1+1/r}+{2}^{-1/r}-\frac{r}{r+1}\right)}^{r+1}}.$
(2.28)

Note that

$\begin{array}{r}{\int }_{m-1/2}^{+\mathrm{\infty }}\frac{{x}^{r}}{{\left(\frac{r}{r+1}{x}^{1+1/r}+{2}^{-1/r}-\frac{r}{r+1}\right)}^{r+1}}\\ \phantom{\rule{1em}{0ex}}={\left(\frac{1+r}{r}\right)}^{r}\frac{{2}^{1/r}}{r+1-{2}^{1/r}r}\left\{1-{\left[1+\left(\frac{r+1}{{2}^{1/r}r}-1\right){\left(m-1/2\right)}^{-1-1/r}\right]}^{-r}\right\},\end{array}$
(2.29)
$\phantom{\rule{1em}{0ex}}0<{\left(m-1/2\right)}^{-1-1/r}\le {2}^{1+1/r}\le 4.$
(2.30)

From Lemma 2.5 and (2.30) one has

$\begin{array}{r}{\left(\frac{r+1}{r}\right)}^{r}\left(1-\frac{c}{{\left(m-1/2\right)}^{1+1/r}}\right)\frac{1}{{\left(m-1/2\right)}^{1+1/r}}\\ \phantom{\rule{1em}{0ex}}\ge {\left(\frac{1+r}{r}\right)}^{r}\frac{{2}^{1/r}}{r+1-{2}^{1/r}r}\left\{1-{\left[1+\left(\frac{r+1}{{2}^{1/r}r}-1\right){\left(m-1/2\right)}^{-1-1/r}\right]}^{-r}\right\}.\end{array}$
(2.31)

Inequalities (2.27), (2.28), and (2.31) together with (2.29) lead to the conclusion that

$\frac{\partial f\left(\mathbf{b}\right)}{\partial {b}_{m}}>0$
(2.32)

for any $\mathbf{b}=\left({b}_{1},{b}_{2},\dots ,{b}_{N}\right)\in {D}_{m}$ and $m=1,2,\dots ,N$.

It follows from Lemma 2.1 and (2.32) that

$f\left({b}_{1},{b}_{2},\dots ,{b}_{N}\right)\ge f\left({B}_{N},{B}_{N},\dots ,{B}_{N}\right).$
(2.33)

Therefore, inequality (2.24) follows from (2.25) and (2.33). □

Lemma 2.7 Let $r\ge 1$, $c=\left(r+1-{2}^{1/r}r\right)/\left[8\left(r+1-{2}^{1/r}r\right)+2\right]$, then

$\left(\frac{r+1}{r}\right)\left(1-{2}^{1+1/r}c\right)>1.$
(2.34)

Proof We clearly see that inequality (2.34) holds for $r=1$. Next, we assume that $r>1$, let $t={2}^{1+1/r}$, then $0 and Lemma 2.5 leads to

$\begin{array}{rl}{\left(\frac{r+1}{r}\right)}^{r}\left(1-{2}^{1+1/r}c\right)-1& \ge \frac{{\left(\frac{r+1}{r}\right)}^{r}}{2\left(r+1-{2}^{1/r}r\right)}\left[1-{\left(1+\frac{2\left(r+1-{2}^{1/r}r\right)}{r}\right)}^{-r}\right]-1\\ \ge \frac{{\left(\frac{r+1}{r}\right)}^{r}}{2\left(r+1-{2}^{1/r}r\right)}\frac{2\left(r+1-{2}^{1/r}r\right)}{1+2\left(r+1-{2}^{1/r}r\right)}-1\\ =\frac{{\left(\frac{r+1}{r}\right)}^{r}}{1+2\left(r+1-{2}^{1/r}r\right)}-1.\end{array}$
(2.35)

Note that

$r+1-{2}^{1/r}r<1-log2$
(2.36)

for all $r\ge 1$. In fact, let $x\ge 1$ and

$f\left(x\right)=x-{2}^{1/x}x+1.$
(2.37)

Then

${f}^{\prime }\left(x\right)=1+\left(\frac{log2}{x}-1\right){2}^{1/x},$
(2.38)
${f}^{″}\left(x\right)=-\frac{{\left(log2\right)}^{2}}{{x}^{3}}{2}^{1/x}<0.$
(2.39)

It follows from (2.38) and (2.39) that

${f}^{\prime }\left(x\right)>\underset{x\to +\mathrm{\infty }}{lim}\left[1+\left(\frac{log2}{x}-1\right){2}^{1/x}\right]=0.$
(2.40)

Equation (2.37) and inequality (2.40) lead to the conclusion that

$f\left(x\right)<\underset{x\to +\mathrm{\infty }}{lim}\left(x-{2}^{1/x}x+1\right)=1-log2.$
(2.41)

From (2.35) and (2.36) together with the fact that ${\left[\left(r+1\right)/r\right]}^{r}\ge 2$ we have

${\left(\frac{r+1}{r}\right)}^{r}\left(1-{2}^{1+1/r}c\right)-1>\frac{2}{1+2\left(1-log2\right)}-1=\frac{2log2-1}{3-2log2}>0.$
(2.42)

Therefore, inequality (2.34) follows from (2.42). □

Lemma 2.8 Let $r\ge 1$, $c=\left(r+1-{2}^{1/r}r\right)/\left[8\left(r+1-{2}^{1/r}r\right)+2\right]$, N is a positive natural number, ${a}_{k}>0$ ($k=1,2,\dots ,N$) and ${B}_{N}={min}_{1\le k\le N}\left\{{\left(k-1/2\right)}^{1/r}{a}_{k}\right\}$, then

$\begin{array}{r}{\left(\frac{r+1}{r}\right)}^{r}\sum _{n=1}^{N}\left(1-\frac{c}{{\left(n-1/2\right)}^{1+1/r}}\right){a}_{n}^{r}-\sum _{n=1}^{N}{\left(\frac{n}{{\sum }_{k=1}^{n}1/{a}_{k}}\right)}^{r}\\ \phantom{\rule{1em}{0ex}}\ge 2{B}_{N}^{r}\left[{\left(\frac{r+1}{r}\right)}^{r}\left(1-{2}^{1+1/r}c\right)-1\right].\end{array}$
(2.43)

Proof Let $m\in \left\{1,2,\dots ,N\right\}$, $f\left(0\right)=0$ and

$f\left(m\right)={\left(\frac{r+1}{r}\right)}^{r}\sum _{n=1}^{m}\left(1-\frac{c}{{\left(n-1/2\right)}^{1+1/r}}\right)\frac{1}{n-1/2}-\sum _{n=1}^{m}{\left(\frac{n}{{\sum }_{k=1}^{n}{\left(k-1/2\right)}^{1/r}}\right)}^{r}.$
(2.44)

Then

$f\left(1\right)=2\left[{\left(\frac{1+r}{r}\right)}^{r}\left(1-{2}^{1+1/r}c\right)-1\right],$
(2.45)
$f\left(m\right)-f\left(m-1\right)=\frac{{\left(\frac{1+r}{r}\right)}^{r}}{m-1/2}\left(1-\frac{c}{{\left(m-1/2\right)}^{1+1/r}}\right)-{\left(\frac{m}{{\sum }_{k=1}^{m}{\left(k-1/2\right)}^{1/r}}\right)}^{r}.$
(2.46)

It follows from Lemma 2.2 and (2.46) together with Remark 2.1 that

$\begin{array}{r}f\left(m\right)-f\left(m-1\right)\\ \phantom{\rule{1em}{0ex}}\ge \frac{{\left(\frac{1+r}{r}\right)}^{r}}{m-1/2}\left(1-\frac{c}{{\left(m-1/2\right)}^{1+1/r}}\right)-{\left(\frac{m}{\frac{r}{r+1}\left({m}^{1+1/r}-1\right)+{2}^{-1/r}}\right)}^{r}\\ \phantom{\rule{1em}{0ex}}\ge \frac{{\left(\frac{1+r}{r}\right)}^{r}}{m-1/2}\left(1-\frac{c}{{\left(m-1/2\right)}^{1+1/r}}\right)-{\left(\frac{m}{\frac{r}{r+1}{m}^{1+1/r}}\right)}^{r}\\ \phantom{\rule{1em}{0ex}}=\frac{{\left(\frac{1+r}{r}\right)}^{r}\left[\left(4\left(r+1-{2}^{1/r}r\right)+1\right){\left(m-1/2\right)}^{1+1/r}-m\left(r+1-{2}^{1/r}r\right)\right]}{m{\left(m-1/2\right)}^{2+1/r}\left[8\left(r+1-{2}^{1/r}r\right)+2\right]}.\end{array}$
(2.47)

Let

$g\left(t\right)=\left[4\left(r+1-{2}^{1/r}r\right)+1\right]{\left(t-1/2\right)}^{1+1/r}-\left(r+1-{2}^{1/r}r\right)t.$
(2.48)

Then

$\begin{array}{rl}g\left(1\right)& =\left[4\left(r+1-{2}^{1/r}r\right)+1\right]{2}^{-1-1/r}-\left(r+1-{2}^{1/r}r\right)\\ >\left({2}^{1-1/r}-1\right)\left(r+1-{2}^{1/r}r\right)\ge 0,\end{array}$
(2.49)
$\begin{array}{rl}{g}^{\prime }\left(t\right)& =\left(1+\frac{1}{r}\right)\left[4\left(r+1-{2}^{1/r}r\right)+1\right]{\left(t-1/2\right)}^{1/r}-\left(r+1-{2}^{1/r}r\right)\\ >\left({2}^{2-1/r}-1\right)\left(r+1-{2}^{1/r}r\right)\ge 0\end{array}$
(2.50)

for $t\ge 1$.

From (2.47)-(2.50) we get

$f\left(1\right)
(2.51)

Therefore, Lemma 2.8 follows easily from Lemma 2.6, (2.44), (2.45), and (2.51). □

## 3 Proof of Theorem 1.1

Let $r=-p$, $c=c\left(r\right)=d\left(-r\right)$ and ${b}_{n}=1/{a}_{n}$ ($n=1,2,\dots$), then $r\ge 1$, $c=\left(r+1-{2}^{1/r}r\right)/\left[8\left(r+1-{2}^{1/r}r\right)+2\right]$, ${b}_{n}>0$ and ${\sum }_{n=1}^{\mathrm{\infty }}{b}_{n}^{r}<+\mathrm{\infty }$.

It follows from Lemmas 2.7 and 2.8 that one has

$\sum _{n=1}^{N}{\left(\frac{n}{{\sum }_{k=1}^{n}1/{b}_{k}}\right)}^{r}\le {\left(\frac{r+1}{r}\right)}^{r}\sum _{n=1}^{N}\left(1-\frac{c}{{\left(n-1/2\right)}^{1+1/r}}\right){b}_{n}^{r}.$
(3.1)

Letting $n\to +\mathrm{\infty }$, (3.1) leads to

$\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{n}{{\sum }_{k=1}^{n}1/{b}_{k}}\right)}^{r}\le {\left(\frac{r+1}{r}\right)}^{r}\sum _{n=1}^{\mathrm{\infty }}\left(1-\frac{c}{{\left(n-1/2\right)}^{1+1/r}}\right){b}_{n}^{r}.$
(3.2)

Therefore, Theorem 1.1 follows immediately from (3.2) and $r=-p$ together with ${b}_{n}=1/{a}_{n}$.

## References

1. Hardy GH, Littlewood JE, Pólya G: Inequalities. Cambridge University Press, Cambridge; 1952.

2. Beesack PR: Hardy’s inequality and its extensions. Pac. J. Math. 1961, 11: 39-61. 10.2140/pjm.1961.11.39

3. Mitrinović DS, Pečarić JE, Fink AM: Inequalities Involving Functions and Their Integrals and Derivatives. Kluwer Academic, Dordrecht; 1991.

4. Yang B-C, Zeng Z-H, Debnath L: On new generalizations of Hardy’s integral inequality. J. Math. Anal. Appl. 1998,217(1):321-327. 10.1006/jmaa.1998.5758

5. Chen C-P, Qi F: Generalization of Hardy’s inequality. Proc. Jangjeon Math. Soc. 2004,7(1):57-61.

6. Pachpatte BG: Mathematical Inequalities. Elsevier, Amsterdam; 2005.

7. Liu H-P, Zhu L: New strengthened Carleman’s inequality and Hardy’s inequality. J. Inequal. Appl. 2007. Article ID 84104, 2007: Article ID 84104

8. Persson L-E, Oguntuase JA: Refinement of Hardy’s Inequality for “All” p. Banach and Function Spaces II. Yokohama Publ., Yokohama; 2008:129-144.

9. Lü Z-X, Gao Y-C, Wei Y-X: Note on the Carleman’s inequality and Hardy’s inequality. Comput. Math. Appl. 2010,59(1):94-97. 10.1016/j.camwa.2009.09.001

10. Liu J-Z, Zhang X-D, Jiang B: Some generalizations and improvements of discrete Hardy’s inequality. Comput. Math. Appl. 2012,63(3):601-607. 10.1016/j.camwa.2011.09.019

11. Yang B-C, Zhu Y-H: An improvement on Hardy’s inequality. Acta Sci. Natur. Univ. Sunyatseni 1998 (in Chinese),37(1):41-44. (in Chinese)

12. Huang Q-L: An improvement for Hardy’s inequality in an interval. Acta Sci. Natur. Univ. Sunyatseni 2000 (in Chinese),39(3):20-24. (in Chinese)

13. Wen J-J, Zhang R-X: A strengthened improvement of Hardy’s inequality. Math. Pract. Theory 2002 (in Chinese),32(3):476-482. (in Chinese)

14. Xu Q, Zhou M-X, Zhang X-M: On a strengthened version of Hardy’s inequality. J. Inequal. Appl. 2012. Article ID 300, 2012: Article ID 300

15. Deng Y-P, Wu S-H, He D:A sharpened version of Hardy’s inequality for parameter $p=5/4$. J. Inequal. Appl. 2013. Article ID 63, 2013: Article ID 63

16. Long NT, Linh NVD: The Carleman’s inequality for a negative power number. J. Math. Anal. Appl. 2001,259(1):219-225. 10.1006/jmaa.2000.7422

17. Zhang X-M, Chu Y-M: A new method to study analytic inequalities. J. Inequal. Appl. 2010. Article ID 698012, 2010: Article ID 698012

## Acknowledgements

The authors would like to express their deep gratitude to the referees for giving many valuable suggestions. This research was supported by the Natural Science Foundation of China under Grants 11171307 and 61374086, and the Natural Science Foundation of Zhejiang Province under Grant Y13A01000.

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Correspondence to Yu-Ming Chu.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

Y-MC provided the main idea and carried out the proof of Lemmas 2.1 and 2.2. QX carried out the proof of Lemmas 2.3-2.5 and Theorem 1.1. X-MZ carried out the proof of Lemmas 2.6-2.8. All authors read and approved the final manuscript.

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Chu, YM., Xu, Q. & Zhang, XM. A note on Hardy’s inequality. J Inequal Appl 2014, 271 (2014). https://doi.org/10.1186/1029-242X-2014-271