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A sharpened version of Hardy’s inequality for parameter p=5/4

Abstract

The purpose of this paper is to investigate a sharpened version of Hardy’s inequality for parameter p=5/4. By evaluating the weight coefficient W(k,5/4), sharpened Hardy’s inequality that contains the best coefficient η 5 / 4 =0.46 is established.

MSC:26D15, 26D20, 26D07.

1 Introduction

Let p>1, 1/p+1/q=1, a n 0 (n=1,2,), 0< n = 1 a n p <. Then

n = 1 ( 1 n k = 1 n a k ) p < q p n = 1 a n p ,
(1)

where q p = ( p p 1 ) p is the best coefficient. Inequality (1) is called Hardy’s inequality which is of great use in the field of modern mathematics (see [1, 2]).

A special case of (1) yields the following inequalities:

n = 1 ( 1 n k = 1 n a k ) 2 <4 n = 1 a n 2 ,
(2)
n = 1 ( 1 n k = 1 n a k ) 3 < 27 8 n = 1 a n 3 .
(3)

In 1998, Yang and Zhu [3] evaluated the weight coefficient W(k,p),

W(k,p)= k 1 1 / p n = k 1 n p ( j = 1 n 1 j 1 / p ) p 1 ,k=1,2,,
(4)

and established an improved version of inequality (2) as follows:

n = 1 ( 1 n k = 1 n a k ) 2 <4 n = 1 ( 1 1 3 n + 5 ) a n 2 .
(5)

With the same approach, that is, evaluating the weight coefficient W(k,p), Huang [47] gave some improvements on Hardy’s inequality for p=3 and p=3/2, i.e.,

n = 1 ( 1 n k = 1 n a k ) 3 < 27 8 n = 1 ( 1 3 19 n 2 / 3 ) a n 3 ,
(6)
n = 1 ( 1 n k = 1 n a k ) 3 / 2 3 3 n = 1 ( 1 1 5 1 n 3 + 3 ) a n 3 / 2 .
(7)

Some further extensions of Hardy’s inequality related to the range of parameter p were given in Huang [7, 8].

In 2005, Yang [9] proved an inequality for the weight coefficient W(k,2)

W(k,2)= k n = k 1 n 2 ( j = 1 n 1 j ) 4 [ 1 1 k ( 1 1 4 W ( 1 , 2 ) ) ]

and established the following inequality:

n = 1 ( 1 n k = 1 n a k ) 2 <4 n = 1 ( 1 θ 2 n ) a n 2 ,
(8)

where θ 2 =1 1 4 W(1,2)=0.13788928 is the best coefficient under the weight coefficient W(k,2).

In 2009, Zhang and Xu made use of the monotonicity theorem [1013] and obtained an improvement of inequality (1):

n = 1 ( 1 n k = 1 n a k ) p ( p p 1 ) p n = 1 ( 1 c p 2 ( n 1 / 2 ) 1 1 / p ) a n p ,
(9)

where

c p = { ( p 1 ) [ 1 2 1 / p ( 1 1 / p ) ] , 1 < p 2 , 1 2 1 1 / p ( 1 1 / p ) p 1 , p > 2 .

By evaluating the weight coefficient W(k,p), and with the help of an inequality-proving package called BOTTEMA [14, 15], He [16] investigated a sharpened version of Hardy’s inequality for pN and obtained the following improved version of inequality (3):

n = 1 ( 1 n k = 1 n a k ) 3 27 8 n = 1 ( 1 θ 3 n 2 / 3 ) a n 3 ,
(10)

where θ 3 =1 8 27 W(1,3)=0.1673 is the best coefficient under the weight coefficient W(k,3).

In addition, in [16] the author wrote the computer program HDISCOVER to accomplish the automated verification of the following inequality for pN (N is the set of natural numbers):

n = 1 ( 1 n k = 1 n a k ) p ( p p 1 ) p n = 1 ( 1 θ p ( 1 ) n 1 1 / p ) a n p ,
(11)

where θ p (1)=1 ( p 1 p ) p W(1,p) is the best coefficient of (11) under the weight coefficient W(k,p).

Recently, based on the program HDISCOVER 2012 written by Deng, He and Wu [17], an automated verification of inequality (11) is achieved for pQ (Q is the set of rational numbers).

For more detailed information of Hardy’s inequality, we refer the interested readers to relevant research papers [10, 12, 1823].

In this paper, by evaluating the weight coefficient W(k,5/4), we establish an improvement of Hardy’s inequality for parameter p=5/4 as follows:

n = 1 ( 1 n k = 1 n a k ) 5 / 4 5 5 / 4 n = 1 ( 1 1 10 1 n 1 / 5 + η 5 / 4 ) a n 5 / 4 ,
(12)

where η 5 / 4 = 5 5 / 4 10 [ 5 5 / 4 W ( 1 , 5 / 4 ) ] 1=0.46 is the best coefficient under the weight coefficient W(k,5/4).

2 Lemmas

To prove the main results in Section 3, we will use the following lemmas.

Lemma 1 (see[22])

If p>1, then for all integers n1, it holds that

p p 1 n 1 1 / p p p 1 + 1 2 + 1 2 n 1 / p j = 1 n 1 j 1 / p p p 1 n 1 1 / p p p 1 + 1 2 + 1 2 n 1 / p + 1 12 p 1 12 p n 1 + 1 / p .
(13)

Lemma 2 (see[3])

If p>1, then for all integers nk1, it holds that

1 ( p 1 ) k p 1 + 1 2 k p < n = k 1 n p < 1 ( p 1 ) k p 1 + 1 2 k p + p 12 k p + 1 .

Lemma 3 Let p>1, 1/p+1/q=1, and let g r , g l be the functions defined by

g r (x)= 6 p + 12 q 1 12 p q x 1 / q + 1 2 q x 1 12 p q x 2 , g l (x)= p + 2 q 2 p q x 1 / q + 1 2 q x ,x[1,+).

Then 1< g r (x)<0, 1< g l (x)<0.

Proof Since p>1, 1/p+1/q=1, hence 1/ x 1 + 1 / p 1/ x 2 for x[1,+).

Further, we have

g r ( x ) = 6 p + 12 q 1 12 p q 2 x 1 + 1 / q 1 2 q x 2 + 1 6 p q x 3 6 p + 12 q 1 12 p q 2 x 2 1 2 q x 2 + 1 6 p q x 3 = ( 5 p x + x + 2 p ) ( p 1 ) 12 p 3 x 3 > 0 ,

and consequently, g r is strictly increasing on [1,+).

Now, from g r (1)=1/p>1 and lim x + g r (x)=0, it follows that g r (x) g r (1)=1/p>1 and g r (x)<0.

Similarly, from

g l ( x ) = p + 2 q 2 p q 2 x 1 + 1 / q 1 2 q x 2 p + 2 q 2 p q 2 x 2 1 2 q x 2 = p 1 2 p 2 x 2 > 0 , g l ( 1 ) = 1 / p > 1 and lim x + g l ( x ) = 0 ,

we deduce that 1< g l (x)<0.

Lemma 3 is proved. □

Lemma 4 Let 1<g(x)<0. If α(0,1], then

( 1 + g ( x ) ) ( 1 + ( α 1 ) g ( x ) + ( α 1 ) ( α 2 ) 2 g 2 ( x ) ) ( 1 + g ( x ) ) α 1 + α g ( x ) + α ( α 1 ) 2 g 2 ( x ) .

If α[1,2], then

( 1 + g ( x ) ) α 1+αg(x)+ α ( α 1 ) 2 g 2 (x).

Proof When α(0,1]. By using the Maclaurin formula

( 1 + g ( x ) ) α = 1 + α g ( x ) + α ( α 1 ) 2 g 2 ( x ) + α ( α 1 ) ( α 2 ) ( 1 + θ g ( x ) ) α 3 6 g 3 ( x ) , θ ( 0 , 1 ) ,

and noticing 1<g(x)<0, we find

1 + θ g ( x ) > 1 + g ( x ) > 0 , α ( α 1 ) ( α 2 ) ( 1 + θ g ( x ) ) α 3 6 g 3 ( x ) 0 , ( α 1 ) ( α 2 ) ( α 3 ) ( 1 + θ g ( x ) ) α 4 6 g 3 ( x ) 0 .

Thus

( 1 + g ( x ) ) α 1 + α g ( x ) + α ( α 1 ) 2 g 2 ( x ) , ( 1 + g ( x ) ) α = ( 1 + g ( x ) ) ( 1 + g ( x ) ) α 1 ( 1 + g ( x ) ) ( 1 + ( α 1 ) g ( x ) + ( α 1 ) ( α 2 ) 2 g 2 ( x ) ) .

When α[1,2]. We have

α ( α 1 ) ( α 2 ) ( 1 + θ g ( x ) ) α 3 6 g 3 (x)0.

Thus

( 1 + g ( x ) ) α 1+αg(x)+ α ( α 1 ) 2 g 2 (x).

The proof of Lemma 4 is complete. □

Lemma 5 Let p>1, 1/p+1/q=1, nk1, and let [x] denote the greatest integer less than or equal to the real number x. Then we have

W ( k , p ) q p 1 k 1 / q n = k [ 1 n 1 + 1 / q ( 1 + g r ( n ) ) [ p ] 1 × ( 1 + ( p [ p ] ) g r ( n ) + ( p [ p ] ) ( p [ p ] 1 ) 2 g r 2 ( n ) ) ] .

Proof By Lemma 1 and the identity pq=p+q, q(p1)=p, it follows that

W ( k , p ) = k 1 1 / p n = k 1 n p ( j = 1 n 1 j 1 / p ) p 1 k 1 1 / p n = k 1 n p ( p p 1 n 1 1 / p p p 1 + 1 2 + 1 2 n 1 / p + 1 12 p 1 12 p n 1 + 1 / p ) p 1 = k 1 / q n = k 1 n p ( q n 1 / q 6 p + 12 q 1 12 p + 1 2 n 1 / p 1 12 p n 1 + 1 / p ) p 1 = k 1 / q n = k 1 n p q p 1 n ( p 1 ) / q ( 1 6 p + 12 q 1 12 p q n 1 / q + 1 2 q n 1 12 p q n 2 ) p 1 = q p 1 k 1 / q n = k 1 n 1 + 1 / q ( 1 + g r ( n ) ) p 1 .

Combining Lemmas 3 and 4, we obtain

W ( k , p ) q p 1 k 1 / q n = k [ 1 n 1 + 1 / q ( 1 + g r ( n ) ) [ p ] 1 ( 1 + g r ( n ) ) p [ p ] ] q p 1 k 1 / q n = k [ 1 n 1 + 1 / q ( 1 + g r ( n ) ) [ p ] 1 × ( 1 + ( p [ p ] ) g r ( n ) + ( p [ p ] ) ( p [ p ] 1 ) 2 g r 2 ( n ) ) ] .

This completes the proof of Lemma 5. □

Lemma 6 Let 1/p+1/q=1, nk1. If p(1,2), then

W ( k , p ) q p 1 k 1 / q n = k [ 1 n 1 + 1 / q ( 1 + g l ( n ) ) [ p ] × ( 1 + ( p [ p ] 1 ) g l ( n ) + ( p [ p ] 1 ) ( p [ p ] 2 ) 2 g l 2 ( n ) ) ] .

If p[2,+), then

W ( k , p ) q p 1 k 1 / q n = k [ 1 n 1 + 1 / q ( 1 + g l ( n ) ) [ p ] 2 × ( 1 + ( p [ p ] + 1 ) g l ( n ) + ( p [ p ] + 1 ) ( p [ p ] ) 2 g l 2 ( n ) ) ] .

Proof Since pq=p+q, q(p1)=p, using Lemma 1 gives

W ( k , p ) = k 1 1 / p n = k 1 n p ( j = 1 n 1 j 1 / p ) p 1 k 1 1 / p n = k 1 n p ( p p 1 n 1 1 / p p p 1 + 1 2 + 1 2 n 1 / p ) p 1 = k 1 / q n = k 1 n p ( q n 1 / q p + 2 q 2 p + 1 2 n 1 / p ) p 1 = k 1 / q n = k 1 n p q p 1 n ( p 1 ) / q ( 1 p + 2 q 2 p q n 1 / q + 1 2 q n ) p 1 = q p 1 k 1 / q n = k 1 n 1 + 1 / q ( 1 + g l ( n ) ) p 1 .

When p(1,2). From Lemmas 3 and 4, we have

W ( k , p ) q p 1 k 1 / q n = k 1 n 1 + 1 / q ( 1 + g l ( n ) ) [ p ] 1 ( 1 + g l ( n ) ) p [ p ] q p 1 k 1 / q n = k [ 1 n 1 + 1 / q ( 1 + g l ( n ) ) [ p ] × ( 1 + ( p [ p ] 1 ) g l ( n ) + ( p [ p ] 1 ) ( p [ p ] 2 ) 2 g l 2 ( n ) ) ] .

When p[2,+). Using Lemmas 3 and 4, we obtain

W ( k , p ) q p 1 k 1 / q n = k 1 n 1 + 1 / q ( 1 + g l ( n ) ) [ p ] 2 ( 1 + g l ( n ) ) p [ p ] + 1 q p 1 k 1 / q n = k [ 1 n 1 + 1 / q ( 1 + g l ( n ) ) [ p ] 2 × ( 1 + ( p [ p ] + 1 ) g l ( n ) + ( p [ p ] + 1 ) ( p [ p ] ) 2 g l 2 ( n ) ) ] .

Lemma 6 is proved. □

Lemma 7 (see[3])

Let p>1, a n 0 (n=1,2,), 0< n = 1 a n p <. Then

n = 1 ( 1 n k = 1 n a k ) p k = 1 [ k 1 1 / p n = k 1 n p ( j = 1 n 1 j 1 / p ) p 1 a k p ] = k = 1 W(k,p) a k p .

3 Main results

Theorem 1 For an arbitrary natural number k, the following inequality holds true:

W(k,5/4)< R 5 / 4 (k),

where

R 5 / 4 ( k ) = 5 1 / 4 ( 5 133 240 k 1 / 5 17 , 689 144 , 000 k 2 / 5 + 25 48 k 19 192 k 6 / 5 17 , 689 480 , 000 k 7 / 5 + 467 4 , 224 k 2 + 133 18 , 000 k 11 / 5 + 97 38 , 400 k 3 + 133 60 , 000 k 16 / 5 + 61 504 , 000 k 4 + 19 240 , 000 k 5 ) .

Proof

Using Lemma 5 gives

W(k,5/4) 5 1 / 4 k 1 / 5 n = k [ 1 n 6 / 5 ( 1 + 1 4 g r ( n ) 3 32 g r 2 ( n ) ) ] = 5 1 / 4 k 1 / 5 n = k r 5 / 4 (n),

where

r 5 / 4 ( n ) = 1 n 6 / 5 133 600 n 7 / 5 17 , 689 240 , 000 n 8 / 5 + 1 40 n 11 / 5 + 133 8 , 000 n 12 / 5 41 9 , 600 n 16 / 5 133 60 , 000 n 17 / 5 + 1 4 , 000 n 21 / 5 1 60 , 000 n 26 / 5 .

Hence

W ( k , 5 / 4 ) 5 1 / 4 k 1 / 5 n = k ( 1 n 6 / 5 133 600 n 7 / 5 17 , 689 240 , 000 n 8 / 5 + 1 40 n 11 / 5 + 133 8 , 000 n 12 / 5 41 9 , 600 n 16 / 5 133 60 , 000 n 17 / 5 + 1 4 , 000 n 21 / 5 1 60 , 000 n 26 / 5 ) .

Using Lemma 2 and taking p=6/5,7/5,8/5,11/5,12/5,16/5,17/5,21/5,26/5 in the right-hand side of inequality (13), respectively, we get

n = k 1 n 6 / 5 < 5 k 1 / 5 + 1 2 k 6 / 5 + 1 10 k 11 / 5 , n = k 133 600 n 7 / 5 < 133 240 k 2 / 5 133 1 , 200 k 7 / 5 , , n = k 1 4 , 000 n 21 / 5 < 1 12 , 800 k 16 / 5 + 1 8 , 000 k 21 / 5 + 7 80 , 000 k 26 / 5 , n = k 1 60 , 000 n 26 / 5 < 1 252 , 000 k 21 / 5 1 120 , 000 k 26 / 5 .

Adding up the above inequalities, we obtain

W(k,5/4)< R 5 / 4 (k).

Theorem 1 is proved. □

Theorem 2 For an arbitrary natural number k, the following inequality holds true:

W(k,5/4)> L 5 / 4 (k),

where

L 5 / 4 ( k ) = 5 1 / 4 ( 5 9 16 k 1 / 5 81 640 k 2 / 5 15 , 309 25 , 600 k 3 / 5 + 25 48 k 45 448 k 6 / 5 + 3 , 159 51 , 200 k 7 / 5 15 , 309 64 , 000 k 8 / 5 + 17 1 , 408 k 2 129 5 , 120 k 11 / 5 + 891 12 , 800 k 12 / 5 45 , 927 640 , 000 k 13 / 5 27 102 , 400 k 3 567 64 , 000 k 16 / 5 + 1 12 , 800 k 4 3 , 213 640 , 000 k 21 / 5 ) .

Proof

Utilizing Lemma 6 gives

W ( k , 5 / 4 ) 5 1 / 4 k 1 / 5 n = k [ 1 n 6 / 5 ( 1 + g l ( n ) ) ( 1 3 4 g l ( n ) + 21 32 g l 2 ( n ) ) ] = 5 1 / 4 k 1 / 5 n = k l 5 / 4 ( n ) ,

where

l 5 / 4 ( n ) = 1 n 6 / 5 9 40 n 7 / 5 243 3 , 200 n 8 / 5 15 , 309 32 , 000 n 9 / 5 + 1 40 n 11 / 5 + 27 1 , 600 n 12 / 5 + 5 , 103 32 , 000 n 13 / 5 3 3 , 200 n 16 / 5 567 32 , 000 n 17 / 5 + 21 32 , 000 n 21 / 5 .

Hence

W ( k , 5 / 4 ) 5 1 / 4 k 1 / 5 n = k ( 1 n 6 / 5 9 40 n 7 / 5 243 3 , 200 n 8 / 5 15 , 309 32 , 000 n 9 / 5 + 1 40 n 11 / 5 + 27 1 , 600 n 12 / 5 + 5 , 103 32 , 000 n 13 / 5 3 3 , 200 n 16 / 5 567 32 , 000 n 17 / 5 + 21 32 , 000 n 21 / 5 ) .

Using Lemma 2 and taking p=6/5,7/5,8/5,9/5,11/5,12/5,13/5,16/5,17/5,21/5 in the left-hand side of inequality (13), respectively, we get

n = k 1 n 6 / 5 > 5 k 1 / 5 + 1 2 k 6 / 5 , n = k 9 40 n 7 / 5 > 9 16 k 2 / 5 9 80 k 7 / 5 21 800 k 12 / 5 , , n = k 567 32 , 000 n 17 / 5 > 189 25 , 600 k 12 / 5 567 64 , 000 k 17 / 5 3 , 213 640 , 000 k 22 / 5 , n = k 21 32 , 000 n 21 / 5 > 21 102 , 400 k 16 / 5 + 21 64 , 000 k 21 / 5 .

Adding up the above inequalities, we obtain

W(k,5/4)> L 5 / 4 (k).

Theorem 2 is proved. □

Theorem 3 Let a n 0 (n=1,2,), 0< n = 1 a n 5 / 4 <. Then

n = 1 ( 1 n k = 1 n a k ) 5 / 4 5 5 / 4 n = 1 ( 1 1 10 1 n 1 / 5 + η 5 / 4 ) a n 5 / 4 ,
(14)

where η 5 / 4 = 5 5 / 4 10 [ 5 5 / 4 W ( 1 , 5 / 4 ) ] 1=0.46 is the best possible under the weight coefficient W(k,5/4).

Proof

By Lemma 7, we have

n = 1 ( 1 n k = 1 n a k ) 5 / 4 k = 1 W(k,5/4) a k 5 / 4 .

Therefore, to prove inequality (14), it suffices to show that

W(k,5/4) 5 5 / 4 ( 1 1 10 1 k 1 / 5 + η 5 / 4 ) .
(15)

Obviously, inequality (15) becomes an equality for k=1. In what follows, we will assume that k2.

By Theorem 1 W(k,5/4)< R 5 / 4 (k), we need only to prove that

R 5 / 4 (k) 5 5 / 4 ( 1 1 10 1 k 1 / 5 + η 5 / 4 ) .

Note that

η 5 / 4 = 5 5 / 4 10 [ 5 5 / 4 W ( 1 , 5 / 4 ) ] 1=0.44> 11 25 ,

it suffices to show

R 5 / 4 (k) 5 5 / 4 ( 1 1 10 1 k 1 / 5 + 11 / 25 ) .
(16)

Substituting k= x 5 in (16), inequality (16) becomes

R 5 / 4 ( x 5 ) 5 5 / 4 ( 1 1 10 1 x + 11 / 25 ) ,where x 2 5 ,

which is equivalent to the following inequality:

5 1 / 4 ( 5 133 240 x 17 , 689 144 , 000 x 2 + 25 48 x 5 19 192 x 6 17 , 689 480 , 000 x 7 + 467 4 , 224 x 10 + 133 18 , 000 x 11 + 97 38 , 400 x 15 + 133 60 , 000 x 16 + 61 504 , 000 x 20 + 19 240 , 000 x 25 ) 5 5 / 4 ( 1 1 10 1 x + 11 / 25 ) 5 133 240 x 17 , 689 144 , 000 x 2 + 25 48 x 5 19 192 x 6 17 , 689 480 , 000 x 7 + 467 4 , 224 x 10 + 133 18 , 000 x 11 + 97 38 , 400 x 15 + 133 60 , 000 x 16 + 61 504 , 000 x 20 + 19 240 , 000 x 25 5 1 2 1 x + 11 / 25 133 240 x 17 , 689 144 , 000 x 2 + 25 48 x 5 19 192 x 6 17 , 689 480 , 000 x 7 + 467 4 , 224 x 10 + 133 18 , 000 x 11 + 97 38 , 400 x 15 + 133 60 , 000 x 16 + 61 504 , 000 x 20 + 19 240 , 000 x 25 + 1 2 1 x + 11 / 25 0 f ( x ) 221 , 760 , 000 x 25 ( 25 x + 11 ) 0 ,
(17)

where

f ( x ) = 300 , 300 , 000 x 25 + 2 , 032 , 838 , 500 x 24 + 299 , 651 , 660 x 23 2 , 887 , 500 , 000 x 21 721 , 875 , 000 x 20 + 445 , 702 , 950 x 19 + 89 , 895 , 498 x 18 612 , 937 , 500 x 16 310 , 656 , 500 x 15 18 , 024 , 160 x 14 14 , 004 , 375 x 11 18 , 451 , 125 x 10 5 , 407 , 248 x 9 671 , 000 x 6 295 , 240 x 5 438 , 900 x 193 , 116 .

From the hypothesis x 2 5 >1.14, we have

300 , 300 , 000 x 25 + 2 , 032 , 838 , 500 x 24 + 299 , 651 , 660 x 23 2 , 887 , 500 , 000 x 21 721 , 875 , 000 x 20 + 445 , 702 , 950 x 19 + 89 , 895 , 498 x 18 > ( 300 , 300 , 000 × 1.14 4 + 2 , 032 , 838 , 500 × 1.14 3 + 299 , 651 , 660 × 1.14 2 2 , 887 , 500 , 000 ) x 21 721 , 875 , 000 x 20 + 445 , 702 , 950 x 19 + 89 , 895 , 498 x 18 = 1 , 020 , 861 , 716 x 21 721 , 875 , 000 x 20 + 445 , 702 , 950 x 19 + 89 , 895 , 498 x 18 = [ ( 1 , 020 , 861 , 716 x 721 , 875 , 000 ) x 2 + 445 , 702 , 950 x + 89 , 895 , 498 ] x 18 > [ ( 1 , 020 , 861 , 716 × 1.14 721 , 875 , 000 ) × 1.14 2 + 445 , 702 , 950 × 1.14 + 89 , 895 , 498 ] x 18 = 1 , 172 , 299 , 661 x 18 .

Further, we have

f ( x ) > 1 , 172 , 299 , 661 x 18 612 , 937 , 500 x 16 310 , 656 , 500 x 15 18 , 024 , 160 x 14 14 , 004 , 375 x 11 18 , 451 , 125 x 10 5 , 407 , 248 x 9 671 , 000 x 6 295 , 240 x 5 438 , 900 x 193 , 116 > 1 , 172 , 299 , 661 x 18 612 , 937 , 500 x 18 310 , 656 , 500 x 18 18 , 024 , 160 x 18 14 , 004 , 375 x 18 18 , 451 , 125 x 18 5 , 407 , 248 x 18 671 , 000 x 18 295 , 240 x 18 438 , 900 x 18 193 , 116 x 18 = 191 , 220 , 497 x 18 > 0 .

Consequently, inequality (17) holds true, and inequality (14) is proved.

Let us now show that η 5 / 4 = 5 5 / 4 10 [ 5 5 / 4 W ( 1 , 5 / 4 ) ] 1=0.46 is the best possible under the weight coefficient W(k,5/4).

Consider inequality (14) in a general form as

W(k,5/4) 5 5 / 4 ( 1 1 10 1 k 1 / 5 + η 5 / 4 ) .
(18)

Putting k=1 in (18) yields

η 5 / 4 5 5 / 4 10 [ 5 5 / 4 W ( 1 , 5 / 4 ) ] .

Thus the best possible value for η 5 / 4 in (18) should be η min = 5 5 / 4 10 [ 5 5 / 4 W ( 1 , 5 / 4 ) ] .

This completes the proof of Theorem 3. □

Remark 1 From the definition of W(k,p) and in the same way as in [17], we can establish the following accurate estimates of W(1,5/4):

6.965042829<W(1,5/4)<6.967740323.
(19)

Further, the approximation of η 5 / 4 can be derived as follows:

η 5 / 4 = 5 5 / 4 10 [ 5 5 / 4 W ( 1 , 5 / 4 ) ] 1=0.46.

Remark 2 For p=5/4, inequality (11) can be written as

n = 1 ( 1 n k = 1 n a k ) 5 / 4 5 5 / 4 n = 1 ( 1 1 ( 1 5 ) 5 / 4 W ( 1 , 5 / 4 ) n 1 / 5 ) a n 5 / 4 .
(20)

It is easy to observe that

1 10 1 n 1 / 5 + η 5 / 4 > 1 10 1 n 1 / 5 + 4 , 711 / 10 , 000 > 7 100 n 1 / 5

and

1 ( 1 5 ) 5 / 4 W(1,5/4)<1 ( 1 5 ) 5 / 4 ×6.967740323=0.06808< 7 100 ,

hence

1 ( 1 5 ) 5 / 4 W ( 1 , 5 / 4 ) n 1 / 5 < 1 10 1 n 1 / 5 + η 5 / 4 .

This implies that inequality (14) is stronger than inequality (11).

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Acknowledgements

The present investigation was supported, in part, by the Natural Science Foundation of Fujian province of China under Grant 2012J01014 and, in part, by the Foundation of Scientific Research Project of Fujian Province Education Department of China under Grant JK2012049.

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Correspondence to Shanhe Wu.

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YD finished the proof and the writing work. SW gave YD some advice on the proof and writing. DH gave YD lots of help in revising the paper. All authors read and approved the final manuscript.

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Deng, Y., Wu, S. & He, D. A sharpened version of Hardy’s inequality for parameter p=5/4. J Inequal Appl 2013, 63 (2013). https://doi.org/10.1186/1029-242X-2013-63

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Keywords

  • Hardy’s inequality
  • weight coefficient
  • sharpened inequality
  • exponential parameter
  • best coefficient