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Sharp bounds by the power mean for the generalized Heronian mean

Journal of Inequalities and Applications20122012:129

https://doi.org/10.1186/1029-242X-2012-129

Received: 20 February 2012

Accepted: 7 June 2012

Published: 7 June 2012

Abstract

In this article, we answer the question: For p, ω with ω > 0 and p(ω - 2) ≠ 0, what are the greatest value r1 = r1(p, ω) and the least value r2 = r2(p, ω) such that the double inequality M r 1 a , b < H p , ω a , b < M r 2 a , b holds for all a, b > 0 with ab? Here Hp,ω(a, b) and M r (a, b) denote the generalized Heronian mean and r th power mean of two positive numbers a and b, respectively.

2010 Mathematics Subject Classification: 26E60.

Keywords

generalized Heronian meanpower meanHeronian mean.

1 Introduction

In the recent past, the bivariate means have been the subject of intensive research. In particular, many remarkable inequalities can be found in the literature [126].

The power mean M r (a, b) of order r of two positive numbers a and b is defined by
M r ( a , b ) = a r + b r 2 1 / r , r 0 , a b , r = 0 .
(1.1)
It is well-known that M r (a, b) is continuous and strictly increasing with respect to r for fixed a, b > 0 with ab. Let A(a, b) = (a + b)/2, G ( a , b ) = a b , H(a, b) = 2ab/(a + b), I(a, b) = 1/e(b b /a a )1/(b-a)(ba), I(a, b) = a (b = a), and L(a, b) = (b-a)/(log b- log a) (ba), L(a, b) = a (b = a) be the arithmetic, geometric, harmonic, identric, and logarithmic means of two positive numbers a and b, respectively. Then
min { a , b } H ( a , b ) = M - 1 ( a , b ) G ( a , b ) = M 0 ( a , b ) L ( a , b ) I ( a , b ) A ( a , b ) = M 1 ( a , b ) max { a , b }
(1.2)

for all a, b > 0, and each inequality becomes equality if and only if a = b.

The classical Heronian mean He(a, b) of two positive numbers a and b is defined by ([27], see also [28])
H e ( a , b ) = 2 3 A ( a , b ) + 1 3 G ( a , b ) = a + a b + b 3 .
(1.3)
In [27], Alzer and Janous established the following sharp double inequality (see also [[28], p. 350]):
M log 2 / log 3 ( a , b ) < H e ( a , b ) < M 2 / 3 ( a , b )

for all a, b > 0 with ab.

Mao [29] proved that
M 1 / 3 ( a , b ) < 1 3 A ( a , b ) + 2 3 G ( a , b ) < M 1 / 2 ( a , b )

for all a, b > 0 with ab, and M1/ 3(a, b) is the best possible lower power mean bound for the sum 1 3 A ( a , b ) + 2 3 G ( a , b ) .

For any α (0, 1), Janous [30] found the greatest value p and the least value q such that M p (a, b) < αA(a, b) + (1 - α)G(a, b) < M q (a, b) for all a, b > 0 with ab.

The following sharp bounds for L, I, (LI)1/ 2and (L + I)/ 2 in terms of power mean are given in [10, 2125, 31, 32]:
M 0 ( a , b ) < L ( a , b ) < M 1 / 3 ( a , b ) , M 2 / 3 ( a , b ) < I ( a , b ) < M log 2 ( a , b ) , M 0 ( a , b ) < L ( a , b ) I ( a , b ) < M 1 / 2 ( a , b ) , M log 2 / ( 1 + log 2 ) ( a , b ) < 1 2 [ L ( a , b ) + I ( a , b ) ] < M 1 / 2 ( a , b )

for all a, b > 0 with ab.

In [6, 7] the authors established the following sharp inequalities:
M - 1 / 3 ( a , b ) < 2 3 G ( a , b ) + 1 3 H ( a , b ) < M 0 ( a , b ) , M - 2 / 3 ( a , b ) < 1 3 G ( a , b ) + 2 3 H ( a , b ) < M 0 ( a , b ) , M 0 ( a , b ) < A α ( a , b ) L 1 - α ( a , b ) < M ( 1 + 2 α ) / 3 ( a , b ) , M 0 ( a , b ) < G α ( a , b ) L 1 - α ( a , b ) < M ( 1 - α ) / 3 ( a , b )

for all for all a, b > 0 with ab and α (0, 1).

For ω ≥ 0 and p the generalized Heronian mean Hp,ω(a, b) of two positive numbers a and b was introduced in [33] as follows:
H p , ω ( a , b ) = [ a p + ω ( a b ) p / 2 + b p ω + 2 ] 1 / p , p 0 , a b , p = 0 .
(1.4)

It is not difficult to verify that Hp,ω(a, b) is continuous with respect to p for fixed a, b > 0 and ω ≥ 0, strictly increasing with respect to p for fixed a, b > 0 with ab and ω ≥ 0, strictly decreasing with respect to ω ≥ 0 for fixed a, b > 0 with ab and p > 0 and strictly increasing with respect to ω ≥ 0 for fixed a, b > 0 with ab and p < 0.

From (1.1) and (1.3) together with (1.4) we clearly see that Hp,0(a, b) = M p (a, b), H p , 2 ( a , b ) = M p 2 ( a , b ) , H0,ω(a, b) = M0(a, b) and H1,1(a, b) = H e (a, b) for all a, b > 0 and ω ≥ 0.

The purpose of this article is to answer the question: For p, ω with ω > 0 and p(ω - 2) ≠ 0, what are the greatest value r1 = r1(p, ω) and the least value r2 = r2(p, ω) such that the double inequality M r 1 ( a , b ) < H p , ω ( a , b ) < M r 2 ( a , b ) holds for all a, b > 0 with ab?

2 Main result

In order to establish our main results we need the following Lemma 2.1.

Lemma 2.1. (see [30]). (ω + 2)2> 2ω+2for ω (0, 2), and (ω + 2)2< 2ω+2for ω (2, +).

Theorem 2.1. For all a, b > 0 with ab we have
M 2 ω + 2 p ( a , b ) < H p , ω ( a , b ) < M log 2 log ( ω + 2 ) p ( a , b )
for (p, ω) {(p, ω): p > 0, ω > 2} {(p, ω): p < 0, 0 < ω < 2} and
M 2 ω + 2 p ( a , b ) > H p , ω ( a , b ) > M log 2 log ( ω + 2 ) p ( a , b )

for (p, ω) {(p, ω): p > 0, 0 < ω < 2} {(p, ω): p < 0, ω > 2}, and the parameters 2 ω + 2 p and log 2 log ( ω + 2 ) p are the best possible in either case.

Proof. Without loss of generality, we can assume that a > b and put t = a b > 1 .

Firstly, we compare the value of M 2 ω + 2 p a , b with that of Hp,ω(a, b). From (1.1) and (1.4) we have
log [ M 2 ω + 2 p ( a , b ) ] - log [ H p , ω ( a , b ) ] = ω + 2 2 p log 1 + t 2 ω + 2 p 2 - 1 p log 1 + ω t p 2 + t p ω + 2 .
(2.1)
Let
f ( t ) = ω + 2 2 p log 1 + t 2 p 2 + ω 2 - 1 p log 1 + ω t p 2 + t p ω + 2 .
(2.2)
Then simple computations lead to
f ( 1 ) = 0 ,
(2.3)
f ( t ) = t 2 p ω + 2 g ( t ) 2 t ( 1 + t 2 p ω + 2 ) ( 1 + ω t p 2 + t p ) , g ( t ) = - 2 t ω p ω + 2 + ω t p 2 - ω t ω - 2 2 ( ω + 2 ) p + 2 ,
(2.4)
g ( 1 ) = 0 ,
(2.5)
g ( t ) = ω p t ( ω - 2 ) p 2 ( ω + 2 ) - 1 h ( t ) ,
(2.6)
h ( t ) = - 2 ω + 2 t p 2 + 1 2 t 2 p ω + 2 - ω - 2 2 ( ω + 2 ) , h ( 1 ) = 0 ,
(2.7)
h ( t ) = p ω + 2 t 2 p ω + 2 - 1 [ 1 - t ( ω - 2 ) p 2 ( ω + 2 ) ] .
(2.8)

We divide the comparison into two cases.

Case 1. If (p, ω) {(p, ω): p > 0, ω > 2} {(p, ω): p < 0, 0 < ω < 2}, then from (2.8) we clearly see that
h ( t ) < 0
(2.9)

for t > 1.

Therefore, M 2 ω + 2 p ( a , b ) < H p , ω ( a , b ) follows from (2.1)-(2.7) and (2.9).

Case 2. If (p, ω) {(p, ω): p > 0, 0 < ω < 2} {(p, ω): p < 0, ω > 2}, then (2.8) leads to
h ( t ) > 0
(2.10)

for t > 1.

Therefore, M 2 ω + 2 p ( a , b ) > H p , ω ( a , b ) follows from (2.1)-(2.7) and (2.10).

Secondly, we compare the value of M log 2 log ( ω + 2 ) p ( a , b ) with that of Hp,ω(a, b). From (1.1) and (1.4) we have
log [ M log 2 log ( ω + 2 ) p ( a , b ) ] - log [ H p , ω ( a , b ) ] = log ( ω + 2 ) p log 2 log 1 + t log 2 log ( ω + 2 ) p 2 - 1 p log 1 + ω t p 2 + t p ω + 2 .
(2.11)
Let
F ( t ) = log ( ω + 2 ) p log 2 log 1 + t log 2 log ( ω + 2 ) p 2 - 1 p log 1 + ω t p 2 + t p ω + 2 .
(2.12)
Then simple computations lead to
F ( 1 ) = lim t + F ( t ) = 0 ,
(2.13)
F ( t ) = t log 2 log ( ω + 2 ) p G ( t ) t ( 1 + t log 2 log ( ω + 2 ) p ) ( 1 + ω t p 2 + t p ) ,
(2.14)
G ( t ) = - t ( 1 - log 2 log ( ω + 2 ) ) p + ω 2 t p 2 - ω 2 t 1 2 ( 1 - 2 log 2 log ( ω + 2 ) ) p + 1 ,
(2.15)
G ( 1 ) = 0 ,
(2.16)
G ( t ) = p t 1 2 ( 1 - 2 log 2 log ( ω + 2 ) ) p - 1 H ( t ) ,
(2.17)
H ( t ) = ( log 2 log ( ω + 2 ) - 1 ) t p 2 + ω 4 t log 2 log ( ω + 2 ) p - ω 4 ( 1 - 2 log 2 log ( ω + 2 ) ) ,
(2.18)
H ( 1 ) = ( ω + 2 ) log 2 2 log ( ω + 2 ) - 1 ,
(2.19)
H ( t ) = log 2 - log ( ω + 2 ) 2 log ( ω + 2 ) p [ t 1 2 ( 1 - 2 log 2 log ( ω + 2 ) ) p - ω log 2 2 ( log ( ω + 2 ) - log 2 ) ] × t log 2 log ( ω + 2 ) p - 1 .
(2.20)

We divide the comparison into four cases.

Case A. If p > 0 and ω > 2, then from (2.15) and (2.18)-(2.20) together with Lemma 2.1 we clearly see that
lim t + G ( t ) = - ,
(2.21)
lim t + H ( t ) = - ,
(2.22)
H ( 1 ) > 0 ,
(2.23)
and there exists a1> 1 such that
H ( t ) > 0
(2.24)
for t [1, a1) and
H ( t ) < 0
(2.25)

for t (a1, +).

From (2.24) and (2.25) we know that H(t) is strictly increasing in [1, a1] and strictly decreasing in [a1, +). Then (2.22) and (2.23) together with the monotonicity of H(t) imply that there exists a2> 1 such that H(t) > 0 for t [1, a2) and H(t) < 0 for t (a2, +). It follows from (2.17) that G(t) is strictly increasing in [1, a2] and strictly decreasing in [a2, +).

From (2.16) and (2.21) together with the monotonicity of G(t) we know that there exists a3> 1 such that G(t) > 0 for t (1, a3) and G(t) < 0 for t (a3, +). Then (2.14) leads to that F (t) is strictly increasing in [1, a3] and strictly decreasing in [a3, +).

Therefore, M log 2 log ( ω + 2 ) ( a , b ) > H p , ω ( a , b ) follows from (2.11)-(2.13) and the monotonicity of F (t).

Case B. If p > 0 and 0 < ω < 2, then (2.15) and (2.18)-(2.20) together with Lemma 2.1 lead to
lim t + G ( t ) = + ,
(2.26)
lim t + H ( t ) = + ,
(2.27)
H ( 1 ) < 0 ,
(2.28)
and there exists b1> 1 such that
H ( t ) < 0
(2.29)
for t [1, b1) and
H ( t ) > 0
(2.30)

for t (b1, +).

From (2.27)-(2.30) we clearly see that there exists b2> 1 such that H(t) < 0 for t [1, b2) and H(t) > 0 for t (b2, +). Then (2.17) implies that G(t) is strictly decreasing in [1, b2] and strictly increasing in [b2, +). It follows from (2.16) and (2.26) together with the monotonicity of G(t) that there exists b3> 1 such that G(t) < 0 for t (1, b3) and G(t) > 0 for t (b3, +). Then (2.14) leads to that F(t) is strictly decreasing in [1, b3] and strictly increasing in [b3, +).

Therefore, M log 2 log ( ω + 2 ) ( a , b ) < H p , ω ( a , b ) follows from (2.11)-(2.13) and the monotonicity of F (t).

Case C. If p < 0 and ω > 2, then it follows from (2.15) and (2.18)-(2.20) together with Lemma 2.1 that
lim t + G ( t ) = 1 ,
(2.31)
lim t + H ( t ) = ω 4 ( 2 log 2 log ( ω + 2 ) - 1 ) < 0 ,
(2.32)
H ( 1 ) > 0 ,
(2.33)
H ( t ) < 0
(2.34)

for t [1, +).

From (2.32)-(2.34) we clearly see that there exists c1> 1 such that H(t) > 0 for t [1, c1) and H(t) < 0 for t (c1, +). Then (2.17) implies that G(t) is strictly decreasing in [1, c1] and strictly increasing in [c1, +).

It follows from (2.16) and (2.31) together with the monotonicity of G(t) that there exists c2> 1 such that G(t) < 0 for t (1, c2) and G(t) > 0 for t (c2, +). Then (2.14) leads to that F (t) is strictly decreasing in [1, c2] and strictly increasing in [c2, +).

Therefore, M log 2 log ( ω + 2 ) ( a , b ) < H p , ω ( a , b ) follows from (2.11)-(2.13) and the monotonicity of F (t).

Case D. If p < 0 and 0 < ω < 2, then (2.15) and (2.18)-(2.20) together with Lemma 2.1 lead to
lim t + G ( t ) = - ,
(2.35)
lim t + H ( t ) = ω 4 ( 2 log 2 log ( ω + 2 ) - 1 ) > 0 ,
(2.36)
H ( 1 ) < 0 ,
(2.37)
H ( t ) > 0
(2.38)

for t > 1.

From (2.17) and (2.36)-(2.38) we clearly see that there exists d1> 1 such that G(t) is strictly increasing in [1, d1] and strictly decreasing in [d1, +). It follows from (2.14), (2.16), (2.35) and the monotonicity of G(t) that there exists d2> 1 such that F (t) is strictly increasing in [1, d2] and strictly decreasing in [d2, +).

Therefore, M log 2 log ( ω + 2 ) ( a , b ) > H p , ω ( a , b ) follows from (2.11)-(2.13) and the monotonicity of F(t).

Thirdly, we prove that the parameter 2 ω + 2 p is the best possible in either case.

For any p, r with pr ≠ 0, ω ≥ 0 and x > 0, one has
log [ M r ( 1 , 1 + x ) ] - log [ H p , ω ( 1 , 1 + x ) ] = 1 r log 1 + ( 1 + x ) r 2 - 1 p log 1 + ω ( 1 + x ) p 2 + ( 1 + x ) p ω + 2 .
(2.39)
Let x → 0, then the Taylor expansion leads to
1 r log 1 + ( 1 + x ) r 2 - 1 p log 1 + ω ( 1 + x ) p 2 + ( 1 + x ) p ω + 2 = ( ω + 2 ) r - 2 p 4 ( ω + 2 ) x 2 + o ( x 2 ) .
(2.40)

If (p, ω) {(p, ω): p > 0, ω > 2} {(p, ω): p < 0, 0 < ω < 2}, then equations (2.39) and (2.40) imply that for any r > 2 ω + 2 p there exists δ1 = δ1(r, p, ω) > 0 such that M r (1, 1 + x) > Hp,ω(1, 1 + x) for x (0, δ1).

If (p, ω) {(p, ω): p > 0, 0 < ω < 2} {(p, ω): p < 0, ω > 2}, then from (2.39) and (2.40) we know that for any r < 2 ω + 2 p there exists δ2 = δ2(r, p, ω) > 0 such that M r (1, 1 + x) < Hp, ω(1, 1 + x) for x (0, δ2).

Finally, we prove that the parameter log 2 log ( ω + 2 ) p is the optimal parameter in either case.

For any p, r with pr > 0, ω ≥ 0 and x > 0 we have
lim x + [ log M r ( 1 , x ) - log H p , ω ( 1 , x ) ] = 1 p log ( ω + 2 ) - 1 r log 2 .
(2.41)

If (p, ω) {(p, ω): p > 0, ω > 2} {(p, ω): p < 0, 0 < ω < 2}, then equation (2.41) implies that for any r < log 2 log ( ω + 2 ) p there exists X1 = X1(r, p, ω) > 1 such that M r (1, x) < Hp, ω(1, x) for x (X1, +).

If (p, ω) ω {(p, ω): p > 0, 0 < ω < 2} {(p, ω): p < 0, ω > 2}, then equation (2.41) leads to that for any r > log 2 log ( ω + 2 ) p there exists X2 = X2(r, p, ω) > 1 such that M r (1, x) > Hp, ω(1, x) for x (X2, +).

Declarations

Acknowledgements

This research was supported by the Natural Science Foundation of China under Grants 11071069 and 11171307, the Natural Science Foundation of Hunan Province under Grant 09JJ6003, and the Innovation Team Foundation of the Department of Education of Zhejiang Province under Grant T200924.

Authors’ Affiliations

(1)
Department of Mathematics, Huzhou Teachers College, Huzhou, China
(2)
School of Mathematical Science, Anhui University, Hefei, China

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© Li et al; licensee Springer. 2012

This article is published under license to BioMed Central Ltd. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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