# A best-possible double inequality between Seiffert and harmonic means

## Abstract

In this paper, we establish a new double inequality between the Seiffert and harmonic means.

The achieved results is inspired by the papers of Sándor (Arch. Math., 76, 34-40, 2001) and Hästö (Math. Inequal. Appl., 7, 47-53, 2004), and the methods from Wang et al. (J. Math. Inequal., 4, 581-586, 2010). The inequalities we obtained improve the existing corresponding results and, in some sense, are optimal.

2010 Mathematics Subject Classification: 26E60.

## 1 Introduction

For a, b > 0 with ab, the Seiffert mean P(a, b) was introduced by Seiffert  as follows:

$P ( a , b ) = a - b 4 arctan a ∕ b - π .$
(1.1)

Recently, the bivariate mean values have been the subject of intensive research. In particular, many remarkable inequalities for the Seiffert mean can be found in the literature .

Let H(a, b) = 2ab/(a+b), $G ( a , b ) = a b$, L(a, b) = (a - b)/(log a - log b), I(a, b) = 1/e(bb /aa )1/(b-a), A(a, b) = (a+b)/2, C(a, b) = (a2+b2)/(a+b), and M p (a, b) = ((ap + bp )/2)1/p(p ≠ 0) and $M 0 ( a , b ) = a b$ be the harmonic, geometric, logarithmic, identric, arithmetic, contraharmonic, and p-th power means of two different positive numbers a and b, respectively. Then, it is well known that

$min { a , b } < H ( a , b ) = M - 1 ( a , b ) < G ( a , b ) = M 0 ( a , b ) < L ( a , b ) < I ( a , b ) < A ( a , b ) = M 1 ( a , b ) < C ( a , b ) < max { a , b } .$

For all a, b > 0 with ab, Seiffert  established that L(a, b) < P(a, b) < I(a, b); Jagers  proved that M1/2(a, b) < P (a, b) < M2/3(a, b) and M2/3(a, b) is the best-possible upper power mean bound for the Seiffert mean P(a, b); Seiffert  established that P(a, b) > A(a, b)G(a, b)/L(a, b) and P(a, b) > 2 A(a, b)/π; Sándor  presented that $( A ( a , b ) + G ( a , b ) ) ∕2 and $A 2 ( a , b ) G ( a , b ) 3 ; Hästö  proved that P(a, b) > Mlog 2/ log π(a, b) and Mlog 2/ log π(a, b) is the best-possible lower power mean bound for the Seiffert mean P(a, b).

Very recently, Wang and Chu  found the greatest value α and the least value β such that the double inequality Aα (a, b)H1-α(a, b) < P(a, b) < Aβ (a, b)H1-β(a, b) holds for a, b > 0 with ab; For any α (0, 1), Chu et al.  presented the best-possible bounds for Pα (a, b)G1(a, b) in terms of the power mean; In , the authors proved that the double inequality αA(a, b) + (1 - α)H(a, b) < P(a, b) < βA(a, b) + (1 - β)H(a, b) holds for all a, b > 0 with ab if and only if α ≤ 2/π and β ≥ 5/6; Liu and Meng  proved that the inequalities

$α 1 C ( a , b ) + ( 1 - α 1 ) G ( a , b ) < P ( a , b ) < β 1 C ( a , b ) + ( 1 - β 1 ) G ( a , b )$

and

$α 2 C ( a , b ) + ( 1 - α 2 ) H ( a , b ) < P ( a , b ) < β 2 C ( a , b ) + ( 1 - β 2 ) H ( a , b )$

hold for all a, b > 0 with ab if and only if α1 ≤ 2/9, β1 ≥ 1/π, α2 ≤ 1/π and β2 ≥ 5/12.

For fixed a, b > 0 with ab and x [0, 1/2], let

$h ( x ) = H ( x a + ( 1 - x ) b , x b + ( 1 - x ) a ) .$

Then, it is not difficult to verify that h(x) is continuous and strictly increasing in [0, 1/2]. Note that h(0) = H(a, b) < P(a, b) and h(1/2) = A(a, b) > P(a, b). Therefore, it is natural to ask what are the greatest value α and least value β in (0, 1/2) such that the double inequality H(αa + (1 - α)b, αb + (1 - α)a) < P(a, b) < H(βa + (1 - β)b, βb + (1 - β)a) holds for all a, b > 0 with ab. The main purpose of this paper is to answer these questions. Our main result is the following Theorem 1.1.

Theorem 1.1. If α, β (0, 1/2), then the double inequality

$H ( α a + ( 1 - α ) b , α b + ( 1 - α ) a ) < P ( a , b ) < H ( β a + ( 1 - β ) b , β b + ( 1 - β ) a )$

holds for all a, b > 0 with ab if and only if $α≤ ( 1 - 1 - 2 ∕ π ) ∕2$ and $β≥ ( 6 - 6 ) ∕12$.

## 2 Proof of Theorem 1.1

Proof of Theorem 1.1. Let $λ= ( 1 - 1 - 2 ∕ π ) ∕2$ and $μ= ( 6 - 6 ) ∕12$. We first prove that inequalities

$P ( a , b ) > H ( λ a + ( 1 - λ ) b , λ b + ( 1 - λ ) a )$
(2.1)

and

$P ( a , b ) < H ( μ a + ( 1 - μ ) b , μ b + ( 1 - μ ) a )$
(2.2)

hold for all a, b > 0 with ab.

Without loss of generality, we assume that a > b. Let $t= a ∕ b >1$ and p (0, 1/2); then, from (1.1), one has

$H ( p a + ( 1 - p ) b , p b + ( 1 - p ) a ) - P ( a , b ) = 2 [ p t 2 + ( 1 - p ) ] [ ( 1 - p ) t 2 + p ] t 2 + 1 - t 2 - 1 4 arctan t - π = 2 [ p t 2 + ( 1 - p ) ] [ ( 1 - p ) t 2 + p ] ( t 2 + 1 ) ( 4 arctan t - π ) × 4 arctan t - t 4 - 1 2 [ p t 2 + ( 1 - p ) ] [ ( 1 - p ) t 2 + p ] - π .$
(2.3)

Let

$f ( t ) = 4 arctan t - t 4 - 1 2 [ p t 2 + ( 1 - p ) ] [ ( 1 - p ) t 2 + p ] - π ,$
(2.4)

(2.5)
$lim t → + ∞ f ( t ) = π - 1 2 p ( 1 - p )$
(2.6)

and

$f ′ ( t ) = f 1 ( t ) ( t 2 + 1 ) [ p ( 1 - p ) t 4 + ( 2 p 2 - 2 p + 1 ) t 2 + p ( 1 - p ) ] 2 ,$
(2.7)

where

$f 1 ( t ) = 4 p 2 ( 1 - p ) 2 t 8 - ( 2 p 2 - 2 p + 1 ) t 7 + 8 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t 6 + ( 2 p 2 - 2 p - 1 ) t 5 + 4 ( 6 p 4 - 1 2 p 3 + 1 0 p 2 - 4 p + 1 ) t 4 + ( 2 p 2 - 2 p - 1 ) t 3 + 8 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t 2 - ( 2 p 2 - 2 p + 1 ) t + 4 p 2 ( 1 - p ) 2 .$
(2.8)

Note that

$f 1 ( 1 ) = 0 ,$
(2.9)
$lim t → + ∞ f 1 ( t ) = + ∞ ,$
(2.10)
$f ′ 1 ( t ) = 3 2 p 2 ( 1 - p ) 2 t 7 - 7 ( 2 p 2 - 2 p + 1 ) t 6 + 4 8 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t 5 + 5 ( 2 p 2 - 2 p - 1 ) t 4 + 1 6 ( 6 p 4 - 1 2 p 3 + 1 0 p 2 - 4 p + 1 ) t 3 + 3 ( 2 p 2 - 2 p - 1 ) t 2 + 1 6 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t - ( 2 p 2 - 2 p + 1 ) ,$
$f 1 ′ ( 1 ) = 0 ,$
(2.11)
$lim t → + ∞ f 1 ′ ( t ) = + ∞ .$
(2.12)

Let $f 2 ( t ) = f 1 ″ ( t ) ∕2$, $f 3 ( t ) = f 2 ′ ( t ) ∕3$, $f 4 ( t ) = f 3 ′ ( t ) ∕4$, $f 5 ( t ) = f 4 ′ ( t ) ∕5$, $f 6 ( t ) = f 5 ′ ( t ) ∕6$ and $f 7 ( t ) = f 6 ′ ( t ) ∕7$. Then, simple computations lead to

$f 2 ( t ) = 1 1 2 p 2 ( 1 - p ) 2 t 6 - 2 1 ( 2 p 2 - 2 p + 1 ) t 5 + 1 2 0 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t 4 + 1 0 ( 2 p 2 - 2 p - 1 ) t 3 + 2 4 ( 6 p 4 - 1 2 p 3 + 1 0 p 2 - 4 p + 1 ) t 2 + 3 ( 2 p 2 - 2 p - 1 ) t + 8 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) ,$
$f 2 ( 1 ) = - 2 ( 2 4 p 2 - 2 4 p + 5 ) ,$
(2.13)
$lim t → + ∞ f 2 ( t ) = + ∞ ,$
(2.14)
$f 3 ( t ) = 2 2 4 p 2 ( 1 - p ) 2 t 5 - 3 5 ( 2 p 2 - 2 p + 1 ) t 4 + 1 6 0 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t 3 + 1 0 ( 2 p 2 - 2 p - 1 ) t 2 + 1 6 ( 6 p 4 - 1 2 p 3 + 1 0 p 2 - 4 p + 1 ) t + ( 2 p 2 - 2 p - 1 ) ,$
$f 3 ( 1 ) = - 6 ( 2 4 p 2 - 2 4 p + 5 ) ,$
(2.15)
$lim t → + ∞ f 3 ( t ) = + ∞ ,$
(2.16)
$f 4 ( t ) = 2 8 0 p 2 ( 1 - p ) 2 t 4 - 3 5 ( 2 p 2 - 2 p + 1 ) t 3 + 1 2 0 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t 2 + 5 ( 2 p 2 - 2 p - 1 ) t + 4 ( 6 p 4 - 1 2 p 3 + 1 0 p 2 - 4 p + 1 ) ,$
$f 4 ( 1 ) = 4 ( 1 6 p 4 - 3 2 p 3 - 2 5 p 2 + 4 1 p - 9 ) ,$
(2.17)
$lim t → + ∞ f 4 ( t ) = + ∞ ,$
(2.18)
$f 5 ( t ) = 2 2 4 p 2 ( 1 - p ) 2 t 3 - 2 1 ( 2 p 2 - 2 p + 1 ) t 2 + 4 8 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) t + ( 2 p 2 - 2 p - 1 ) ,$
$f 5 ( 1 ) = 2 ( 6 4 p 4 - 1 2 8 p 3 + 2 0 p 2 + 4 4 p - 1 1 ) ,$
(2.19)
$lim t → + ∞ f 5 ( t ) = + ∞ ,$
(2.20)
$f 6 ( t ) = 1 1 2 p 2 ( 1 - p ) 2 t 2 - 7 ( 2 p 2 - 2 p + 1 ) t + 8 p ( 1 - p ) ( 2 p 2 - 2 p + 1 ) ,$
(2.21)
$f 6 ( 1 ) = 9 6 p 4 - 1 9 2 p 3 + 7 4 p 2 + 2 2 p - 7 ,$
(2.22)
$lim t → + ∞ f 6 ( t ) = + ∞ ,$
(2.23)
$f 7 ( t ) = 3 2 p 2 ( 1 - p ) 2 t - ( 2 p 2 - 2 p + 1 )$
(2.24)

and

$f 7 ( 1 ) = 3 2 p 4 - 6 4 p 3 + 3 0 p 2 + 2 p - 1 .$
(2.25)

We divide the proof into two cases.

Case 1. $p=λ= ( 1 - 1 - 2 ∕ π ) ∕2$. Then equations (2.6), (2.13), (2.15), (2.17), (2.19), (2.22) and (2.25) become

$lim t → + ∞ f ( t ) = 0 ,$
(2.26)
$f 2 ( 1 ) = - 2 ( 5 π - 1 2 ) π < 0 ,$
(2.27)
$f 3 ( 1 ) = - 6 ( 5 π - 1 2 ) π < 0 ,$
(2.28)
$f 4 ( 1 ) = - 2 ( 1 8 π 2 - 4 1 π - 8 ) π 2 < 0 ,$
(2.29)
$f 5 ( 1 ) = - 2 ( 1 1 π 2 - 2 2 π - 1 6 ) π 2 < 0 ,$
(2.30)
$f 6 ( 1 ) = - 7 π 2 - 1 1 π - 2 4 π 2 < 0$
(2.31)

and

$f 7 ( 1 ) = π + 8 - π 2 π 2 > 0 .$
(2.32)

From (2.24), we clearly see that f7(t) is strictly increasing in [1, +∞), and then (2.32) leads to the conclusion that f7(t) > 0 for t [1, +∞). Thus, f6(t) is strictly increasing in [1, +∞).

It follows from (2.23) and (2.31) together with the monotonicity of f6(t) that there exists t1> 1 such that f6(t) < 0 for t (1, t1) and f6(t) > 0 for t (t1, +∞). Thus, f5(t) is strictly decreasing in [1, t1] and strictly increasing in [t1, +∞).

From (2.20) and (2.30), together with the piecewise monotonicity of f5(t), we clearly see that there exists t2> t1> 1 such that f4(t) is strictly decreasing in [1, t2] and strictly increasing in [t2, +∞). Then, equation (2.18) and inequality (2.29) lead to the conclusion that there exists t3> t2> 1 such that f3(t) is strictly decreasing in [1, t3] and strictly increasing in [t3, +∞).

It follows from (2.16) and (2.28) together with the piecewise monotonicity of f3(t) we conclude that there exists t4> t3> 1 such that f2(t) is strictly decreasing in [1, t4] and strictly increasing in [t4, +∞). Then, equation (2.14) and inequality (2.27) lead to the conclusion that there exists t5> t4> 1 such that $f 1 ′ ( t )$ is strictly decreasing in [1, t5] and strictly increasing in [t5, +∞).

From equations (2.11) and (2.12), together with the piecewise monotonicity of $f 1 ′ ( t )$, we know that there exists t6> t5> 1 such that f1(t) is strictly decreasing in [1, t6] and strictly increasing in [t6, +∞). Then, equations (2.7)-(2.10) lead to the conclusion that there exists t7> t6> 1 such that f(t) is strictly decreasing in [1, t7] and strictly increasing in [t7, +∞).

Therefore, inequality (2.1) follows from equations (2.3)-(2.5) and (2.26) together with the piecewise monotonicity of f(t).

Case 2. $p=μ= ( 6 - 6 ) ∕12$. Then, equations (2.13), (2.15), (2.17), (2.19) and (2.21) become

(2.33)
(2.34)
$f 4 ( 1 ) = 1 7 1 8 > 0 ,$
(2.35)
$f 5 ( 1 ) = 1 7 9 > 0$
(2.36)

and

$f 6 ( t ) = 1 3 6 ( 1 7 5 t 2 - 1 4 7 t + 3 5 ) > 0$
(2.37)

for t > 1.

From inequality (2.37), we know that f5(t) is strictly increasing in [1, +∞), and then inequality (2.36) leads to the conclusion that f5(t) > 0 for t [1, +∞). Thus, f4(t) is strictly increasing in [1, +∞).

It follows from inequality (2.35) and the monotonicity of f4(t) that f3(t) is strictly increasing in [1, +∞).

Therefore, inequality (2.2) follows easily from equations (2.3)-(2.5), (2.7), (2.9), (2.11), (2.33), and (2.34) together with the monotonicity of f3(t).

Next, we prove that $λ= ( 1 - 1 - 2 ∕ π ) ∕2$ is the best-possible parameter such that inequality (2.1) holds for all a, b > 0 with ab. In fact, if $( 1 - 1 - 2 ∕ π ) ∕2=λ, then equation (2.6) leads to

$lim t → + ∞ f ( t ) = π - 1 2 p ( 1 - p ) > 0 .$
(2.38)

Inequality (2.38) implies that there exists T = T(p) > 1 such that

$f ( t ) >0$
(2.39)

for t (T, +∞).

From equations (2.3) and (2.4), together with inequality (2.39), we clearly see that P(a, b) < H(pa + (1 - p)b, pb + (1 - p)a) for a/b (T2, +∞).

Finally, we prove that $μ= ( 6 - 6 ) ∕12$ is the best-possible parameter such that inequality (2.2) holds for all a, b > 0 with ab. In fact, if $0, then equation (2.13) leads to

$f 2 ( 1 ) = - 2 ( 2 4 p 2 - 2 4 p + 5 ) < 0 .$
(2.40)

Inequality (2.40) implies that there exists δ = δ (p) > 0 such that

$f 2 ( t ) <0$
(2.41)

for t (1, 1 + δ).

Therefore, P(a, b) > H(pa + (1 - p)b, pb + (1 - p)a) for a/b (1, (1 + δ)2) follows from equations (2.3)-(2.5), (2.7), (2.9), and (2.11) together with inequality (2.41).

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## Acknowledgements

This study is partly supported by the Natural Science Foundation of China (Grant no. 11071069), the Natural Science Foundation of Hunan Province (Grant no. 09JJ6003), and the Innovation Team Foundation of the Department of Education of Zhejiang Province(Grant no. T200924).

## Author information

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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 in this paper. M-KW carried out the proof of inequality (2.1) in this paper. Z-KW carried out the proof of inequality (2.2) in this paper. All authors read and approved the final manuscript.

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Chu, Y., Wang, M. & Wang, Z. A best-possible double inequality between Seiffert and harmonic means. J Inequal Appl 2011, 94 (2011). https://doi.org/10.1186/1029-242X-2011-94 