Open Access

Optimal bounds for the Neuman-Sándor mean in terms of the first Seiffert and quadratic means

Journal of Inequalities and Applications20132013:552

https://doi.org/10.1186/1029-242X-2013-552

Received: 12 July 2013

Accepted: 18 October 2013

Published: 22 November 2013

Abstract

In this paper, we find the least value α and the greatest value β such that the double inequality

P α ( a , b ) Q 1 α ( a , b ) < M ( a , b ) < P β ( a , b ) Q 1 β ( a , b )

holds true for all a , b > 0 with a b , where P ( a , b ) , M ( a , b ) and Q ( a , b ) are the first Seiffert, Neuman-Sándor and quadratic means of a and b, respectively.

MSC:26E60.

Keywords

Neuman-Sándor meanfirst Seiffert meanquadratic mean

1 Introduction

Let u, v and w be the bivariate means such that u ( a , b ) < w ( a , b ) < v ( a , b ) for all a , b > 0 with a b . The problems of finding the best possible parameters α and β such that the inequalities α u ( a , b ) + ( 1 α ) v ( a , b ) < w ( a , b ) < β u ( a , b ) + ( 1 β ) v ( a , b ) and u α ( a , b ) v 1 α ( a , b ) < w ( a , b ) < u β ( a , b ) v 1 β ( a , b ) hold for all a , b > 0 with a b have attracted the interest of many mathematicians.

For a , b > 0 with a b , the first Seiffert mean P ( a , b ) [1], the Neuman-Sándor mean M ( a , b ) [2], the quadratic mean Q ( a , b ) are defined by
P ( a , b ) = a b 4 arctan ( a / b ) π , M ( a , b ) = a b 2 sinh 1 ( a b a + b ) , Q ( a , b ) = a 2 + b 2 2 ,
(1.1)

respectively. In here, sinh 1 ( x ) = log ( x + x 2 + 1 ) is the inverse hyperbolic sine function.

Recently, the means P, M and Q have been the subject of intensive research. In particular, many remarkable inequalities for these means can be found in the literature [314]. The first Seiffert mean P ( a , b ) can be rewritten as (see [[2], Eq. (2.4)])
P ( a , b ) = a b 2 arcsin [ ( a b ) / ( a + b ) ] .
(1.2)
Let H ( a , b ) = 2 a b / ( a + b ) , L ( a , b ) = ( b a ) / ( log b log a ) , A ( a , b ) = ( a + b ) / 2 , T ( a , b ) = ( a b ) / [ 2 arctan ( ( a b ) / ( a + b ) ) ] and C ( a , b ) = ( a 2 + b 2 ) / ( a + b ) be the harmonic, logarithmic, arithmetic, second Seiffert and contra-harmonic means of a and b, respectively. Then it is known that the inequalities
H ( a , b ) < L ( a , b ) < P ( a , b ) < A ( a , b ) < M ( a , b ) < T ( a , b ) < Q ( a , b ) < C ( a , b )

hold for all a , b > 0 with a b .

Neuman and Sándor [2, 15] proved that the inequalities
π 4 log ( 1 + 2 ) T ( a , b ) < M ( a , b ) < A ( a , b ) log ( 1 + 2 ) , 2 T 2 ( a , b ) Q 2 ( a , b ) < M ( a , b ) < T 2 ( a , b ) Q ( a , b ) , H ( T ( a , b ) , A ( a , b ) ) < M ( a , b ) < L ( A ( a , b ) , Q ( a , b ) ) , T ( a , b ) > H ( M ( a , b ) , Q ( a , b ) ) , M ( a , b ) < A 2 ( a , b ) P ( a , b ) , A 2 / 3 ( a , b ) Q 1 / 3 ( a , b ) < M ( a , b ) < 2 A ( a , b ) + Q ( a , b ) 3 , A ( a , b ) T ( a , b ) < M ( a , b ) < A 2 ( a , b ) + T 2 ( a , b ) , A ( x , y ) A ( 1 x , 1 y ) < M ( x , y ) M ( 1 x , 1 y ) < T ( x , y ) T ( 1 x , 1 y ) , 1 A ( 1 x , 1 y ) 1 A ( x , y ) < 1 M ( 1 x , 1 y ) 1 M ( x , y ) < 1 T ( 1 x , 1 y ) 1 T ( x , y ) , A ( x , y ) A ( 1 x , 1 y ) < M ( x , y ) M ( 1 x , 1 y ) < T ( x , y ) T ( 1 x , 1 y )

hold for all a , b > 0 and x , y ( 0 , 1 / 2 ] with a b and x y .

Li et al. [16] proved that the double inequality L p 0 ( a , b ) < M ( a , b ) < L 2 ( a , b ) holds for all a , b > 0 with a b , where L p ( a , b ) = [ ( b p + 1 a p + 1 ) / ( ( p + 1 ) ( b a ) ) ] 1 / p ( p 1 , 0 ), L 0 ( a , b ) = 1 / e ( b b / a a ) 1 / ( b a ) and L 1 ( a , b ) = ( b a ) / ( log b log a ) is the p th generalized logarithmic mean of a and b, and p 0 = 1.843 is the unique solution of the equation ( p + 1 ) 1 / p = 2 log ( 1 + 2 ) .

In [13], Neuman proved that the double inequalities
Q α ( a , b ) A 1 α ( a , b ) < M ( a , b ) < Q β ( a , b ) A 1 β ( a , b )
(1.3)
and
C λ ( a , b ) A 1 λ ( a , b ) < M ( a , b ) < C μ ( a , b ) A 1 μ ( a , b )
(1.4)

hold for all a , b > 0 with a b if α 1 / 3 , β 2 [ log ( 2 + 2 ) log 3 ] / log 2 , λ 1 / 6 and μ [ log ( 2 + 2 ) log 3 ] / log 2 .

Jiang and Qi [17, 18] gave the best possible parameters α, β, t 1 and t 2 in ( 0 , 1 / 2 ) such that the inequalities
Q ( α a + ( 1 α ) b , α b + ( 1 α ) a ) < M ( a , b ) < Q ( β a + ( 1 β ) b , β b + ( 1 β ) a ) , Q t 1 , p ( a , b ) < M ( a , b ) < Q t 2 , p ( a , b )

hold for all a , b > 0 with a b and p 1 / 2 , where Q t , p ( a , b ) = C p ( t a + ( 1 t ) b , t b + ( 1 t ) a ) A 1 p ( a , b ) .

Inspired by inequalities (1.3) and (1.4), in this paper, we present the optimal upper and lower bounds for the Neuman-Sándor mean M ( a , b ) in terms of the geometric convex combinations of the first Seiffert mean P ( a , b ) and the quadratic mean Q ( a , b ) . All numerical computations are carried out using Mathematica software.

2 Lemmas

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

Lemma 2.1 The double inequality
x + x 3 3 2 x 5 15 < 1 + x 2 sinh 1 ( x ) < x + x 3 3 2 x 5 15 + 8 x 7 105
(2.1)

holds for x ( 0 , 1 ) .

Proof To show inequality (2.1), it suffices to prove that
ω 1 ( x ) = 1 + x 2 sinh 1 ( x ) ( x + x 3 3 2 x 5 15 ) > 0
(2.2)
and
ω 2 ( x ) = 1 + x 2 sinh 1 ( x ) ( x + x 3 3 2 x 5 15 + 8 x 7 105 ) < 0
(2.3)

for x ( 0 , 1 ) .

From the expressions of ω 1 ( x ) and ω 2 ( x ) , we get
ω 1 ( 0 ) = ω 2 ( 0 ) = 0 ,
(2.4)
ω 1 ( x ) = x ω 1 ( x ) 1 + x 2 , ω 2 ( x ) = x ω 2 ( x ) 1 + x 2 ,
(2.5)
where
ω 1 ( x ) = sinh 1 ( x ) ( x 2 x 3 3 ) 1 + x 2 , ω 2 ( x ) = sinh 1 ( x ) ( x 2 x 3 3 + 8 x 5 15 ) 1 + x 2 , ω 1 ( 0 ) = ω 2 ( 0 ) = 0 ,
(2.6)
ω 1 ( x ) = 8 x 4 3 1 + x 2 > 0
(2.7)
and
ω 2 ( x ) = 16 x 6 5 1 + x 2 < 0
(2.8)

for x ( 0 , 1 ) .

Therefore, inequality (2.2) follows from (2.4)-(2.7), and inequality (2.3) follows from (2.4)-(2.6) and (2.8). □

Lemma 2.2 The inequality
x 3 1 + x 2 > [ sinh 1 ( x ) ] 3

holds for x ( 0 , 1 ) .

Proof Let x ( 0 , 1 ) , then from (1.3) we have
M ( 1 + x , 1 x ) > A 2 / 3 ( 1 + x , 1 x ) Q 1 / 3 ( 1 + x , 1 x ) .
(2.9)

Therefore, Lemma 2.2 follows from (2.9). □

Lemma 2.3 The inequality
1 x 2 arcsin ( x ) > x x 3 3 x 5 3
(2.10)
holds for x ( 0 , 0.7 ) , and the inequality
1 x 2 arcsin ( x ) < x x 3 3 2 x 5 15
(2.11)

holds for x ( 0 , 1 ) , where arcsin ( x ) is the inverse sine function.

Proof Let
φ 1 ( x ) = 1 x 2 arcsin ( x ) x + x 3 3 + x 5 3 ,
(2.12)
φ 2 ( x ) = 1 x 2 arcsin ( x ) x + x 3 3 + 2 x 5 15 .
(2.13)
Then simple computations lead to
φ 1 ( 0 ) = φ 2 ( 0 ) = 0 ,
(2.14)
φ 1 ( x ) = x φ 1 ( x ) 1 x 2 , φ 2 ( x ) = x φ 2 ( x ) 1 x 2 ,
(2.15)
where
φ 1 ( x ) = ( x + 5 x 3 3 ) 1 x 2 arcsin ( x ) , φ 2 ( x ) = ( x + 2 x 3 3 ) 1 x 2 arcsin ( x ) .
Note that
φ 1 ( 0 ) = φ 2 ( 0 ) = 0 , φ 1 ( 0.7 ) = 0.1327 ,
(2.16)
φ 1 ( x ) = x 2 ( 9 20 x 2 ) 3 1 x 2 ,
(2.17)
φ 2 ( x ) = 8 x 4 3 1 x 2 < 0
(2.18)

for x ( 0 , 1 ) .

From (2.17) we clearly see that φ 1 ( x ) is strictly increasing on ( 0 , 3 5 / 10 ] and strictly decreasing on [ 3 5 / 10 , 0.7 ) . This in conjunction with (2.16) implies that
φ 1 ( x ) > 0
(2.19)

for x ( 0 , 0.7 ) .

Therefore, inequality (2.10) follows from (2.12), (2.14), (2.15) and (2.19), and inequality (2.11) follows from (2.12) and (2.14)-(2.16) together with (2.18). □

Lemma 2.4 Let
Φ ( x ) = 1 1 + x 2 sinh 1 ( x ) 1 x ( 1 + x 2 ) .
Then the inequality
Φ ( x ) > 2 x 3 34 x 3 45 + 754 x 5 945 x 7
(2.20)
holds for x ( 0 , 0.7 ) , and
Φ ( x ) < 2 x 3 34 x 3 45 + 4 x 5 5
(2.21)

holds for x ( 0 , 1 ) .

Proof To show inequalities (2.20) and (2.21), it suffices to prove that
ϕ 1 ( x ) : = x ( 1 + x 2 ) sinh 1 ( x ) [ Φ ( x ) ( 2 x 3 34 x 3 45 + 754 x 5 945 x 7 ) ] = x 1 + x 2 sinh 1 ( x ) x ( 1 + x 2 ) sinh 1 ( x ) ( 2 x 3 34 x 3 45 + 754 x 5 945 x 7 ) > 0
(2.22)
for x ( 0 , 0.7 ) , and
ϕ 2 ( x ) : = x ( 1 + x 2 ) sinh 1 ( x ) [ Φ ( x ) ( 2 x 3 34 x 3 45 + 4 x 5 5 ) ] = x 1 + x 2 sinh 1 ( x ) x ( 1 + x 2 ) sinh 1 ( x ) ( 2 x 3 34 x 3 45 + 4 x 5 5 ) < 0
(2.23)

for x ( 0 , 1 ) .

From the expressions of ϕ 1 ( x ) and ϕ 2 ( x ) , one has
ϕ 1 ( 0 ) = ϕ 2 ( 0 ) = 0 ,
(2.24)
ϕ 1 ( x ) = x 945 1 + x 2 ϕ 1 ( x ) , ϕ 2 ( x ) = 2 x 45 1 + x 2 ϕ 2 ( x ) ,
(2.25)
where
ϕ 1 ( x ) = x ( 1 , 260 + 84 x 2 40 x 4 + 191 x 6 + 945 x 8 ) ϕ 1 ( x ) = 2 ( 630 168 x 2 + 120 x 4 764 x 6 4 , 725 x 8 ) 1 + x 2 sinh 1 ( x ) ,
(2.26)
ϕ 2 ( x ) = x ( 18 x 6 + x 4 2 x 2 30 ) ϕ 2 ( x ) = + 2 ( 15 4 x 2 + 3 x 4 + 72 x 6 ) 1 + x 2 sinh 1 ( x ) .
(2.27)
Note that
630 168 x 2 + 120 x 4 764 x 6 4 , 725 x 8 > 630 168 × ( 0.7 ) 2 764 × ( 0.7 ) 6 4 , 725 × ( 0.7 ) 8 = 185.4 > 0
(2.28)

for x ( 0 , 0.7 ) .

Lemma 2.1 and equations (2.26)-(2.28) lead to
ϕ 1 ( x ) > x ( 1 , 260 + 84 x 2 40 x 4 + 191 x 6 + 945 x 8 ) 2 ( 630 168 x 2 + 120 x 4 764 x 6 4 , 725 x 8 ) ( x + x 3 3 2 x 5 15 + 8 x 7 105 ) = x 7 105 ( 157 , 311 + 1 , 151 , 003 x 2 + 307 , 438 x 4 120 , 076 x 6 + 75 , 600 x 8 ) > 0
(2.29)
for x ( 0 , 0.7 ) , and
ϕ 2 ( x ) > x ( 18 x 6 + x 4 2 x 2 30 ) + 2 ( 15 4 x 2 + 3 x 4 + 72 x 6 ) ( x + x 3 3 2 x 5 15 ) = x 5 15 ( 5 + 2 , 476 x 2 + 708 x 4 288 x 6 ) > 0
(2.30)

for x ( 0 , 1 ) .

Therefore, inequality (2.22) follows from (2.24), (2.25) and (2.29), and inequality (2.23) follows from (2.24), (2.25) and (2.30). □

Lemma 2.5 Let
ϒ ( x ) = 1 x ( 1 + x 2 ) 1 1 x 2 arcsin ( x ) .
Then the inequality
ϒ ( x ) > 4 x 3 + 34 x 3 45 3 x 5 2
(2.31)
holds for x ( 0 , 0.7 ) , and
ϒ ( x ) < 4 x 3 + 34 x 3 45 8 x 5 9
(2.32)

holds for x ( 0 , 1 ) .

Proof Let
ϵ 1 ( x ) : = x ( 1 + x 2 ) 1 x 2 arcsin ( x ) [ ϒ ( x ) + ( 4 x 3 34 x 3 45 + 3 x 5 2 ) ] = 1 x 2 arcsin ( x ) x ( 1 + x 2 ) + x ( 1 + x 2 ) 1 x 2 arcsin ( x ) ( 4 x 3 34 x 3 45 + 3 x 5 2 )
(2.33)
and
ϵ 2 ( x ) : = x ( 1 + x 2 ) 1 x 2 arcsin ( x ) [ ϒ ( x ) + ( 4 x 3 34 x 3 45 + 8 x 5 9 ) ] = 1 x 2 arcsin ( x ) x ( 1 + x 2 ) + x ( 1 + x 2 ) 1 x 2 arcsin ( x ) ( 4 x 3 34 x 3 45 + 8 x 5 9 ) .
(2.34)
An easy calculation gives rise to
ϵ 1 ( 0 ) = ϵ 2 ( 0 ) = 0 ,
(2.35)
ϵ 1 ( x ) = x 90 ( 1 x 2 ) ϵ 1 ( x ) , ϵ 2 ( x ) = x 45 ( 1 x 2 ) ϵ 2 ( x ) ,
(2.36)
where
ϵ 1 ( x ) = x ( 150 202 x 2 15 x 4 68 x 6 + 135 x 8 ) ϵ 1 ( x ) = + ( 150 152 x 2 + 142 x 4 + 611 x 6 1 , 215 x 8 ) 1 x 2 arcsin ( x ) ,
(2.37)
ϵ 2 ( x ) = x ( 1 x 2 ) ( 75 26 x 2 6 x 4 40 x 6 ) ϵ 2 ( x ) = + ( 75 76 x 2 94 x 4 + 278 x 6 360 x 8 ) 1 x 2 arcsin ( x ) .
(2.38)
Note that
150 152 x 2 + 142 x 4 + 611 x 6 1 , 215 x 8 > 150 152 × ( 0.7 ) 2 1 , 215 × ( 0.7 ) 8 = 5.477 > 0
(2.39)

for x ( 0 , 0.7 ) .

It follows from (2.10), (2.37) and (2.39) that
ϵ 1 ( x ) > x ( 150 202 x 2 15 x 4 68 x 6 + 135 x 8 ) + ( 150 152 x 2 + 142 x 4 + 611 x 6 1 , 215 x 8 ) ( x x 3 3 x 5 3 ) = x 5 3 [ 1 , 183 4 + 709 ( 1 4 x 4 ) + 2 , 047 x 2 ( 1 2 x 2 ) + 604 x 6 + 1 , 215 x 8 ] > 0
(2.40)

for x ( 0 , 0.7 ) .

We claim that
ϵ 2 ( x ) < 0
(2.41)
for x ( 0 , 1 ) . Indeed, let q ( x ) = 75 76 x 2 94 x 4 + 278 x 6 360 x 8 , then q ( 0.8009 ) = 0.000171  , q ( 0.80091 ) = 0.00356 and
q ( x ) = 4 x [ 38 + 10 , 759 x 2 320 + 720 x 2 ( x 2 139 480 ) ] < 0
for x ( 0 , 1 ) . Therefore, there exists unique x 0 = 0.80090 ( 0 , 1 ) such that q ( x ) > 0 for x ( 0 , x 0 ) and q ( x ) 0 for [ x 0 , 1 ) . This in conjunction with (2.11) and (2.38) leads to
ϵ 2 ( x ) < x ( 1 x 2 ) ( 75 26 x 2 6 x 4 40 x 6 ) + ( 75 76 x 2 94 x 4 + 278 x 6 360 x 8 ) ( x x 3 3 2 x 5 15 ) = 2 x 5 15 [ 1 , 897 , 305 , 741 27 , 436 , 644 + 2 , 619 ( x 2 2 , 651 5 , 238 ) 2 + 2 x 4 ( 1 x 2 ) ( 491 + 180 x 2 ) ] < 0

for x ( 0 , x 0 ) and ϵ 2 ( x ) x ( 1 x 2 ) ( 75 26 x 2 6 x 4 40 x 6 ) < 0 for x [ x 0 , 1 ) .

Therefore, inequality (2.31) follows from (2.33), (2.35), (2.36) and (2.40), and inequality (2.32) follows from (2.33)-(2.36) and (2.41). □

Lemma 2.6 Let
μ ( x ) = 1 + 3 x 2 ( x + x 3 ) 2 1 ( 1 + x 2 ) [ sinh 1 ( x ) ] 2 x ( 1 + x 2 ) 3 / 2 sinh 1 ( x ) .

Then μ ( x ) < 0.2 for x [ 0.7 , 1 ) .

Proof Let
μ 1 ( x ) = 1 x 2 1 [ sinh 1 ( x ) ] 2 , μ 2 ( x ) = 2 1 + x 2 x sinh 1 ( x ) .
Then
μ ( x ) = μ 1 ( x ) 1 + x 2 + μ 2 ( x ) ( 1 + x 2 ) 3 / 2 .
(2.42)
Lemma 2.2 together with x > sinh 1 ( x ) gives μ 1 ( x ) < 0 and
μ 1 ( x ) = 2 x 3 [ sinh 1 ( x ) ] 3 [ x 3 1 + x 2 ( sinh 1 ( x ) ) 3 ] > 0
for x ( 0 , 1 ) . This in turn implies that
[ μ 1 ( x ) 1 + x 2 ] = μ 1 ( x ) ( 1 + x 2 ) 2 x μ 1 ( x ) ( 1 + x 2 ) 2 > 0
(2.43)

for x ( 0 , 1 ) .

On the other hand, from the expression of μ 2 ( x ) , we get
μ 2 ( 1 ) = 0.2796 > 0 ,
(2.44)
μ 2 ( x ) = 2 x ( 1 + x 2 ) 3 / 2 + μ 2 ( x ) [ sinh 1 ( x ) ] 2 ,
(2.45)
where
μ 2 ( x ) = x 1 + x 2 sinh 1 ( x ) ,
(2.46)
μ 2 ( 0 ) = 0 ,
(2.47)
μ 2 ( x ) = x 2 ( 1 + x 2 ) 3 / 2 < 0
(2.48)

for x ( 0 , 1 ) .

From (2.44)-(2.48) we clearly see that μ 2 ( x ) < 0 and μ 2 ( x ) > 0 for x ( 0 , 1 ) . This in turn implies that
[ μ 2 ( x ) ( 1 + x 2 ) 3 / 2 ] = μ 2 ( x ) ( 1 + x 2 ) 3 / 2 3 x 1 + x 2 μ 2 ( x ) ( 1 + x 2 ) 3 < 0
(2.49)

for x ( 0 , 1 ) .

Equation (2.42) and inequalities (2.43) and (2.49) lead to the conclusion that
μ ( x ) μ 1 ( 1 ) 2 + μ 2 ( 0.7 ) [ 1 + ( 0.7 ) 2 ] 3 / 2 = 0.167 < 0.2

for x [ 0.7 , 1 ) . □

Lemma 2.7 Let
ν ( x ) = 1 + 3 x 2 ( x + x 3 ) 2 + 1 ( 1 x 2 ) arcsin 2 ( x ) x ( 1 x 2 ) 3 / 2 arcsin ( x ) .

Then ν ( x ) < 1.48 for x [ 0.7 , 1 ) .

Proof Differentiating ν ( x ) yields
ν ( x ) = ( x + x 3 ) 3 arcsin ( x ) ν 1 ( x ) + ( 1 x 2 ) ν 2 ( x ) x 3 ( 1 x 2 ) 5 / 2 ( 1 + x 2 ) 3 arcsin 3 ( x ) ,
(2.50)
where
ν 1 ( x ) = 3 x 1 x 2 ( 1 + 2 x 2 ) arcsin ( x ) ,
(2.51)
ν 2 ( x ) = 2 ( 1 + 3 x 2 + 6 x 4 ) [ 1 x 2 arcsin ( x ) ] 3 2 ( x + x 3 ) 3 .
(2.52)
Equation (2.51) leads to
ν 1 ( 0.7 ) = 0.03558 ,
(2.53)
ν 1 ( x ) = 2 8 x 2 4 x 1 x 2 arcsin ( x ) 1 x 2 < 0
(2.54)

for x [ 0.7 , 1 ) .

Therefore,
ν 1 ( x ) < 0
(2.55)

for x [ 0.7 , 1 ) follows from (2.53) and (2.54).

It follows from (2.52) and (2.11) that
ν 2 ( x ) < 2 ( 1 + 3 x 2 + 6 x 4 ) ( x x 3 3 ) 3 2 ( x + x 3 ) 3 = 2 x 5 27 ( 27 9 x 2 + 163 x 4 51 x 6 + 6 x 8 ) < 0
(2.56)

for x [ 0.7 , 1 ) .

Equation (2.50) together with inequalities (2.55) and (2.56) leads to the conclusion that ν ( x ) is strictly decreasing on [ 0.7 , 1 ) . This in turn implies that
ν ( x ) ν ( 0.7 ) = 1.48798 < 1.48

for x [ 0.7 , 1 ) . □

Lemma 2.8 Let λ 0 = [ 2 log ( log ( 1 + 2 ) ) + log 2 ] / [ 2 log π log 2 ] = 0.2760  , and Θ ( x ) = Φ ( x ) + λ 0 ϒ ( x ) , where Φ ( x ) and ϒ ( x ) are defined as in Lemmas 2.4 and 2.5, respectively. Then the function Θ ( x ) is strictly decreasing on [ 0.7 , 1 ) .

Proof Let μ ( x ) and ν ( x ) be defined as in Lemmas 2.6 and 2.7, respectively. Then differentiating Θ ( x ) yields
Θ ( x ) = Φ ( x ) + λ 0 ϒ ( x ) = μ ( x ) + λ 0 ν ( x ) < 0.2 1.48 λ 0 = 0.208 < 0

for x [ 0.7 , 1 ) . This in turn implies that Θ ( x ) is strictly decreasing on [ 0.7 , 1 ) . □

3 Main result

Theorem 3.1 The double inequality
P α ( a , b ) Q 1 α ( a , b ) < M ( a , b ) < P β ( a , b ) Q 1 β ( a , b )

holds for all a , b > 0 with a b if and only if α 1 / 2 and β [ 2 log ( log ( 1 + 2 ) ) + log 2 ] / [ 2 log π log 2 ] = 0.2760  .

Proof Since P ( a , b ) , M ( a , b ) and Q ( a , b ) are symmetric and homogeneous of degree 1, without loss of generality, we assume that a > b . Let p ( 0 , 1 ) , λ 0 = [ 2 log ( log ( 1 + 2 ) ) + log 2 ] / [ 2 log π log 2 ] and x = ( a b ) / ( a + b ) . Then x ( 0 , 1 ) ,
P ( a , b ) A ( a , b ) = x arcsin ( x ) , M ( a , b ) A ( a , b ) = x sinh 1 ( x ) , Q ( a , b ) A ( a , b ) = 1 + x 2 , log [ Q ( a , b ) ] log [ M ( a , b ) ] log [ Q ( a , b ) ] log [ P ( a , b ) ] = log ( 1 + x 2 ) 2 log x + 2 log [ sinh 1 ( x ) ] log ( 1 + x 2 ) 2 log x + 2 log [ arcsin ( x ) ]
(3.1)
and
lim x 0 + log ( 1 + x 2 ) 2 log x + 2 log [ sinh 1 ( x ) ] log ( 1 + x 2 ) 2 log x + 2 log [ arcsin ( x ) ] = 1 2 ,
(3.2)
lim x 1 log ( 1 + x 2 ) 2 log x + 2 log [ sinh 1 ( x ) ] log ( 1 + x 2 ) 2 log x + 2 log [ arcsin ( x ) ] = λ 0 .
(3.3)
The difference between the convex combination of log [ P ( a , b ) ] , log [ Q ( a , b ) ] and log [ M ( a , b ) ] is given by
p log [ P ( a , b ) ] + ( 1 p ) log [ Q ( a , b ) ] log [ M ( a , b ) ] = p log [ x arcsin ( x ) ] + 1 p 2 log ( 1 + x 2 ) log [ x sinh 1 ( x ) ] : = D p ( x ) .
(3.4)
Equation (3.4) leads to
D p ( 0 + ) = 0 , D p ( 1 ) = log [ 2 log ( 1 + 2 ) ] p log ( π 2 ) , D λ 0 ( 1 ) = 0 ,
(3.5)
D p ( x ) = p 1 x 2 arcsin ( x ) ( 1 p ) x ( 1 + x 2 ) + 1 1 + x 2 sinh 1 ( x ) = Φ ( x ) + p ϒ ( x ) ,
(3.6)

where Φ ( x ) and ϒ ( x ) are defined as in Lemmas 2.4 and 2.5, respectively.

From Lemmas 2.4 and 2.5, we clearly see that
D 1 / 2 ( x ) = Φ ( x ) + 1 2 ϒ ( x ) < 2 x 3 34 x 3 45 + 4 x 5 5 1 2 ( 4 x 3 34 x 3 45 + 8 x 5 9 ) = 16 x 3 45 ( 17 16 x 2 ) < 0
(3.7)
for x ( 0 , 1 ) , and
D λ 0 ( x ) = Φ ( x ) + λ 0 ϒ ( x ) > 2 x 3 34 x 3 45 + 754 x 5 945 x 7 λ 0 ( 4 x 3 34 x 3 45 + 3 x 5 2 ) = x [ 2 ( 1 2 λ 0 ) 3 34 ( 1 λ 0 ) 45 x 2 + ( 754 945 3 λ 0 2 ) x 4 x 6 ] : = x F λ 0 ( x ) > 0
(3.8)

for x ( 0 , 0.7 ) .

Note that
F λ 0 ( 0 ) = 2 ( 1 2 λ 0 ) / 3 > 0 , F λ 0 ( 0.7 ) = 0.00513 > 0
(3.9)
and
F λ 0 ( x ) = 30 [ ( x 2 1 , 508 2 , 835 λ 0 9 , 450 ) 2 + 2 , 224 , 136 + 4 , 052 , 160 λ 0 8 , 037 , 225 λ 0 2 89 , 302 , 500 ] < 0
(3.10)

for x ( 0 , 0.7 ) .

Inequalities (3.8)-(3.10) lead to the conclusion that
D λ 0 ( x ) > 0
(3.11)

for x ( 0 , 0.7 ) .

It follows from Lemma 2.8 and (3.6) that D λ 0 ( x ) is strictly decreasing in [ 0.7 , 1 ) . Then from (3.11) and D λ 0 ( 0.7 ) = 0.0626 together with D λ 0 ( 1 ) = , we know that there exists x ( 0.7 , 1 ) such that D λ 0 ( x ) is strictly increasing on ( 0 , x ] and strictly decreasing on [ x , 1 ) . This in conjunction with (3.5) implies that
D λ 0 ( x ) > 0
(3.12)

for x ( 0 , 1 ) .

Equations (3.4), (3.5), (3.7) and (3.12) lead to the conclusion that
M ( a , b ) < P λ 0 ( a , b ) Q 1 λ 0 ( a , b )
(3.13)
and
M ( a , b ) > P 1 / 2 ( a , b ) Q 1 / 2 ( a , b ) .
(3.14)

Therefore, Theorem 3.1 follows from (3.13) and (3.14) together with the following statements:

  • If α < 1 / 2 , then (3.1) and (3.2) imply that there exists δ 1 ( 0 , 1 ) such that M ( a , b ) < P α ( a , b ) Q 1 α ( a , b ) for all a , b > 0 with ( a b ) / ( a + b ) ( 0 , δ 1 ) .

  • If β > λ 0 , then (3.1) and (3.3) imply that there exists δ 2 ( 0 , 1 ) such that M ( a , b ) > P β ( a , b ) Q 1 β ( a , b ) for all a , b > 0 with ( a b ) / ( a + b ) ( 1 δ 2 , 1 ) .

 □

Declarations

Acknowledgements

This research was supported by the Natural Science Foundation of China under Grants 11071069 and 11171307, and the Natural Science Foundation of Zhejiang Province under Grants LY13H070004 and LY13A010004.

Authors’ Affiliations

(1)
School of Mathematics and Computation Science, Hunan City University
(2)
College of Nursing, Huzhou Teachers College

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© Gong et al.; licensee Springer. 2013

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