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Sharp bounds for Neuman means in terms of one-parameter family of bivariate means

Journal of Inequalities and Applications20142014:468

https://doi.org/10.1186/1029-242X-2014-468

Received: 12 September 2014

Accepted: 12 November 2014

Published: 26 November 2014

Abstract

We present the best possible parameters p 1 , p 2 , p 3 , p 4 , q 1 , q 2 , q 3 , q 4 [ 0 , 1 ] such that the double inequalities G p 1 ( a , b ) < S H A ( a , b ) < G q 1 ( a , b ) , Q p 2 ( a , b ) < S C A ( a , b ) < Q q 2 ( a , b ) , H p 3 ( a , b ) < S A H ( a , b ) < H q 3 ( a , b ) , C p 4 ( a , b ) < S A C ( a , b ) < C q 4 ( a , b ) hold for all a , b > 0 with a b , where S H A , S C A , S A H , S A C are the Neuman means, and G p , Q p , H p , C p are the one-parameter means.

MSC:26E60.

Keywords

Neuman meansone-parameter meanharmonic meangeometric meanarithmetic meanquadratic meancontraharmonic mean

1 Introduction

Let a , b > 0 with a b . Then the Schwab-Borchardt mean S B ( a , b ) [13], and the Neuman means S H A ( a , b ) , S A H ( a , b ) , S C A ( a , b ) , and S A C ( a , b ) [4, 5] of a and b are given by
S B ( a , b ) = b 2 a 2 cos 1 ( a / b ) ( a < b ) , S B ( a , b ) = a 2 b 2 cosh 1 ( a / b ) ( a > b ) , S H A ( a , b ) = S B [ H ( a , b ) , A ( a , b ) ] , S A H ( a , b ) = S B [ A ( a , b ) , H ( a , b ) ] , S C A ( a , b ) = S B [ C ( a , b ) , A ( a , b ) ] , S A C ( a , b ) = S B [ A ( a , b ) , C ( a , b ) ] ,

respectively. Here, cos 1 ( x ) and cosh 1 ( x ) = log ( x + x 2 1 ) are, respectively, the inverse cosine and inverse hyperbolic cosine functions, and H ( a , b ) = 2 a b / ( a + b ) , A ( a , b ) = ( a + b ) / 2 , and C ( a , b ) = ( a 2 + b 2 ) / ( a + b ) are, respectively, the classical harmonic, arithmetic, and contraharmonic means of a and b.

Let v = ( a b ) / ( a + b ) ( 1 , 1 ) , and p ( 0 , ) , q ( 0 , π / 2 ) , r ( 0 , log ( 2 + 3 ) ) , and s ( 0 , π / 3 ) be the parameters such that 1 / cosh ( p ) = cos ( q ) = 1 v 2 and cosh ( r ) = 1 / cosh ( s ) = 1 + v 2 . Then the following explicit formulas were found by Neuman [4]:
S A H ( a , b ) = A ( a , b ) tanh ( p ) p , S H A ( a , b ) = A ( a , b ) sin ( q ) q ,
(1.1)
S C A ( a , b ) = A ( a , b ) sinh ( r ) r , S A C ( a , b ) = A ( a , b ) tan ( s ) s .
(1.2)
Let p [ 0 , 1 ] and N be a bivariate symmetric mean. Then the one-parameter bivariate mean N p ( a , b ) was defined by Neuman [6] as follows:
N p ( a , b ) = N [ ( 1 + p ) 2 a + ( 1 p ) 2 b , ( 1 + p ) 2 b + ( 1 p ) 2 a ] .
(1.3)
Recently, the Neuman means S A H , S H A , S C A , and S A C , and the one-parameter bivariate mean N p have been the subject of intensive research. He et al. [7] found the greatest values α 1 , α 2 [ 0 , 1 / 2 ] , and α 3 , α 4 [ 1 / 2 , 1 ] , and the least values β 1 , β 2 [ 0 , 1 / 2 ] , and β 3 , β 4 [ 1 / 2 , 1 ] such that the double inequalities
H [ α 1 a + ( 1 α 1 ) b , α 1 b + ( 1 α 1 ) a ] < S A H ( a , b ) < H [ β 1 a + ( 1 β 1 ) b , β 1 b + ( 1 β 1 ) a ] , H [ α 2 a + ( 1 α 2 ) b , α 2 b + ( 1 α 2 ) a ] < S H A ( a , b ) < H [ β 2 a + ( 1 β 2 ) b , β 2 b + ( 1 β 2 ) a ] , C [ α 3 a + ( 1 α 3 ) b , α 3 b + ( 1 α 3 ) a ] < S C A ( a , b ) < C [ β 3 a + ( 1 β 3 ) b , β 3 b + ( 1 β 3 ) a ] , C [ α 4 a + ( 1 α 4 ) b , α 4 b + ( 1 α 4 ) a ] < S A C ( a , b ) < C [ β 4 a + ( 1 β 4 ) b , β 4 b + ( 1 β 4 ) a ]

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

In [4, 5], Neuman proved that the inequalities
H ( a , b ) < S A H ( a , b ) < L ( a , b ) < S H A ( a , b ) < P ( a , b ) , T ( a , b ) < S C A ( a , b ) < Q ( a , b ) < S A C ( a , b ) < C ( a , b ) , H 1 / 3 ( a , b ) A 2 / 3 ( a , b ) < S H A ( a , b ) < 1 3 H ( a , b ) + 2 3 A ( a , b ) , C 1 / 3 ( a , b ) A 2 / 3 ( a , b ) < S C A ( a , b ) < 1 3 C ( a , b ) + 2 3 A ( a , b ) , A 1 / 3 ( a , b ) H 2 / 3 ( a , b ) < S A H ( a , b ) < 1 3 A ( a , b ) + 2 3 H ( a , b ) , A 1 / 3 ( a , b ) C 2 / 3 ( a , b ) < S A C ( a , b ) < 1 3 A ( a , b ) + 2 3 C ( a , b )
(1.4)

hold for all a , b > 0 with a b , where L ( a , b ) = ( a b ) / ( log a log b ) , P ( a , b ) = ( a b ) / [ 2 arcsin ( ( a b ) / ( a + b ) ) ] , Q ( a , b ) = ( a 2 + b 2 ) / 2 , and T ( a , b ) = ( a b ) / [ 2 arctan ( ( a b ) / ( a + b ) ) ] are, respectively, the logarithmic, first Seiffert, quadratic, and second Seiffert means of a and b.

Qian and Chu [8] proved that the double inequalities
α 1 A ( a , b ) + ( 1 α 1 ) G ( a , b ) < S H A ( a , b ) < β 1 A ( a , b ) + ( 1 β 1 ) G ( a , b ) , α 2 A ( a , b ) + ( 1 α 2 ) Q ( a , b ) < S C A ( a , b ) < β 2 A ( a , b ) + ( 1 β 2 ) Q ( a , b )

hold for all a , b > 0 with a b if and only if α 1 1 / 3 , β 1 π / 2 , α 2 1 / 3 , and β 2 [ 2 log ( 2 + 3 ) 3 ] / [ ( 2 1 ) log ( 2 + 3 ) ] = 0.2394 , where G ( a , b ) = a b is the geometric mean of a and b.

In [9], the authors proved that the double inequalities
α 1 [ H ( a , b ) 3 + 2 A ( a , b ) 3 ] + ( 1 α 1 ) H 1 / 3 ( a , b ) A 2 / 3 ( a , b ) < S H A ( a , b ) < β 1 [ H ( a , b ) 3 + 2 A ( a , b ) 3 ] + ( 1 β 1 ) H 1 / 3 ( a , b ) A 2 / 3 ( a , b ) , α 2 [ C ( a , b ) 3 + 2 A ( a , b ) 3 ] + ( 1 α 2 ) C 1 / 3 ( a , b ) A 2 / 3 ( a , b ) < S C A ( a , b ) < β 2 [ C ( a , b ) 3 + 2 A ( a , b ) 3 ] + ( 1 β 2 ) C 1 / 3 ( a , b ) A 2 / 3 ( a , b ) , α 3 [ A ( a , b ) 3 + 2 H ( a , b ) 3 ] + ( 1 α 3 ) A 1 / 3 ( a , b ) H 2 / 3 ( a , b ) < S A H ( a , b ) < β 3 [ A ( a , b ) 3 + 2 H ( a , b ) 3 ] + ( 1 β 3 ) A 1 / 3 ( a , b ) H 2 / 3 ( a , b ) , α 4 [ A ( a , b ) 3 + 2 C ( a , b ) 3 ] + ( 1 α 4 ) A 1 / 3 ( a , b ) C 2 / 3 ( a , b ) < S A C ( a , b ) < β 4 [ A ( a , b ) 3 + 2 C ( a , b ) 3 ] + ( 1 β 4 ) A 1 / 3 ( a , b ) C 2 / 3 ( a , b )

hold for all a , b > 0 with a b if and only if α 1 4 / 5 , β 1 3 / π , α 2 3 [ 2 3 log ( 2 + 3 ) 3 ] / [ ( 3 2 3 4 ) log ( 2 + 3 ) ] = 0.7528 , β 2 4 / 5 , α 3 0 , β 3 4 / 5 , α 4 4 / 5 , and β 4 3 ( 3 3 4 3 π ) / [ ( 5 3 4 3 ) π ] = 0.8400 .

Let p , p i , q i , α j , β j [ 0 , 1 ] ( i , j = 1 , 2 , , 8 ). Then Neuman [6, 10] proved that the inequalities
H p 1 ( a , b ) < P ( a , b ) < H q 1 ( a , b ) , G p 2 ( a , b ) < P ( a , b ) < G q 2 ( a , b ) , Q p 3 ( a , b ) < T ( a , b ) < Q q 3 ( a , b ) , C p 4 ( a , b ) < T ( a , b ) < C q 4 ( a , b ) , Q p 5 ( a , b ) < M ( a , b ) < Q q 5 ( a , b ) , C p 6 ( a , b ) < M ( a , b ) < C q 6 ( a , b ) , H p 7 ( a , b ) < L ( a , b ) < H q 7 ( a , b ) , G p 8 ( a , b ) < L ( a , b ) < G q 8 ( a , b ) , α 1 A ( a , b ) + ( 1 α 1 ) G p ( a , b ) < P p ( a , b ) < β 1 A ( a , b ) + ( 1 β 1 ) G p ( a , b ) , α 2 Q p ( a , b ) + ( 1 α 2 ) A ( a , b ) < T p ( a , b ) < β 2 Q p ( a , b ) + ( 1 β 2 ) A ( a , b ) , α 3 Q p ( a , b ) + ( 1 α 3 ) A ( a , b ) < M p ( a , b ) < β 3 Q p ( a , b ) + ( 1 β 3 ) A ( a , b ) , α 4 A ( a , b ) + ( 1 α 4 ) G p ( a , b ) < L p ( a , b ) < β 4 A ( a , b ) + ( 1 β 4 ) G p ( a , b ) , A α 5 ( a , b ) G p 1 α 5 ( a , b ) < P p ( a , b ) < A β 5 ( a , b ) G p 1 β 5 ( a , b ) , Q p α 6 ( a , b ) A 1 α 6 ( a , b ) < T p ( a , b ) < Q p β 6 ( a , b ) A 1 β 6 ( a , b ) , Q p α 7 ( a , b ) A 1 α 7 ( a , b ) < M p ( a , b ) < Q p β 7 ( a , b ) A 1 β 7 ( a , b ) , A α 8 ( a , b ) G p 1 α 8 ( a , b ) < L p ( a , b ) < A β 8 ( a , b ) G p 1 β 8 ( a , b ) ,

hold for all a , b > 0 with a b if and only if p 1 1 2 / π , q 1 6 / 6 , p 2 1 4 / π 2 , q 2 3 / 3 , p 3 16 / π 2 1 , q 3 6 / 3 , p 4 4 / π 1 , q 4 3 / 3 , p 5 1 / log 2 ( 1 + 2 ) 1 , q 5 3 / 3 , p 6 1 / log ( 1 + 2 ) 1 , q 6 6 / 6 , p 7 = 1 , q 7 3 / 3 , p 8 = 1 , q 8 6 / 3 , α 1 2 / π , β 1 2 / 3 , α 2 ( 4 π ) / [ ( 2 1 ) π ] , β 2 2 / 3 , α 3 [ 1 log ( 1 + 2 ) ] / [ ( 2 1 ) log ( 1 + 2 ) ] , β 3 1 / 3 , α 4 = 0 , β 4 1 / 3 , α 5 2 / 3 , β 5 = 1 , α 6 2 / 3 , β 6 ( 4 log 2 2 log π ) / log 2 , α 7 1 / 3 , β 7 log [ log ( 1 + 2 ) ] / log [ cosh ( log ( 1 + 2 ) ) ] , α 8 1 / 3 , β 8 = 1 , where M ( a , b ) = ( a b ) / [ 2 sinh 1 ( ( a b ) / ( a + b ) ) ] is the Neuman-Sándor mean of a and b.

The main purpose of this paper is to present the best possible parameters p 1 , p 2 , p 3 , p 4 , q 1 , q 2 , q 3 , q 4 on the interval [ 0 , 1 ] such that the double inequalities
G p 1 ( a , b ) < S H A ( a , b ) < G q 1 ( a , b ) , Q p 2 ( a , b ) < S C A ( a , b ) < Q q 2 ( a , b ) , H p 3 ( a , b ) < S A H ( a , b ) < H q 3 ( a , b ) , C p 4 ( a , b ) < S A C ( a , b ) < C q 4 ( a , b )

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

2 Main results

Theorem 2.1 Let p 1 , q 1 [ 0 , 1 ] . Then the double inequality
G p 1 ( a , b ) < S H A ( a , b ) < G q 1 ( a , b )
(2.1)

holds for all a , b > 0 with a b if and only if p 1 6 / 3 and q 1 1 4 / π 2 .

Proof Without loss of generality, we assume that a > b . Let v = ( a b ) / ( a + b ) , λ = v 2 v 2 , x = 1 λ 2 and p [ 0 , 1 ] . Then v , λ , x ( 0 , 1 ) , and (1.1) and (1.3) lead to
S H A ( a , b ) G p ( a , b ) = A ( a , b ) [ λ arcsin ( λ ) 1 p 2 ( 1 1 λ 2 ) ] = A ( a , b ) 1 p 2 ( 1 1 λ 2 ) arcsin ( λ ) F ( x ) ,
(2.2)
where
F ( x ) = 1 x 2 1 p 2 ( 1 x ) arcsin ( 1 x 2 ) ,
(2.3)
F ( 0 ) = 1 1 p 2 π 2 , F ( 1 ) = 0 ,
(2.4)
F ( x ) = ( 1 x ) f ( x ) 2 1 x 2 ( p 2 x + 1 p 2 ) 3 / 2 [ 2 ( p 2 x + 1 p 2 ) 3 / 2 + p 2 x + 2 ( 1 p 2 ) x + p 2 ] ,
(2.5)
where
f ( x ) = p 4 x 3 + ( 4 p 6 + 3 p 4 4 p 2 ) x 2 + ( 8 p 6 + 9 p 4 + 4 p 2 4 ) x + ( 4 p 6 11 p 4 + 12 p 2 4 ) ,
(2.6)
f ( x ) = 3 p 4 x 2 + 2 ( 4 p 6 + 3 p 4 4 p 2 ) x + ( 8 p 6 + 9 p 4 + 4 p 2 4 ) .
(2.7)

We divide the discussion into two cases.

Case 1 p = 6 / 3 . Then (2.6) becomes
f ( x ) = 4 27 ( 1 x ) ( 3 x 2 + 4 x + 2 ) .
(2.8)
From (2.5) and (2.8) we clearly see that F ( x ) is strictly decreasing on [ 0 , 1 ] , then (2.4) leads to the conclusion that
F ( x ) > 0
(2.9)

for all x ( 0 , 1 ) .

Therefore,
S H A ( a , b ) > G 6 / 3 ( a , b )
(2.10)

for all a , b > 0 with a b follows from (2.2) and (2.9).

Case 2 p = 1 4 / π 2 . Then numerical computations lead to
4 p 6 + 3 p 4 4 p 2 = 3 π 6 56 π 4 + 240 π 2 256 π 6 < 0 ,
(2.11)
8 p 6 + 9 p 4 + 4 p 2 4 = π 6 + 8 π 4 240 π 2 + 512 π 6 < 0 ,
(2.12)
f ( 0 ) = 4 p 6 11 p 4 + 12 p 2 4 = π 6 8 π 4 + 16 π 2 256 π 6 > 0 ,
(2.13)
f ( 1 ) = 4 ( 3 p 2 2 ) = 4 ( 12 π 2 ) π 2 < 0 .
(2.14)

It follows from (2.7) and (2.11) together with (2.12) that f ( x ) is strictly decreasing on [ 0 , 1 ] . Then inequalities (2.13) and (2.14) together with (2.5) lead to the conclusion that there exists λ 1 ( 0 , 1 ) such that F ( x ) is strictly decreasing on [ 0 , λ 1 ] and strictly increasing on [ λ 1 , 1 ] .

Note that inequality (2.4) becomes
F ( 0 ) = F ( 1 ) = 0 .
(2.15)
From (2.2), (2.15), and the piecewise monotonicity of F ( x ) we clearly see that the inequality
S H A ( a , b ) < G 1 4 / π 2 ( a , b )
(2.16)

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

Note that
lim λ 0 + arcsin 2 ( λ ) λ 2 arcsin ( λ ) 1 1 λ 2 = 6 3 ,
(2.17)
lim λ 1 arcsin 2 ( λ ) λ 2 arcsin ( λ ) 1 1 λ 2 = 1 4 π 2 .
(2.18)
Therefore, Theorem 2.1 follows from (2.10) and (2.16)-(2.18) together with the fact that inequality (2.1) is equivalent to the inequality (2.19) as follows:
q 1 < arcsin 2 ( λ ) λ 2 arcsin ( λ ) 1 1 λ 2 < p 1 .
(2.19)

 □

Theorem 2.2 Let p 2 , q 2 [ 0 , 1 ] . Then the double inequality
Q p 2 ( a , b ) < S C A ( a , b ) < Q q 2 ( a , b )
(2.20)

holds for all a , b > 0 with a b if and only if p 2 6 / 3 and q 2 3 / log 2 ( 2 + 3 ) 1 = 0.8542 .

Proof Without loss of generality, we assume that a > b . Let v = ( a b ) / ( a + b ) , μ = v 2 + v 2 , x = 1 + μ 2 , and p [ 0 , 1 ] . Then v ( 0 , 1 ) , μ ( 0 , 3 ) , x ( 1 , 2 ) , and (1.2) and (1.3) lead to
S C A ( a , b ) Q p ( a , b ) = A ( a , b ) [ μ sinh 1 ( μ ) 1 + p 2 ( 1 + μ 2 1 ) ] = A ( a , b ) 1 + p 2 ( 1 + μ 2 1 ) sinh 1 ( μ ) G ( x ) ,
(2.21)
where
G ( x ) = x 2 1 1 + p 2 ( x 1 ) sinh 1 ( x 2 1 ) , G ( 1 ) = 0 , G ( 2 ) = 3 1 + p 2 log ( 2 + 3 ) ,
(2.22)
G ( x ) = ( x 1 ) f ( x ) 2 x 2 1 ( p 2 x + 1 p 2 ) 3 / 2 [ p 2 x 2 + 2 ( p 2 x + 1 p 2 ) 3 / 2 + 2 ( 1 p 2 ) x + p 2 ] ,
(2.23)

where f ( x ) is defined by (2.6).

We divide the discussion into two cases.

Case 1 p = 6 / 3 . Then it follows from (2.6) that
f ( x ) = 4 27 ( x 1 ) ( 3 x 2 + 4 x + 2 ) < 0
(2.24)

for all x ( 1 , 2 ) .

Therefore,
S C A ( a , b ) > Q 6 / 3 ( a , b )
(2.25)

for all a , b > 0 with a b follows easily from (2.21)-(2.24).

Case 2 p = 3 / log 2 ( 2 + 3 ) 1 . Then numerical computations lead to
4 p 6 + 3 p 4 4 p 2 = 0.2329 > 0 ,
(2.26)
8 p 6 + 9 p 4 + 4 p 2 4 = 0.6027 > 0 ,
(2.27)
3 p 4 p 2 1 = 0.1322 < 0 ,
(2.28)
f ( 1 ) = 4 ( 3 p 2 2 ) = 0.7567 > 0 ,
(2.29)
f ( 2 ) = 4 p 6 + 11 p 4 + 4 p 2 12 = 1.669 < 0 .
(2.30)
It follows from (2.7) and (2.26)-(2.28) that
f ( x ) < 3 p 4 x 2 + 2 ( 4 p 6 + 3 p 4 4 p 2 ) x 2 + ( 8 p 6 + 9 p 4 + 4 p 2 4 ) x 2 = 4 ( 3 p 4 p 2 1 ) x 2 < 0
(2.31)

for x ( 1 , 2 ) .

Equation (2.23) and inequalities (2.29)-(2.31) lead to the conclusion that there exists λ 2 ( 1 , 2 ) such that G ( x ) is strictly decreasing on [ 0 , λ 2 ] and strictly increasing on [ λ 2 , 1 ] .

Note that (2.22) becomes
G ( 1 ) = G ( 2 ) = 0 .
(2.32)
Therefore,
S C A ( a , b ) < Q 3 / log 2 ( 2 + 3 ) 1 ( a , b )
(2.33)

for all a , b > 0 with a b follows from (2.21) and (2.32) together with the piecewise monotonicity of G ( x ) .

Note that
lim μ 0 + μ 2 [ sinh 1 ( μ ) ] 2 sinh 1 ( μ ) 1 + μ 2 1 = 6 3 ,
(2.34)
lim μ 3 μ 2 [ sinh 1 ( μ ) ] 2 sinh 1 ( μ ) 1 + μ 2 1 = 3 log 2 ( 2 + 3 ) 1 .
(2.35)
Therefore, Theorem 2.2 follows from (2.25) and (2.33)-(2.35) together with the fact that inequality (2.20) is equivalent to the inequality (2.36) as follows:
p 2 < μ 2 [ sinh 1 ( μ ) ] 2 sinh 1 ( μ ) 1 + μ 2 1 < q 2 .
(2.36)

 □

Theorem 2.3 Let p 3 , q 3 [ 0 , 1 ] . Then the double inequality
H p 3 ( a , b ) < S A H ( a , b ) < H q 3 ( a , b )
(2.37)

holds for all a , b > 0 with a b if and only if p 3 = 1 and q 3 6 / 3 .

Proof Without loss of generality, we assume that a > b . Let v = ( a b ) / ( a + b ) , λ = v 2 v 2 , x = 1 λ 2 and p [ 0 , 1 ] . Then v , λ , x ( 0 , 1 ) , and (1.1) and (1.3) lead to
S A H ( a , b ) H p ( a , b ) = A ( a , b ) [ λ tanh 1 ( λ ) + p 2 ( 1 1 λ 2 ) 1 ] = A ( a , b ) [ 1 p 2 ( 1 1 λ 2 ) ] tanh 1 ( λ ) H ( x ) ,
(2.38)
where
H ( x ) = 1 x 2 p 2 x + ( 1 p 2 ) tanh 1 ( 1 x 2 ) , H ( 1 ) = 0 ,
(2.39)
H ( x ) = 1 x x 1 x 2 [ p 2 x + ( 1 p 2 ) ] 2 g ( x ) ,
(2.40)
where
g ( x ) = ( p 4 + p 2 1 ) x p 4 + 2 p 2 1 .
(2.41)

We divide the discussion into two cases.

Case 1 p = 6 / 3 . Then (2.41) leads to
g ( x ) = 1 9 ( 1 x ) < 0
(2.42)

for x ( 0 , 1 ) .

Therefore,
S A H ( a , b ) < H 6 / 3 ( a , b )
(2.43)

for all a , b > 0 with a b follows easily from (2.38)-(2.40) and (2.42).

Case 2 p = 1 . Then it follows from (1.3) and (1.4) that
S A H ( a , b ) > H ( a , b ) = H 1 ( a , b )
(2.44)

for all a , b > 0 with a b .

Note that
lim λ 0 + tanh 1 ( λ ) λ tanh 1 ( λ ) ( 1 1 λ 2 ) = 6 3 ,
(2.45)
lim λ 1 tanh 1 ( λ ) λ tanh 1 ( λ ) ( 1 1 λ 2 ) = 1 .
(2.46)
Therefore, Theorem 2.3 follows from (2.43)-(2.46) and the fact that inequality (2.37) is equivalent to
q 3 < tanh 1 ( λ ) λ tanh 1 ( λ ) ( 1 1 λ 2 ) < p 3 .

 □

Theorem 2.4 Let p 4 , q 4 [ 0 , 1 ] . Then the double inequality
C p 4 ( a , b ) < S A C ( a , b ) < C q 4 ( a , b )
(2.47)

holds for all a , b > 0 with a b if and only if p 4 3 3 / π 1 and q 4 6 / 3 .

Proof Without loss of generality, we assume that a > b . Let v = ( a b ) / ( a + b ) , μ = v 2 + v 2 , x = 1 + μ 2 , and p [ 0 , 1 ] . Then v ( 0 , 1 ) , μ ( 0 , 3 ) , x ( 1 , 2 ) , and (1.2) and (1.3) lead to
S A C ( a , b ) C p ( a , b ) = A ( a , b ) [ μ arctan ( μ ) p 2 ( 1 + μ 2 1 ) 1 ] = A ( a , b ) [ 1 + p 2 ( 1 + μ 2 1 ) ] arctan ( μ ) J ( x ) ,
(2.48)
where
J ( x ) = x 2 1 p 2 x + ( 1 p 2 ) arctan ( x 2 1 ) , J ( 1 ) = 0 , J ( 2 ) = 3 p 2 + 1 π 3 ,
(2.49)
J ( x ) = x 1 x x 2 1 [ p 2 x + ( 1 p 2 ) ] 2 g ( x ) ,
(2.50)

where g ( x ) is defined by (2.41).

We divide the discussion into two cases.

Case 1 p = 6 / 3 . Then (2.41) leads to
g ( x ) = 1 9 ( x 1 ) > 0
(2.51)

for x ( 1 , 2 ) .

Therefore,
S A C ( a , b ) < C 6 / 3 ( a , b )
(2.52)

for all a , b > 0 with a b follows easily from (2.48)-(2.51).

Case 2 p = 3 3 / π 1 . Then numerical computations lead to
p 4 + p 2 1 = 27 π 2 3 3 π π 2 > 0 ,
(2.53)
g ( 1 ) = 3 p 2 2 = 9 3 5 π π < 0 ,
(2.54)
g ( 2 ) = p 4 + 4 p 2 3 = 27 6 π 2 + 6 3 π π 2 > 0 .
(2.55)

From (2.41) and (2.50) together with (2.53)-(2.55) we clearly see that there exists λ 3 ( 1 , 2 ) such that J ( x ) is strictly increasing on [ 1 , λ 3 ] and strictly decreasing on [ λ 3 , 2 ] .

Note that (2.49) becomes
J ( 1 ) = J ( 2 ) = 0 .
(2.56)
It follows from (2.56) and the piecewise monotonicity of J ( x ) that
J ( x ) > 0
(2.57)

for all x ( 1 , 2 ) .

Therefore,
S A C ( a , b ) > C 3 3 / π 1 ( a , b )
(2.58)

for all a , b > 0 with a b follows from (2.48) and (2.58).

Note that
lim μ 0 + μ arctan ( μ ) arctan ( μ ) ( 1 + μ 2 1 ) = 6 3 ,
(2.59)
lim μ 1 μ arctan ( μ ) arctan ( μ ) ( 1 + μ 2 1 ) = 3 3 π 1 .
(2.60)
Therefore, Theorem 2.4 follows from (2.52) and (2.58)-(2.60) together with the fact that inequality (2.47) is equivalent to
p 4 < μ arctan ( μ ) arctan ( μ ) ( 1 + μ 2 1 ) < q 4 .

 □

Declarations

Acknowledgements

The research was supported by the Natural Science Foundation of China under Grants 61374086 and 11171307, the Natural Science Foundation of the Open University of China under Grant Q1601E-Y and the Natural Science Foundation of Zhejiang Broadcast and TV University under Grant XKT-13Z04.

Authors’ Affiliations

(1)
School of Mathematics and Computation Sciences, Hunan City University, Yiyang, China
(2)
School of Distance Education, Huzhou Broadcast and TV University, Huzhou, China

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© Shao et al.; licensee Springer. 2014

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/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

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