Skip to main content

Optimal bounds for Neuman means in terms of geometric, arithmetic and quadratic means

Abstract

In this paper, we present sharp bounds for the two Neuman means S H A and S C A derived from the Schwab-Borchardt mean in terms of convex combinations of either the weighted arithmetic and geometric means or the weighted arithmetic and quadratic means, and the mean generated either by the geometric or by the quadratic mean.

MSC:26E60.

1 Introduction

Let a,b>0 with ab, then the Schwab-Borchardt mean SB(a,b) is defined by

SB(a,b)={ b 2 a 2 cos 1 ( a / b ) , a < b , a 2 b 2 cosh 1 ( a / b ) , a > b ,
(1.1)

where cos 1 (x) and cosh 1 (x)=log(x+ x 2 1 ) are the inverse cosine and inverse hyperbolic cosine functions, respectively.

It is well known that SB(a,b) is strictly increasing in both a and b, nonsymmetric and homogeneous of degree 1 with respect to a and b. Many symmetric bivariate means are special cases of the Schwab-Borchardt mean, for example,

P ( a , b ) = a b 2 sin 1 [ ( a b ) / ( a + b ) ] = SB ( G , A ) is the first Seiffert mean , T ( a , b ) = a b 2 tan 1 [ ( a b ) / ( a + b ) ] = SB ( A , Q ) is the second Seiffert mean , M ( a , b ) = a b 2 sinh 1 [ ( a b ) / ( a + b ) ] = SB ( Q , A ) is the Neuman-Sándor mean , L ( a , b ) = a b 2 tanh 1 [ ( a b ) / ( a + b ) ] = SB ( A , G ) is the logarithmic mean ,

where G(a,b)= a b , A(a,b)=(a+b)/2 and Q(a,b)= ( a 2 + b 2 ) / 2 denote the classical geometric mean, arithmetic mean and quadratic mean of a and b, respectively. The Schwab-Borchardt mean SB(a,b) was investigated in [1, 2].

Let H(a,b)=2ab/(a+b), C(a,b)=( a 2 + b 2 )/(a+b) be the harmonic and contraharmonic means of two positive numbers a and b, respectively. Then it is well known that

H ( a , b ) < G ( a , b ) < L ( a , b ) < P ( a , b ) < A ( a , b ) < M ( a , b ) < T ( a , b ) < Q ( a , b ) < C ( a , b )
(1.2)

for a,b>0 with ab.

Recently, the Schwab-Borchardt mean and its special cases have been the subject of intensive research. Neuman and Sándor [3, 4] proved that the inequalities

P ( a , b ) > 2 π A ( a , b ) , A ( a , b ) log ( 1 + 2 ) > M ( a , b ) > π 4 log ( 1 + 2 ) T ( a , b ) , T ( A ( a , b ) , G ( a , b ) ) < P ( a , b ) , T ( a , b ) > T ( A ( a , b ) , Q ( a , b ) ) , L ( a , b ) < L ( A ( a , b ) , G ( a , b ) ) , M ( a , b ) < L ( A ( a , b ) , Q ( a , b ) ) , L ( a , b ) > H ( P ( a , b ) , G ( a , b ) ) , P ( a , b ) > H ( L ( a , b ) , A ( a , b ) ) , M ( a , b ) > H ( T ( a , b ) , A ( a , b ) ) , T ( a , b ) > H ( M ( a , b ) , Q ( a , b ) ) , G ( a , b ) P ( a , b ) < L 2 ( a , b ) < G 2 ( a , b ) + P 2 ( a , b ) 2 , L ( a , b ) A ( a , b ) < P 2 ( a , b ) < L 2 ( a , b ) + A 2 ( a , b ) 2 , A ( a , b ) T ( a , b ) < M 2 ( a , b ) < A 2 ( a , b ) + T 2 ( a , b ) 2 , M ( a , b ) Q ( a , b ) < T 2 ( a , b ) < M 2 ( a , b ) + Q 2 ( a , b ) 2 , Q 1 / 3 ( a , b ) A 2 / 3 ( a , b ) < M ( a , b ) < 1 3 Q ( a , b ) + 2 3 A ( a , b )

hold for all a,b>0 with ab. In [5], the author proved that the double inequalities

αQ(a,b)+(1α)A(a,b)<M(a,b)<βQ(a,b)+(1β)A(a,b)

and

λC(a,b)+(1λ)A(a,b)<M(a,b)<μC(a,b)+(1μ)A(a,b)

hold for all a,b>0 with ab if and only if α[1log(1+ 2 )]/[( 2 1)log(1+ 2 )]=0.3249 , β1/3, λ[1log(1+ 2 )]/log(1+ 2 )=0.1345 and μ1/6. Chu and Long [6] found that the double inequality

M p (a,b)<M(a,b)<qI(a,b)

holds for all a,b>0 with ab if and only if plog2/log[2log(1+ 2 )]=1.224 and qe/[2log(1+ 2 )]=1.5420 , where M p (a,b)= [ ( a p + b p ) / 2 ] 1 / p (p0) and M 0 (a,b)= a b is the p th power mean of a and b. Zhao et al. [7] presented the least values α 1 , α 2 , α 3 and the greatest values β 1 , β 2 , β 3 such that the double inequalities

α 1 H ( a , b ) + ( 1 α 1 ) Q ( a , b ) < M ( a , b ) < β 1 H ( a , b ) + ( 1 β 1 ) Q ( a , b ) , α 2 G ( a , b ) + ( 1 α 2 ) Q ( a , b ) < M ( a , b ) < β 2 G ( a , b ) + ( 1 β 2 ) Q ( a , b )

and

α 3 H(a,b)+(1 α 3 )C(a,b)<M(a,b)< β 3 H(a,b)+(1 β 3 )C(a,b)

hold for all a,b>0 with ab.

Very recently, the bivariate means S A H , S H A , S C A and S A C derived from the Schwab-Borchardt mean have been defined by Neuman [8, 9] as follows:

S A H =SB(A,H), S H A =SB(H,A), S C A =SB(C,A), S A C =SB(A,C).
(1.3)

We call the means S A H , S H A , S C A and S A C given in (1.3) the Neuman means. Moreover, let v=(ab)/(a+b)(1,1), then the following explicit formulas for S A H , S H A , S A C and S C A have been found by Neuman [8]:

S A H =A tanh ( p ) p , S H A =A sin ( q ) q ,
(1.4)
S C A =A sinh ( r ) r , S A C =A tan ( s ) s ,
(1.5)

where p, q, r and s are defined implicitly as sech(p)=1 v 2 , cos(q)=1 v 2 , cosh(r)=1+ v 2 and sec(s)=1+ v 2 , respectively. Clearly, p(0,), q(0,π/2), r(0,log(2+ 3 )) and s(0,π/3).

In [8], Neuman proved that the inequalities

H(a,b)< S A H (a,b)<L(a,b)< S H A (a,b)<P(a,b),
(1.6)
T(a,b)< S C A (a,b)<Q(a,b)< S A C (a,b)<C(a,b)
(1.7)

hold for a,b>0 with ab.

He et al. [10] found the greatest values α 1 , α 2 [0,1/2], α 3 , α 4 [1/2,1] and the least values β 1 , β 2 [0,1/2], 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 )

and

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 ab.

It follows from (1.2) and (1.6) together with (1.7) that

G(a,b)< S H A (a,b)<A(a,b)< S C A (a,b)<Q(a,b)
(1.8)

for all a,b>0 with ab.

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

f(x)=G [ x a + ( 1 x ) b , x b + ( 1 x ) a ] ,
(1.9)
g(y)=Q [ y a + ( 1 y ) b , y b + ( 1 y ) a ] .
(1.10)

Then it is not difficult to verify that f(x) and g(y) are continuous and strictly increasing on [0,1/2] and [1/2,1], respectively. Note that

f(0)=G(a,b)< S H A (a,b)<A(a,b)=f(1/2),
(1.11)
g(1/2)=A(a,b)< S C A (a,b)<Q(a,b)=g(1).
(1.12)

Motivated by (1.8)-(1.12), in the article we present the best possible parameters α 1 , α 2 , β 1 , β 2 R, α 3 , β 3 [0,1/2] and α 4 , β 4 [1/2,1] such 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 ) , G [ α 3 a + ( 1 α 3 ) b , α 3 b + ( 1 α 3 ) a ] < S H A ( a , b ) < G [ β 3 a + ( 1 β 3 ) b , β 3 b + ( 1 β 3 ) a ] , Q [ α 4 a + ( 1 α 4 ) b , α 4 b + ( 1 α 4 ) a ] < S C A ( a , b ) < Q [ β 4 a + ( 1 β 4 ) b , β 4 b + ( 1 β 4 ) a ]

hold for all a,b>0 with ab.

Our main results are the following Theorems 1.1-1.4. All numerical computations are carried out using Mathematica software.

Theorem 1.1 The double inequality

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)

holds for all a,b>0 with ab if and only if α 1 1/3 and β 1 2/π.

Theorem 1.2 The two-sided inequality

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)

holds true for all a,b>0 with ab if and only if α 2 1/3 and β 2 [ 2 log(2+ 3 ) 3 ]/[( 2 1)log(2+ 3 )]=0.2390 .

Theorem 1.3 Let α 3 , β 3 [0,1/2], then the double inequality

G [ α 3 a + ( 1 α 3 ) b , α 3 b + ( 1 α 3 ) a ] < S H A (a,b)<G [ β 3 a + ( 1 β 3 ) b , β 3 b + ( 1 β 3 ) a ]

holds for all a,b>0 with ab if and only if α 3 1/2 6 /6=0.09175 and β 3 1/2 π 2 4 /(2π)=0.1144 .

Theorem 1.4 Let α 4 , β 4 [1/2,1], then the two-sided inequality

Q [ α 4 a + ( 1 α 4 ) b , α 4 b + ( 1 α 4 ) a ] < S C A (a,b)<Q [ β 4 a + ( 1 β 4 ) b , β 4 b + ( 1 β 4 ) a ]

holds true for all a,b>0 with ab if and only if α 4 1/2+ 6 /6=0.9082 and β 4 1/2+ 3 / [ log ( 2 + 3 ) ] 2 1 /2=0.9271 .

2 Two lemmas

In order to prove our main results, we need two lemmas, which we present in this section.

Lemma 2.1 Let pR and

f(x)=(1p) x 3 + ( 2 p 2 + 5 p 1 ) x 2 + ( 2 p 2 + p 1 ) x+p1.
(2.1)

Then the following statements are true:

  1. (1)

    If p=1/3, then f(x)<0 for all x(0,1) and f(x)>0 for all x(1, 2 );

  2. (2)

    If p=2/π, then there exists λ 1 (0,1) such that f(x)<0 for x(0, λ 1 ) and f(x)>0 for x( λ 1 ,1);

  3. (3)

    If p=[ 2 log(2+ 3 ) 3 ]/[( 2 1)log(2+ 3 )], then there exists λ 2 (1, 2 ) such that f(x)<0 for x(1, λ 2 ) and f(x)>0 for x( λ 2 , 2 ).

Proof For part (1), if p=1/3, then (2.1) becomes

f(x)= 2 9 (x1) ( 3 x 2 + 5 x + 3 ) .
(2.2)

Therefore, part (1) follows easily from (2.2).

For part (2), if p=2/π, then simple computations lead to

2 p 2 +5p1= π 2 + 10 π 8 π 2 >0,
(2.3)
2 p 2 +p1= π 2 + 2 π + 8 π 2 >0,
(2.4)
f(0)= π 2 π <0,
(2.5)
f(1)= 2 ( 6 π ) π >0,
(2.6)
f (x)=3(1p) x 2 +2 ( 2 p 2 + 5 p 1 ) x+ ( 2 p 2 + p 1 ) .
(2.7)

It follows from (2.3) and (2.4) together with (2.7) that f(x) is strictly increasing on (0,1). Therefore, part (2) follows from (2.5) and (2.6) together with the monotonicity of f(x).

For part (3), if p=[ 2 log(2+ 3 ) 3 ]/[( 2 1)log(2+ 3 )]=0.2390 , then numerical computations lead to

2 p 2 +5p1=0.0810>0,
(2.8)
f(1)=0.5656<0,
(2.9)
f( 2 )=0.6388>0.
(2.10)

It follows from (2.7) and (2.8) that

f (x)>3(1p)+2 ( 2 p 2 + 5 p 1 ) + ( 2 p 2 + p 1 ) =2p(4p)>0
(2.11)

for x(1, 2 ).

Therefore, part (3) follows easily from (2.9)-(2.11). □

Lemma 2.2 Let pR and

g ( x ) = ( 2 p 1 ) 4 x 3 + ( 256 p 6 + 768 p 5 1 , 008 p 4 + 736 p 3 296 p 2 + 56 p 3 ) x 2 + ( 512 p 6 1 , 536 p 5 + 1 , 776 p 4 992 p 3 + 248 p 2 8 p 1 ) x + ( 256 p 6 + 768 p 5 784 p 4 + 288 p 3 24 p 2 + 8 p 1 ) .
(2.12)

Then the following statements are true:

  1. (1)

    If p=1/2 6 /6, then g(x)<0 for all x(0,1);

  2. (2)

    If p=1/2+ 6 /6, then g(x)>0 for all x(1,2);

  3. (3)

    If p=1/2 π 2 4 /(2π), then there exists λ 3 (0,1) such that g(x)<0 for x(0, λ 3 ) and g(x)>0 for x( λ 3 ,1);

  4. (4)

    If p=1/2+ 3 / [ log ( 2 + 3 ) ] 2 1 /2, then there exists λ 4 (1,2) such that g(x)<0 for x(1, λ 4 ) and g(x)>0 for x( λ 4 ,2).

Proof For parts (1) and (2), if p=1/2 6 /6 or p=1/2+ 6 /6, then (2.12) becomes

g(x)= 4 27 (x1) ( 3 x 2 + 4 x + 2 ) .
(2.13)

Therefore, parts (1) and (2) follow from (2.13).

For part (3), if p=1/2 π 2 4 /(2π), then numerical computations show that

256 p 6 + 768 p 5 1 , 008 p 4 + 736 p 3 296 p 2 + 56 p 3 = 3 π 6 + 56 π 4 240 π 2 + 256 π 6 > 0 ,
(2.14)
512 p 6 1 , 536 p 5 + 1 , 776 p 4 992 p 3 + 248 p 2 8 p 1 = π 6 8 π 4 + 240 π 2 512 π 6 > 0 ,
(2.15)
g(0)= π 6 + 8 π 4 16 π 2 + 256 π 6 <0,
(2.16)
g(1)= 4 ( 12 π 2 ) π 2 >0,
(2.17)
g ( x ) = 3 ( 2 p 1 ) 4 x 2 + 2 ( 256 p 6 + 768 p 5 1 , 008 p 4 + 736 p 3 296 p 2 + 56 p 3 ) x g ( x ) = + ( 512 p 6 1 , 536 p 5 + 1 , 776 p 4 992 p 3 + 248 p 2 8 p 1 ) .
(2.18)

From (2.14), (2.15) and (2.18) we clearly see that g(x) is strictly increasing on (0,1). Therefore, part (3) follows from (2.16) and (2.17) together with the monotonicity of g(x).

For part (4), if p=1/2+ 3 / [ log ( 2 + 3 ) ] 2 1 /2, then numerical computations lead to

256 p 6 +768 p 5 1,008 p 4 +736 p 3 296 p 2 +56p3=0.2329<0,
(2.19)
512 p 6 1,536 p 5 +1,776 p 4 992 p 3 +248 p 2 8p1=0.6027<0,
(2.20)
g(1)=0.7567<0,
(2.21)
g(2)=1.6692>0,
(2.22)
48 p 4 +96 p 3 68 p 2 +20p1=0.1322>0.
(2.23)

It follows from (2.18), (2.19), (2.20) and (2.23) that

g ( x ) > 3 ( 2 p 1 ) 4 x 2 + 2 ( 256 p 6 + 768 p 5 1 , 008 p 4 + 736 p 3 296 p 2 + 56 p 3 ) x 2 + ( 512 p 6 1 , 536 p 5 + 1 , 776 p 4 992 p 3 + 248 p 2 8 p 1 ) x 2 = 4 ( 48 p 4 + 96 p 3 68 p 2 + 20 p 1 ) x 2 > 0
(2.24)

for x(1,2).

Therefore, part (4) follows from (2.21) and (2.22) together with (2.24). □

3 Proofs of Theorems 1.1-1.4

Proof of Theorem 1.1 Without loss of generality, we assume that a>b. Let v=(ab)/(a+b), λ=v 2 v 2 , x= 1 λ 2 4 and p{1/3,2/π}. Then v,λ,x(0,1) and (1.4) leads to

S H A ( a , b ) G ( a , b ) A ( a , b ) G ( a , b ) = λ ( 1 λ 2 ) 1 / 4 sin 1 ( λ ) [ 1 ( 1 λ 2 ) 1 / 4 ] sin 1 ( λ ) ,
(3.1)
S H A ( a , b ) [ p A ( a , b ) + ( 1 p ) G ( a , b ) ] = A ( a , b ) [ λ sin 1 ( λ ) ( 1 p ) ( 1 λ 2 ) 1 / 4 p ] = A ( a , b ) [ p + ( 1 p ) ( 1 λ 2 ) 1 / 4 ] sin 1 ( λ ) F ( x ) ,
(3.2)

where

F ( x ) = 1 x 4 ( 1 p ) x + p sin 1 ( 1 x 4 ) , F ( 0 ) = 1 p π 2 ,
(3.3)
F(1)=0
(3.4)

and

F (x)= 1 x 1 x 4 [ ( 1 p ) x + p ] 2 f(x),
(3.5)

where f(x) is defined as in Lemma 2.1.

We divide the proof into two cases.

Case 1: p=1/3. Then from Lemma 2.1(1) and (3.5) we clearly see that F(x) is strictly decreasing on (0,1). Therefore,

S H A (a,b)> 1 3 A(a,b)+ 2 3 G(a,b)
(3.6)

for all a,b>0 with ab follows from (3.2) and (3.4) together with the monotonicity of F(x).

Case 2: p=2/π. Then from (3.3), (3.5) and Lemma 2.1(2) we know that

F(0)=0
(3.7)

and there exists λ 1 (0,1) such that F(x) is strictly decreasing on (0, λ 1 ] and strictly increasing on [ λ 1 ,1). Therefore,

S H A (a,b)< 2 π A(a,b)+ ( 1 2 π ) G(a,b)
(3.8)

for all a,b>0 with ab follows from (3.2) and (3.4) together with (3.7) and the piecewise monotonicity of F(x).

Note that

lim λ 0 + λ ( 1 λ 2 ) 1 / 4 sin 1 ( λ ) [ 1 ( 1 λ 2 ) 1 / 4 ] sin 1 ( λ ) = 1 3
(3.9)

and

lim λ 1 λ ( 1 λ 2 ) 1 / 4 sin 1 ( λ ) [ 1 ( 1 λ 2 ) 1 / 4 ] sin 1 ( λ ) = 2 π .
(3.10)

Therefore, Theorem 1.1 follows from (3.6) and (3.8) together with the following statements.

  • If α>1/3, then equations (3.1) and (3.9) imply that there exists small enough δ>0 such that S H A (a,b)<αA(a,b)+(1α)G(a,b) for all a>b>0 with b/a(1δ,1).

  • If β<2/π, then equations (3.1) and (3.10) imply that there exists large enough M>1 such that S H A (a,b)>βA(a,b)+(1β)G(a,b) for all a>b>0 with a/b(M,+).

 □

Proof of Theorem 1.2 Without loss of generality, we assume that a>b. Let v=(ab)/(a+b), μ=v 2 + v 2 , x= 1 + μ 2 4 and p{[ 2 log(2+ 3 ) 3 ]/[( 2 1)log(2+ 3 )],1/3}. Then v(0,1), μ(0, 3 ), x(1, 2 ) and (1.5) leads to

S C A ( a , b ) Q ( a , b ) A ( a , b ) Q ( a , b ) = μ ( 1 + μ 2 ) 1 / 4 sinh 1 ( μ ) [ 1 ( 1 + μ 2 ) 1 / 4 ] sinh 1 ( μ ) ,
(3.11)
S C A ( a , b ) [ p A ( a , b ) + ( 1 p ) Q ( a , b ) ] = A ( a , b ) [ μ sinh 1 ( μ ) ( 1 p ) ( 1 + μ 2 ) 1 / 4 p ] = A ( a , b ) [ ( 1 p ) ( 1 + μ 2 ) 1 / 4 + p ] sinh 1 ( μ ) G ( x ) ,
(3.12)

where

G ( x ) = x 4 1 ( 1 p ) x + p sinh 1 ( x 4 1 ) , G ( 1 ) = 0 ,
(3.13)
G( 2 )= 3 2 ( 2 1 ) p log(2+ 3 ),
(3.14)
G (x)= x 1 x 4 1 [ ( 1 p ) x + p ] 2 f(x),
(3.15)

where f(x) is defined as in Lemma 2.1.

We divide the proof into two cases.

Case 1: p=[ 2 log(2+ 3 ) 3 ]/[( 2 1)log(2+ 3 )]=0.2390 . Then from (3.14) and (3.15) together with Lemma 2.1(3) we clearly see that there exists λ 2 (1, 2 ) such that G(x) is strictly decreasing on (1, λ 2 ] and strictly increasing on [ λ 2 , 2 ), and

G( 2 )=0.
(3.16)

Therefore,

S C A ( a , b ) < 2 log ( 2 + 3 ) 3 ( 2 1 ) log ( 2 + 3 ) A ( a , b ) + 3 log ( 2 + 3 ) ( 2 1 ) log ( 2 + 3 ) Q ( a , b )
(3.17)

for all a,b>0 with ab follows easily from (3.12) and (3.13) together with (3.16) and the piecewise monotonicity of G(x).

Case 2: p=1/3. Then Lemma 2.1(1) and (3.15) lead to the conclusion that G(x) is strictly increasing on (1, 2 ). Therefore,

S C A (a,b)> 1 3 A(a,b)+ 2 3 Q(a,b)
(3.18)

for all a,b>0 with ab follows from (3.12) and (3.13) together with the monotonicity of G(x).

Note that

lim μ 0 + μ ( 1 + μ 2 ) 1 / 4 sinh 1 ( μ ) [ 1 ( 1 + μ 2 ) 1 / 4 ] sinh 1 ( μ ) = 1 3
(3.19)

and

lim μ 3 μ ( 1 + μ 2 ) 1 / 4 sinh 1 ( μ ) [ 1 ( 1 + μ 2 ) 1 / 4 ] sinh 1 ( μ ) = 2 log ( 2 + 3 ) 3 ( 2 1 ) log ( 2 + 3 ) .
(3.20)

Therefore, Theorem 1.2 follows from (3.11) and (3.17)-(3.20). □

Proof of Theorem 1.3 Without loss of generality, we assume that a>b. Let v=(ab)/(a+b), λ=v 2 v 2 , x= 1 λ 2 and p[0,1/2]. Then v,λ,x(0,1) and (1.4) leads to

G [ p a + ( 1 p ) b , p b + ( 1 p ) a ] S H A ( a , b ) = A ( a , b ) [ 1 ( 1 2 p ) 2 ( 1 1 λ 2 ) λ sin 1 ( λ ) ] = A ( a , b ) 1 ( 1 2 p ) 2 ( 1 1 λ 2 ) sin 1 ( λ ) H ( x ) ,
(3.21)

where

H ( x ) = sin 1 ( 1 x 2 ) 1 x 2 ( 1 2 p ) 2 x ( 1 2 p ) 2 + 1 , H ( 1 ) = 0 ,
(3.22)
H(0)= π 2 1 1 ( 1 2 p ) 2
(3.23)

and

H (x)= h ( x ) 2 1 x 2 [ ( 1 2 p ) 2 x ( 1 2 p ) 2 + 1 ] 3 / 2 ,
(3.24)

where

h ( x ) = ( 1 2 p ) 2 x 2 + 2 [ 1 ( 1 2 p ) 2 ] x + ( 1 2 p ) 2 2 [ ( 1 2 p ) 2 x ( 1 2 p ) 2 + 1 ] 3 / 2 = ( x 1 ) g ( x ) ( 1 2 p ) 2 x 2 + 2 [ 1 ( 1 2 p ) 2 ] x + ( 1 2 p ) 2 + 2 [ ( 1 2 p ) 2 x ( 1 2 p ) 2 + 1 ] 3 / 2 ,
(3.25)

where g(x) is defined as in Lemma 2.2.

We divide the proof into four cases.

Case 1: p=1/2 6 /6. Then Lemma 2.2(1) and (3.24) together with (3.25) lead to the conclusion that H(x) is strictly increasing on (0,1). Therefore,

S H A (a,b)>G [ ( 1 2 6 6 ) a + ( 1 2 + 6 6 ) b , ( 1 2 6 6 ) b + ( 1 2 + 6 6 ) a ]

for all a,b>0 with ab follows easily from (3.21) and (3.22) together with the monotonicity of H(x).

Case 2: 1/2 6 /6<p1/2. Let q= ( 1 2 p ) 2 and λ 0 + , then 0q<2/3 and power series expansions lead to

1 ( 1 2 p ) 2 ( 1 1 λ 2 ) λ sin 1 λ = 1 q ( 1 1 λ 2 ) sin 1 λ λ sin 1 λ = 1 sin 1 λ [ ( 1 6 q 4 ) λ 3 + o ( λ 3 ) ] .
(3.26)

Equations (3.21) and (3.26) imply that there exists small enough δ 1 >0 such that S H A (a,b)<G[pa+(1p)b,pb+(1p)a] for all a,b>0 with b/a(1 δ 1 ,1).

Case 3: p=1/2 π 2 4 /(2π). Then from Lemma 2.2(3) and (3.23)-(3.25) we clearly see that there exists λ 3 (0,1) such that H(x) is strictly increasing on (0, λ 3 ] and strictly decreasing on [ λ 3 ,1), and

H(0)=0.
(3.27)

Therefore,

S H A ( a , b ) < G [ ( 1 2 π 2 4 2 π ) a + ( 1 2 + π 2 4 2 π ) b , ( 1 2 π 2 4 2 π ) b + ( 1 2 + π 2 4 2 π ) a ]

for all a,b>0 with ab follows easily from (3.21) and (3.22) together with (3.27) and the piecewise monotonicity of H(x).

Case 4: 0p<1/2 π 2 4 /(2π). Then

lim λ 1 [ 1 ( 1 2 p ) 2 ( 1 1 λ 2 ) λ sin 1 ( λ ) ] = 1 ( 1 2 p ) 2 2 π <0.
(3.28)

Equation (3.21) and inequality (3.28) imply that there exists large enough M 1 >1 such that S H A (a,b)>G[pa+(1p)b,pb+(1p)a] for all a,b>0 with a/b( M 1 ,+). □

Proof of Theorem 1.4 Without loss of generality, we assume that a>b. Let v=(ab)/(a+b), μ=v 2 + v 2 , x= 1 + μ 2 and p[1/2,1]. Then v(0,1), μ(0, 3 ), x(1,2) and (1.5) leads to

Q [ p a + ( 1 p ) b , p b + ( 1 p ) a ] S C A ( a , b ) = A ( a , b ) [ 1 + ( 1 2 p ) 2 ( 1 + μ 2 1 ) μ sinh 1 ( μ ) ] = A ( a , b ) 1 + ( 1 2 p ) 2 ( 1 + μ 2 1 ) sinh 1 ( μ ) J ( x ) ,
(3.29)

where

J ( x ) = sinh 1 ( x 2 1 ) x 2 1 ( 1 2 p ) 2 x ( 1 2 p ) 2 + 1 , J ( 1 ) = 0 ,
(3.30)
J(2)=log(2+ 3 ) 3 1 + ( 1 2 p ) 2 ,
(3.31)
J ( x ) = 2 [ ( 1 2 p ) 2 x ( 1 2 p ) 2 + 1 ] 3 / 2 [ ( 1 2 p ) 2 x 2 + 2 ( 1 ( 1 2 p ) 2 ) x + ( 1 2 p ) 2 ] 2 x 2 1 [ ( 1 2 p ) 2 x ( 1 2 p ) 2 + 1 ] 3 / 2 J ( x ) = 1 2 [ ( 1 2 p ) 2 x ( 1 2 p ) 2 + 1 ] 3 / 2 + [ ( 1 2 p ) 2 x 2 + 2 ( 1 ( 1 2 p ) 2 ) x + ( 1 2 p ) 2 ] J ( x ) = × x 1 2 x 2 1 [ ( 1 2 p ) 2 x ( 1 2 p ) 2 + 1 ] 3 / 2 g ( x ) ,
(3.32)

where g(x) is defined as in Lemma 2.2.

We divide the proof into four cases.

Case 1: p=1/2+ 6 /6. Then Lemma 2.2(2) and (3.32) lead to the conclusion that J(x) is strictly increasing on (1,2). Therefore,

S C A (a,b)>Q [ ( 1 2 + 6 6 ) a + ( 1 2 6 6 ) b , ( 1 2 + 6 6 ) b + ( 1 2 6 6 ) a ]

for all a,b>0 with ab follows easily from (3.29) and (3.30) together with the monotonicity of J(x).

Case 2: 1/2+ 6 /6<p1. Let q= ( 1 2 p ) 2 and μ 0 + , then 1q>2/3 and power series expansions lead to

1 + ( 1 2 p ) 2 ( 1 + μ 2 1 ) μ sinh 1 ( μ ) = 1 + q ( 1 + μ 2 1 ) sinh 1 ( μ ) μ sinh 1 ( μ ) = 1 sinh 1 ( μ ) [ ( 1 4 q 1 6 ) μ 3 + o ( μ 3 ) ] .
(3.33)

Equations (3.29) and (3.33) imply that there exists small enough δ 2 >0 such that S C A (a,b)<Q[pa+(1p)b,pb+(1p)a] for all a,b>0 with b/a(1 δ 2 ,1).

Case 3: p=1/2+ 3 / [ log ( 2 + 3 ) ] 2 1 /2. Then (3.31) and (3.32) together with Lemma 2.2(4) lead to the conclusion that there exists λ 4 (1,2) such that J(x) is strictly increasing on (1, λ 4 ] and strictly decreasing on [ λ 4 ,2), and

J(2)=0.
(3.34)

Therefore,

S C A (a,b)<Q [ p a + ( 1 p ) b , p b + ( 1 p ) a ]

for all a,b>0 with ab follows easily from (3.29) and (3.30) together with (3.34) and the piecewise monotonicity of J(x).

Case 4: 1/2p<1/2+ 3 / [ log ( 2 + 3 ) ] 2 1 /2. Then

lim μ 3 [ 1 + ( 1 2 p ) 2 ( 1 + μ 2 1 ) μ sinh 1 ( μ ) ] = 1 + ( 2 p 1 ) 2 3 log ( 2 + 3 ) < 0 .
(3.35)

Equation (3.29) and inequality (3.35) imply that there exists large enough M 2 >1 such that S C A (a,b)>Q[pa+(1p)b,pb+(1p)a] for all a,b>0 with a/b( M 2 ,+). □

References

  1. Carlson BC: Algorithms involving arithmetic and geometric means. Am. Math. Mon. 1971, 78: 496–505. 10.2307/2317754

    Article  MathSciNet  MATH  Google Scholar 

  2. Borwein JM, Borwein PB: Pi and the AGM: A Study in Analytic Number Theory and Computational Complexity. Wiley, New York; 1987.

    MATH  Google Scholar 

  3. Neuman E, Sándor J: On the Schwab-Borchardt mean. Math. Pannon. 2003,14(2):253–266.

    MathSciNet  MATH  Google Scholar 

  4. Neuman E, Sándor J: On the Schwab-Borchardt mean II. Math. Pannon. 2006,17(1):49–59.

    MathSciNet  MATH  Google Scholar 

  5. Neuman E: A note on a certain bivariate mean. J. Math. Inequal. 2012,6(4):637–643.

    Article  MathSciNet  MATH  Google Scholar 

  6. Chu Y-M, Long B-Y: Bounds of the Neuman-Sándor mean using power and identric mean. Abstr. Appl. Anal. 2013., 2013: Article ID 832591

    Google Scholar 

  7. Zhao T-H, Chu Y-M, Liu B-Y: Optimal bounds for Neuman-Sándor mean in terms of the convex combinations of harmonic, geometric, quadratic, and contraharmonic means. Abstr. Appl. Anal. 2013., 2013: Article ID 302635

    Google Scholar 

  8. Neuman E: On some means derived from the Schwab-Borchardt mean. J. Math. Inequal. 2014,8(1):171–183.

    Article  MathSciNet  MATH  Google Scholar 

  9. Neuman E: On some means derived from the Schwab-Borchardt mean II. J. Math. Inequal. 2014,8(2):361–370.

    MathSciNet  MATH  Google Scholar 

  10. He Z-Y, Chu Y-M, Wang M-K: Optimal bounds for Neuman means in terms of harmonic and contraharmonic means. J. Appl. Math. 2013., 2013: Article ID 807623

    Google Scholar 

Download references

Acknowledgements

The authors would like to express their deep gratitude to the referees for giving many valuable suggestions. 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.

Author information

Authors and Affiliations

Authors

Corresponding author

Correspondence to Yu-Ming Chu.

Additional information

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

W-MQ provided the main idea and carried out the proof of Lemmas 2.1 and 2.2. Y-MC carried out the proof of Theorems 1.1-1.4. All authors read and approved the final manuscript.

Rights and permissions

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

Reprints and permissions

About this article

Check for updates. Verify currency and authenticity via CrossMark

Cite this article

Qian, WM., Chu, YM. Optimal bounds for Neuman means in terms of geometric, arithmetic and quadratic means. J Inequal Appl 2014, 175 (2014). https://doi.org/10.1186/1029-242X-2014-175

Download citation

  • Received:

  • Accepted:

  • Published:

  • DOI: https://doi.org/10.1186/1029-242X-2014-175

Keywords