Two sharp inequalities for bounding the Seiffert mean by the arithmetic, centroidal, and contra-harmonic means

0.12 MB·English

by  Wei-Dong Jiang

TWO SHARP INEQUALITIES FOR BOUNDING THE SEIFFERT
MEAN BY THE ARITHMETIC, CENTROIDAL, AND
CONTRA-HARMONIC MEANS
2
1
0
WEI-DONGJIANG,JIANCAO,ANDFENGQI
2

n
Abstract. Inthepaper,theauthorsfindthebestpossibleconstantsappeared
a
intwoinequalities forbounding the Seiffertmeanbythe linearcombinations
J
ofthearithmetic,centroidal,andcontra-harmonicmeans.
1
3

]
A
1. Introduction
C
Fora,b>0witha=b,theSeiffertmeanT(a,b)andthecentroidalmeanC(a,b)
. 6
h are defined respectively by
t
a a b
m T(a,b)= 2arcta−n a−b (1.1)
a+b
[
and (cid:0) (cid:1)
1 2 a2+ab+b2
v C(a,b)= . (1.2)
3(a+b)
2 (cid:0) (cid:1)
3 It is well known that
4
a+b a2+b2
6 A(a,b)= , G(a,b)=√ab, S(a,b)= ,
1. 2 r 2
0 C(a,b)= a2+b2, M (a,b)= p ap+bp
2 a+b p 2
r
1
for p = 0 are respectively the arithmetic, geometric, root-square, contra-harmonic
: 6
v andp-th power means of two positive numbers a and b, that the p-th power means
Xi Mp(a,b)iscontinuousandstrictlyincreasingwithrespecttop Rforfixeda,b>0
∈
with a=b, and that the inequalities in
r 6
a
G(a,b)=M (a,b)<A(a,b)=M (a,b)<C(a,b)
0 1
<S(a,b)=M (a,b)<C(a,b) (1.3)
2
holdfora,b>0witha=b. Formoreinformationonresultsofmeanvalues,please
6
refer to, for example, [11, 12, 19, 20, 21] and closely related references therein.
In [22], Seiffert proved the double inequality
A(a,b)=M (a,b)<T(a,b)<M (a,b)=S(a,b) (1.4)
1 2
2010 Mathematics Subject Classification. Primary 26E60; Secondary 11H60, 26A48, 26D05,
33B10.
Key words and phrases. Seiffert mean; Arithmetic mean; Centroidal mean; Contra-harmonic
mean;Inequality; Bestconstant.
Thefirstauthor waspartiallysupportedbytheProjectofShandong ProvinceHigherEduca-
tionalScienceandTechnologyProgramundergrantNo. J11LA57.
1
2 W.-D.JIANG,J.CAO,ANDF.QI
fora,b>0witha=b. In[13],Ha¨sto¨showedthatthefunction T(1,x) isincreasing
6 Mp(1,x)
with respect to x (0, ) if p 1. In [3, 5], the authors demonstrated that the
∈ ∞ ≤
double inequalities
α S(a,b)+(1 α )A(a,b)<T(a,b)<β S(a,b)+(1 β )A(a,b) (1.5)
1 1 1 1
− −
and
C α a+(1 α )b,α b+(1 α )a <T(a,b)
2 2 2 2
− −
(cid:0) <(cid:1) C β2a+(1 β2)b,β2b+(1 β2)a (1.6)
− −
hold for a,b>0 with a=b if and only if (cid:0) (cid:1)
6
4 π 2 1 4 3+√3
α − , β , α 1+ 1 , β . (1.7)
1 ≤ √2 1 π 1 ≥ 3 2 ≤ 2 rπ − ! 2 ≥ 6
−
For more(cid:0)informa(cid:1)tion on this topic, please refer to recently published papers [4, 6,
7, 8, 9, 10, 14, 15, 16, 18, 23, 24] and cited references therein.
For positive numbers a,b>0 with a=b, let
6
J(x)=C xa+(1 x)b,xb+(1 x)a (1.8)
− −
on 1,1 . It is not difficult to(cid:0)directly verify that J(x) is c(cid:1)ontinuous and strictly
2
increasing on 1,1 and to notice that
(cid:2) (cid:3) 2
(cid:2)1 (cid:3)
J =A(a,b)<T(a,b) and J(1)=C(a,b)>T(a,b). (1.9)
2
(cid:18) (cid:19)
Therefore,itismuchnaturaltoaskaquestion: Whatarethebestconstantsα 1
≥ 2
and β 1 such that the double inequality
≤
C αa+(1 α)b,αb+(1 α)a <T(a,b)<C βa+(1 β)b,βb+(1 β)a (1.10)
− − − −
hol(cid:0)ds for a,b>0 with a=b? (cid:1) (cid:0) (cid:1)
6
ThefollowingTheorem1.1,thefirstmainresultofthispaper,givesanaffirmative
answer to this question.
Theorem 1.1. For positive numbers a,b > 0 with a = b, the double inequal-
6
ity (1.10) is valid if and only if
1 12
α 1+ 3 and β =1. (1.11)
≤ 2 r π − !
In [17] the author posed an unsolved problem: Find the greatest value α and
1
the least value β such that the double inequality
1
α C(a,b)+(1 α )A(a,b)<T(a,b)<β C(a,b)+(1 β )A(a,b) (1.12)
1 1 1 1
− −
holds for a,b>0 with a=b.
6
The following Theorem 1.2, the second main result of this paper, solves this
problem.
Theorem 1.2. for a,b > 0 with a = b, the double inequality (1.12) holds if and
6
only if α 4 1 and β 1.
1 ≤ π − 1 ≥ 3
TWO SHARP INEQUALITIES FOR BOUNDING SEIFFERT MEAN 3
2. Proof of Theorem 1.1
In this section, we supply a proof of Theorem 1.1.
For simplicity, we denote two numbers in (1.11) by λ and µ respectively.
It is clearthat, in orderto provethe double inequality (1.10), it suffices to show
T(a,b)>C λa+(1 λ)b,λb+(1 λ)a (2.1)
− −
and (cid:0) (cid:1)
T(a,b)<C µa+(1 µ)b,µb+(1 µ)a . (2.2)
− −
Fromdefinitions(1.1)and(1.2)(cid:0)weseethatbothT(a,b)and(cid:1)C(a,b)aresymmetric
and homogenous of degree 1. Hence, without loss of generality, we assume that
a>b. If replacing a >1 by t>1 and letting p 1,1 , then
b ∈ 2
C pa+(1 p)b,pb+(1 p)a T(a,b) (cid:0) (cid:1)
− − −
(cid:0) [pt+(1 p)]2+[pt+(cid:1)(1 p)][p+(1 p)t]+[p+(1 p)t]2
= − − − − bf(t), (2.3)
6(1+t)arctant−1
t+1
where
t 1
f(t)=4arctan −
t+1
(2.4)
3(t2 1)
− .
− [pt+(1 p)]2+[pt+(1 p)][p+(1 p)t]+[p+(1 p)t]2
− − − −
Standard computations lead to
f(1)=0, (2.5)
3
lim f(t)=π , (2.6)
t→∞ − p2 p+1
−
and
f (t)
f′(t)= 1 , (2.7)
h (t)
1
where
f (t)= 4p4 8p3+18p2 14p+1 t4 4 4p4 8p3+9p2 5p+1 t3
1
− − − − −
+(cid:0) 6 4p4 8p3+6p2 2p+1(cid:1) t2 4(cid:0) 4p4 8p3+9p2 5p+1(cid:1) t (2.8)
− − − − −
+4p(cid:0)4 8p3+18p2 14p+1(cid:1), (cid:0) (cid:1)
− −
f (1)=0, (2.9)
1
and
h (t)= [pt+(1 p)]2+[pt+(1 p)][p+(1 p)t]+[p+(1 p)t]2 2 1+t2 .
1
− − − −
Let (cid:8) (cid:9) (cid:0) (cid:1)
f′(t) f′(t) f′(t)
f (t)= 1 , f (t)= 2 , and f (t)= 3 .
2 3 4
4 3 2
Then, by standard argument, we have
f (t)= 4p4 8p3+18p2 14p+1 t3 3 4p4 8p3+9p2 5p+1 t2
2
− − − − − (2.10)
+(cid:0) 3 4p4 8p3+6p2 2p+1(cid:1) t 4(cid:0)p4 8p3+9p2 5p+1 ,(cid:1)
− − − − −
f (1)=0, (2.11)
2 (cid:0) (cid:1) (cid:0) (cid:1)
4 W.-D.JIANG,J.CAO,ANDF.QI
f (t)= 4p4 8p3+18p2 14p+1 t2 2 4p4 8p3+9p2 5p+1 t
3
− − − − − (2.12)
+(cid:0) 4p4 8p3+6p2 2p+1,(cid:1) (cid:0) (cid:1)
− −
f (1)=6p2 6p, (2.13)
3
−
f (t)= 4p4 8p3+18p2 14p+1 t 4p4 8p3+9p2 5p+1 , (2.14)
4
− − − − −
f4(1)=(cid:0)9p2 9p. (cid:1) (cid:0) (cid:1) (2.15)
−
If p=λ, then the quantities (2.6), (2.13), and (2.15) become
lim f(t)=0, (2.16)
t→∞
18
f (1)= 6<0, (2.17)
3 π −
27
f (1)= 9<0, (2.18)
4 π −
and
36+18π 9π2
4p4 8p3+18p2 14p+1= − >0. (2.19)
− − π2
Thus, from (2.8), (2.10), (2.12), (2.14), and (2.19), it is very easy to obtain that
lim f (t)= , (2.20)
t→∞ 1 ∞
lim f (t)= , (2.21)
t→∞ 2 ∞
lim f (t)= , (2.22)
t→∞ 3 ∞
lim f (t)= . (2.23)
t→∞ 4 ∞
From (2.14) and (2.19), it is clear that the function f (t) is strictly increasing on
4
[1, ), and so, by virtue of (2.18) and (2.23), there exists a point t > 1 such
0
∞
that f (t) < 0 on [1,t ) and f (t) > 0 on (t , ). Hence, the function f (t) is
4 0 4 0 3
∞
strictly decreasing on [1,t ] and strictly increasing on [t , ). Similarly, by (2.17)
0 0
∞
and (2.22), there exists a point t > t > 1 such that f (t) is strictly decreasing
1 0 2
on [1,t ] and strictly increasing on [t , ), and, by (2.11) and (2.21), there exists
1 1
∞
a point t > t > 1 such that f (t) is strictly decreasing on [1,t ] and strictly
2 1 1 2
increasing on [t , ). Further, by (2.7), (2.9), and (2.20), there exists a point
2
∞
t >t >1 such that f(t) is strictly decreasing on [1,t ] and strictly increasing on
3 2 3
[t , ). Finally,by (2.3)and(2.16),itisdeducedthatthefunctionf(t)isnegative
3
∞
on (1, ). The inequality (2.1) is thus proved.
∞
If p=µ=1, then the function (2.8) becomes
f (t)=(t 1)4 >0 (2.24)
1
−
for t > 1. Combining this with (2.7) and (2.5) results in that f(t) is strictly
increasing and positive on (1, ). Therefore, the inequality (2.2) is obtained.
∞
Combiningtheinequalities(2.1)and(2.2)withthemonotonicityofJ(x)defined
by (1.8), the double inequality (1.10) is established for all α λ and β 1.
≤ ≥
For any given number p satisfying 1 > p > λ, it is obvious that the limit (2.6)
is positive. This positivity together with (2.3) and (2.4) implies that for 1>p>λ
there exists T =T (p)>1 such that the inequality
0 0
C pa+(1 p)b,pb+(1 p)a >T(a,b)
− −
holds for ab ∈(T0,∞). (cid:0)This tells us that the consta(cid:1)nt λ is the best possible.
TWO SHARP INEQUALITIES FOR BOUNDING SEIFFERT MEAN 5
For 1 < p < µ = 1, the quantity (2.13) is positive. Accordingly, there exists a
2
number δ = δ(p) > 0 such that the function f (t) is negative on (1,1+δ). This
3
negativity together with (2.3), (2.5), (2.7) and (2.9) implies that for any 1 < p <
2
µ=1, there exists δ =δ(p)>0 such that the inequality
T(a,b)>C pa+(1 p)b,pb+(1 p)a
− −
is valid for ab ∈ (T0,∞). Conse(cid:0)quently, the number µ is th(cid:1)e best possible. The
proof of Theorem 1.1 is complete.
3. Proof of Theorem 1.2
In order to prove Theorem 1.2, we need the following Lemmas.
Lemma 3.1. The Bernoulli numbers B for n N have the property
2n
∈
( 1)n−1B = B , (3.1)
2n 2n
− | |
where the Bernoulli numbers B for i 0 are defined by
i
≥
∞ ∞
x B x x2i
= ixi =1 + B , x <2π. (3.2)
ex 1 n! − 2 2i(2i)! | |
− i=0 i=1
X X
Proof. In [2, p. 16 and p. 56], it is listed that for q 1
≥
(2π)2q B
ζ(2q)=( 1)q−1 2q, (3.3)
− (2q)! 2
where ζ is the Riemann zeta function defined by
∞
1
ζ(s)= . (3.4)
ns
n=1
X
From (3.3), the formula (3.1) follows. (cid:3)
Lemma 3.2. For 0< x <π,
| |
∞
1 22n B
cotx= | 2n|x2n−1. (3.5)
x − (2n)!
n=1
X
Proof. This may be derived readily from combining the formula [1, p. 75, 4.3.70]
with the identity (3.1). (cid:3)
Lemma 3.3. For 0< x <π,
| |
∞
1 1 22n(2n 1)B
= + − | 2n|x2(n−1). (3.6)
sin2x x2 (2n)!
n=1
X
Proof. Since
1 d
=csc2x= (cotx),
sin2x −dx
the formula (3.6) follows from differentiating (3.5). (cid:3)
Now we are ready to prove Theorem 1.2. It is easy to see that the double
inequality (1.12) is equivalent to
T(a,b) A(a,b)
α < − <β . (3.7)
1 C(a,b) A(a,b) 1
−
6 W.-D.JIANG,J.CAO,ANDF.QI
Without loss of generality, we assume a>b>0 and let x= a. Then x>1 and
b
x−1 x+1
T(a,b) A(a,b) 2arctanx−1 − 2
− = x+1 .
C(a,b) A(a,b) x2+1 x+1
− x+1 − 2
Let t= x−1. Then t (0,1) and
x+1 ∈
T(a,b) A(a,b) t 1
− = arctant − .
C(a,b) A(a,b) t2
−
Let t=tanθ for θ 0,π . Then
∈ 4
T(a,(cid:0)b) A(cid:1)(a,b) tanθ 1 cotθ 1
− = θ − = +1.
C(a,b) A(a,b) (tanθ)2 θ − sin2θ
−
By Lemmas 3.2 and 3.3, we have
∞ ∞
cotθ 1 22n (2n 1)22n
+1=1 B θ2n−2 − B θ2n−2
θ − sin2θ − (2n)!| 2n| − (2n)! | 2n|
n=1 n=1
X X
∞
n22n+1
=1 B θ2n−2
− (2n)! | 2n|
n=1
X
which is strictly decreasing on 0,π . Moreover, by L’Hoˆspital rule and standard
4
argument, we have
(cid:0) (cid:1)
1 4
lim = and lim = 1.
x→0+ 3 x→(π/4)− π −
The proof of Theorem 1.2 is complete.
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(Jiang)DepartmentofInformationEngineering,WeihaiVocationalUniversity,Wei-
haiCity, ShandongProvince,264210,China
E-mail address: [email protected]
(Cao) Department of Mathematics, Hangzhou Normal University, Hangzhou City,
ZhejiangProvince,310036,China
E-mail address: [email protected], [email protected]
(Qi)SchoolofMathematicsandInformatics,HenanPolytechnicUniversity,Jiaozuo
City, Henan Province, 454010, China; Department of Mathematics, School of Science,
TianjinPolytechnic University,Tianjin City, 300387,China
E-mail address: [email protected], [email protected], [email protected]
URL:http://qifeng618.wordpress.com