Table Of ContentSHARP BOUNDS FOR NEUMAN-SA´NDOR’S MEAN IN TERMS
OF THE ROOT-MEAN-SQUARE
3
1 WEI-DONGJIANGANDFENGQI
0
2
Abstract. Inthepaper,theauthorsfindsharpboundsforNeuman-S´andor’s
n meanintermsoftheroot-mean-square.
a
J
5
1
1. Introduction
]
A Throughout this paper, we assume that the real numbers a and b are positive
and that a=b.
C
6
The second Seiffert’s mean T(a,b) and Neuman-Sa´ndor’s mean M(a,b) are re-
.
h spectively defined in [5, 7] by
t
a a b a b
m T(a,b)= − and M(a,b)= − , (1.1)
2arctan a b 2arcsinh a b
[ a−+b a+−b
while the arithmetic mean and(cid:0)the(cid:1)root-mean-squareare respec(cid:0)tively(cid:1)defined by
1
v a+b a2+b2
7 A(a,b)= and S(a,b)= . (1.2)
2 2
6 r
2 A chain of inequalities between these four means
3
A<M <T <S
.
1
0 were established in [5, 6].
3 In [2], the authors demonstrated that the double inequality
1
: αS(a,b)+(1 α)A(a,b)<T(a,b)<βS(a,b)+(1 β)A(a,b)
v − −
i holds if and only if α 4 π and β 2.
X ≤ (√2−1)π ≥ 3
−
In [3, 4], the authors independently found that the double inequality
r
a
αS(a,b)+(1 α)A(a,b)<M(a,b)<βS(a,b)+(1 β)A(a,b)
− −
holds if and only if α 1−ln(1+√2) and β 1.
≤ (√2 1)ln(1+√2) ≥ 3
−
In [1], the authors discovered that the double inequality
S αa+(1 α)b,αb+(1 α)a <T(a,b)<S βa+(1 β)b,βb+(1 β)a
− − − −
holds(cid:0)if and only if α 1+√16/π2−(cid:1)1 and β 3+√(cid:0)6. (cid:1)
≤ 2 ≥ 6
2010 Mathematics Subject Classification. Primary26E60; Secondary26D99.
Key words and phrases. bound; Seiffert’s mean; root-mean-square; Neuman-S´andor’s mean; in-
equality.
This work was partially supported by the Project of Shandong Province Higher Educational
ScienceandTechnologyProgramundergrantNo. J11LA57.
Thispaperwastypeset usingAMS-LATEX.
1
2 W.-D.JIANGANDF.QI
Motivated by the above double inequalities, we naturally ask a question: What
are the best constants α 1 and β 1 such that the double inequality
≥ 2 ≤
S αa+(1 α)b,αb+(1 α)a <M(a,b)<S βa+(1 β)b,βb+(1 β)a (1.3)
− − − −
hol(cid:0)ds for all a,b>0 with a=b(cid:1)? (cid:0) (cid:1)
6
The aim of this paper is just to give an affirmative answer to this question.
The main result of this paper may be formulated as the following theorem.
Theorem 1.1. The double inequality (1.3) holds true if and only if
1 1 3+√3
α 1+ 1 =0.76... and β =0.78... .
≤ 2( s ln 1+√2 2 − ) ≥ 6
(cid:2) (cid:0) (cid:1)(cid:3)
2. Proof of Theorem 1.1
For simplicity, denote
1 1 3+√3
λ= 1+ 1 and µ= .
2( s ln 1+√2 2 − ) 6
It is clear that, in order to p(cid:2)ro(cid:0)ve the do(cid:1)(cid:3)uble inequality (1.3), it suffices to show
M(a,b)>S λa+(1 λ)b,λb+(1 λ)a (2.1)
− −
and (cid:0) (cid:1)
M(a,b)<S µa+(1 µ)b,µb+(1 µ)a . (2.2)
− −
From definitions in (1.1) and (1.2), we see that both M(a,b) and S(a,b) are
(cid:0) (cid:1)
symmetric and homogeneous of degree 1. Hence, without loss of generality, we
assume that a>b>0. If replacing a >1 by t>1 and letting p 1,1 , then
b ∈ 2
S(pa+(1 p)b,pb+(1 p)a) M(a,b) (cid:0) (cid:1)
− − −
b [pt+(1 p)]2+[p+(1 p)t]2
= − − f(t), (2.3)
2 arcsinht 1
p t+−1
where
t 1 t 1
f(t)=√2 arcsinh − − . (2.4)
t+1 − [pt+(1 p)]2+[p+(1 p)t]2
− −
Standard computations lead to
p
f(1)=0, (2.5)
1
lim f(t)=√2 ln 1+√2 , (2.6)
t − 2p2 2p+1
→∞ −
(cid:0) (cid:1)
and p
f (t)
f (t)= 1 , (2.7)
′
(1+t)√1+t2 [pt+(1 p)]2+[p+(1 p)t]2 3/2
{ − − }
where
f (t)=2 [pt+(1 p)]2+[p+(1 p)t]2 3/2 (1+t)2 1+t2 (2.8)
1
− − −
and (cid:8) (cid:9) p
2 [pt+(1 p)]2+[p+(1 p)t]2 3/2 2 (1+t)2 1+t2 2=(t 1)2g (t) (2.9)
1
− − − −
n (cid:0) (cid:1) o h p i
SHARP BOUNDS FOR NEUMAN-SA´NDOR’S MEAN 3
with
g (t)= 32p6 96p5+144p4 128p3+72p2 24p+3 t4
1
− − −
(cid:0) 2 64p6 192p5+240p4 160p3+48p2 1 t3(cid:1)
− − − −
+6(cid:0)32p6 96p5+112p4 64p3+24p2 8p+(cid:1)1 t2 (2.10)
− − −
2(cid:0)64p6 192p5+240p4 160p3+48p2 1 t (cid:1)
− − − −
+3(cid:0)2p6 96p5+144p4 128p3+72p2 24p+(cid:1)3
− − −
and
g (1)=16 6p2 6p+1 . (2.11)
1
−
Let (cid:0) (cid:1)
g (t) g (t)
g2(t)= 1′2 , g3(t)= 2′6 , and g4(t)=g3′(t).
Then simple computations result in
g (t)=2 32p6 96p5+144p4 128p3+72p2 24p+3 t3
2
− − −
(cid:0)3 64p6 192p5+240p4 160p3+48p2 1 t2(cid:1)
− − − − (2.12)
+6(cid:0)32p6 96p5+112p4 64p3+24p2 8p+(cid:1)1 t
− − −
6(cid:0)4p6 192p5+240p4 160p3+48p2 1 , (cid:1)
− − − −
g2(1)=16(cid:0)6p2 6p+1 , (cid:1) (2.13)
−
g3(t)= 32(cid:0)p6 96p5+1(cid:1)44p4 128p3+72p2 24p+3 t2
− − −
(cid:0) 64p6 192p5+240p4 160p3+48p2 1 t (cid:1) (2.14)
− − − −
+(cid:0)32p6 96p5+112p4 64p3+24p2 8p+(cid:1)1,
− − −
g (1)=16p4 32p3+48p2 32p+5, (2.15)
3
− −
g (t)=2 32p6 96p5+144p4 128p3+72p2 24p+3 t
4
− − − (2.16)
(cid:0) 64p6 192p5+240p4 160p3+48p2 1 , (cid:1)
− − − −
g4(1)=48(cid:0)p4 96p3+96p2 48p+7. (cid:1) (2.17)
− −
When p=λ, the quantities (2.6), (2.11), (2.13), (2.15), and (2.17) become
lim f(t)=0, (2.18)
t
→∞
8 4 ln 1+√2 2 3
g (1)=g (1)= − <0, (2.19)
1 2 − (cid:8) (cid:2) ln(cid:0) 1+√2(cid:1)(cid:3) 2 (cid:9)
7 ln 1+√2 4(cid:2) 4(cid:0) ln 1+(cid:1)√(cid:3)2 2 1
g (1)= − − <0, (2.20)
3 − (cid:2) (cid:0) (cid:1)ln(cid:3) 1+(cid:2)√2(cid:0) 4 (cid:1)(cid:3)
5 l(cid:2)n 1(cid:0)+√2 (cid:1)4(cid:3) 3
g (1)= − <0, (2.21)
4 − (cid:2) l(cid:0)n 1+√2(cid:1)(cid:3) 4
and (cid:2) (cid:0) (cid:1)(cid:3)
2 ln 1+√2 6 1
32p6 96p5+144p4 128p3+72p2 24p+3= − >0. (2.22)
− − − − (cid:2)2(cid:0)ln 1+√(cid:1)2(cid:3) 6
(cid:2) (cid:0) (cid:1)(cid:3)
4 W.-D.JIANGANDF.QI
Consequently,from(2.10),(2.12),(2.14),(2.16),and(2.18),itisveryeasytoobtain
that
lim g (t)= , lim g (t)= , lim g (t)= , lim g (t)= . (2.23)
1 2 3 4
t ∞ t ∞ t ∞ t ∞
→∞ →∞ →∞ →∞
From (2.16) and (2.22), it is immediate to derive that the function g (t) is strictly
4
increasing on [1, ), and so, by virtue of (2.21) and the final limit in (2.23), there
∞
exists a point t > 1 such that g (t) < 0 on [1,t ) and g (t) > 0 on (t , ).
0 4 0 4 0
∞
Hence, the function g (t) is strictly decreasing on [1,t ] and strictly increasing on
3 0
[t , ). Similarly, by (2.20) and the third limit in (2.23), there exists a point
0
∞
t > t > 1 such that g (t) is strictly decreasing on [1,t ] and strictly increasing
1 0 2 1
on [t , ). Further, by (2.19) and the second limit in (2.23), there exists a point
1
∞
t > t > 1 such that g (t) is strictly decreasing on [1,t ] and strictly increasing
2 1 1 2
on [t , ). Thereafter, by (2.7) to (2.9), (2.19), and the first limit in (2.23), there
2
∞
exists a point t >t >1 such that f(t) is strictly decreasing on [1,t ] and strictly
3 2 3
increasing on [t , ). As a result, the inequality (2.1) follows from equations (2.3)
3
∞
to (2.5), and (2.18), together with the piecewise monotonicity of f(t).
When p=µ, the equation (2.10) becomes
5t2+8t+5
g (t)= (t 1)2 >0 (2.24)
1
27 −
fort>1. Byequations(2.7)to(2.10)andtheinequality(2.24),itcanbeconcluded
that f(t) is strictly increasing and positive on [1, ). The inequality (2.2) follows.
∞
ItisnotdifficulttoverifythatthemeanS xa+(1 x)b,xb+(1 x)a iscontinuous
− −
and strictly increasing on 1,1 . From this monotonicity and inequalities (2.1)
2 (cid:0) (cid:1)
and(2.2),onecanconcludethatthedoubleinequality(1.3)holdstrueforallα λ
(cid:2) (cid:3) ≤
and β µ.
≥
For any givennumber 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
S(pa+(1 p)b,pb+(1 p)a)>M(a,b)
− −
holds for a (T , ). This tells us that the constant λ is the best possible.
b ∈ 0 ∞
For 1 <p<µ, from (2.11) one has
2
g (1)=16(6p2 6p+1)<0. (2.25)
1
−
From the inequality (2.25) and the continuity of g (t), there exists a number δ =
1
δ(p) > 0 such that the function g (t) is negative on (1,1+δ). This negativity,
1
together with (2.3), (2.5), (2.7), and (2.10), implies that for any 1 < p < µ, there
2
exists δ =δ(p)>0 such that the inequality
M(a,b)>S(pa+(1 p)b,pb+(1 p)a)
− −
is valid for a (1,1+δ). Consequently, the number µ is the best possible. The
b ∈
proof of Theorem 1.1 is complete.
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SHARP BOUNDS FOR NEUMAN-SA´NDOR’S MEAN 5
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(Jiang)DepartmentofInformationEngineering,WeihaiVocationalCollege,Weihai
City,ShandongProvince,264210,China
E-mail address: [email protected]
(Qi)SchoolofMathematicsandInformatics,HenanPolytechnicUniversity,Jiaozuo
City,Henan Province,454010,China
E-mail address: [email protected], [email protected], [email protected]
URL:http://qifeng618.wordpress.com
