A new proof of the geometric-arithmetic mean inequality by Cauchy's integral formula

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by  Feng Qi

A NEW PROOF OF THE GEOMETRIC-ARITHMETIC MEAN
INEQUALITY BY CAUCHY’S INTEGRAL FORMULA
3
1 FENGQI,XIAO-JINGZHANG,ANDWEN-HUILI
0
2
Abstract. Let a = (a1,a2,...,an) for n ∈ N be a given sequence of posi-
n tivenumbers. Inthepaper, theauthors establish, byusingCauchy’s integral
a formula in the theory of complex functions, an integral representation of the
J
principalbranchofthegeometricmean

8 n 1/n
2 Gn(a+z)= (ak+z)
"kY=1 #
A] forz∈C\(−∞,−min{ak,1≤k≤n}],andthen provideanew proofof the
wellknownGAmeaninequality.
C
.
h
t
a
1. Introduction
m
[ Let a = (a1,a2,...,an) for n ∈ N, the set of all positive integers, be a given
sequence of positive numbers. Then the arithmetic and geometric means An(a)
1
and G (a) of the numbers a ,a ,...,a are defined respectively as
v n 1 2 n
2 n
1
3 A (a)= a (1.1)
n n k
4
k=1
6 X
and
.
1 n 1/n
0 G (a)= a . (1.2)
n k
3 !
k=1
1 Y
It is common knowledge that
:
v
i Gn(a)≤An(a), (1.3)
X
with equality if and only if a =a =···=a .
r 1 2 n
a There has been a large number, presumably over one hundred, of proofs of
the GA mean inequality (1.3) in the mathematical literature. The most complete
information, so far, can be found in the monographs [1, 2, 3, 4, 5, 6] and a lot of
references therein.
In this paper, we establish, by using Cauchy’s integral formula in the theory of
complex functions, an integral representation of the principal branch of the geo-
metric mean
n 1/n
G (a+z)= (a +z) (1.4)
n k
" #
k=1
Y
2010 Mathematics Subject Classification. Primary26E60, 30E20;Secondary26A48,44A20.
Key words and phrases. Integral representation; Cauchy’s integral formula; Arithmetic mean;
Geometricmean;GAmeaninequality;Newproof.
Thispaperwastypeset usingAMS-LATEX.
1
2 F.QI,X.-J.ZHANG,ANDW.-H.LI
for z ∈C\(−∞,−min{a ,1≤k ≤n}], and provide a new proof of the GA mean
k
inequality (1.3).
2. Lemmas
In order to prove our main results, we need the following lemmas.
Lemma 2.1. For z ∈C\(−∞,−min{a ,1≤k ≤n}], the principal branch of the
k
complex function
f (z)=G (a+z)−z, (2.1)
n n
where a+z =(a +z,a +z,...,a +z), meets
1 2 n
lim f (z)=A (a). (2.2)
n n
z→∞
Proof. By L’Hoˆspital’s rule in the theory of complex functions, we have
a
lim f (z)= lim z G 1+ −1
z→∞ n z→∞ n z
(cid:26) (cid:20) (cid:18) (cid:19) (cid:21)(cid:27)
G (1+az)−1 d n 1/n
= lim n = lim (1+a z) =A (a),
z→0 z z→0dz" k # n
k=1
Y
where1+a = 1+a1,1+a2,...,1+an and1+az=(1+a z,1+a z,...,1+a z).
z z z z 1 2 n
Lemma 2.1 is thus proved. (cid:3)
(cid:0) (cid:1)
Lemma 2.2. For z ∈C\(−∞,0] and a=(a ,a ,...,a ) satisfying a ≤a for
1 2 n ℓ ℓ+1
1≤ℓ≤n−1, let
h (z)=G (a−a +z)−z, (2.3)
n n 1
where a−a +z =(z,a −a +z,...,a −a +z). Then the imaginary part of the
1 2 1 n 1
principal branch of h (z) meets
n
n 1/n
ℓπ
|a −a −t| sin , t∈(a −a ,a −a ]
lim ℑh (−t+iε)= k 1 n ℓ 1 ℓ+1 1
n " #
ε→0+ 0,kY=1 t>a −a
n 1
(2.4)
for 1≤ℓ≤n−1. 
Proof. For t=a −a for 1≤ℓ≤n−1, we have
ℓ+1 1
n
1 1
h (−t+iε)=exp ln(a −a −t+iε)+ ln(iε) +t−iε
n n k 1 n
" #
k6=ℓ+1
X
1 n 1 π
=exp ln(a −a −t+iε) exp ln|ε|+ i +t−iε
n k 1 n 2
" k6=ℓ+1 # (cid:20) (cid:18) (cid:19)(cid:21)
X
1 n 1 π
→exp ln(a −a −t) lim exp ln|ε|+ i +t
"nk6=ℓ+1 k 1 #ε→0+ (cid:20)n(cid:18) 2 (cid:19)(cid:21)
X
=t
as ε→0+. Hence, when t=a −a for 1≤ℓ≤n−1, we have
ℓ+1 1
lim ℑh (−t+iε)=0.
n
ε→0+
A NEW PROOF OF THE GEOMETRIC-ARITHMETIC MEAN INEQUALITY 3
For t∈(0,∞)\{a −a ,1≤ℓ≤n−1} and ε>0, we have
ℓ+1 1
h (−t+iε)=G (a−a −t+iε)+t−iε
n n 1
n
1
=exp ln(a −a −t+iε) +t−iε
n k 1
" #
k=1
X
n
1
=exp [ln|a −a −t+iε|+iarg(a −a −t+iε)] +t−iε
n k 1 k 1
( )
k=1
X
n
1 ℓπ
exp ln|a −a −t|+ i +t, t∈(a −a ,a −a )
n k 1 n ℓ 1 ℓ+1 1
 !
→exp 1 Xk=n1ln|a −a −t|+πi +t, t>a −a
n k 1 n 1
!
k=1
n  X1/n ℓπ ℓπ
|a −a −t| cos +isin +t, t∈(a −a ,a −a )
= kY=n1|ak−a1−t|!1/n(cid:18)(cosπn+isinπ)+nt(cid:19), t>aℓ−a1 ℓ+1 1
k 1 n 1
!
as ε→0+kY=. 1As a result, we have
n 1/n ℓπ
|a −a −t| sin , t∈(a −a ,a −a );
lim ℑh (−t+iε)= k 1 n ℓ 1 ℓ+1 1
n !
ε→0+ 0,kY=1 t>a −a .
n 1
The proof of Lemma 2.2 is completed. (cid:3)
3. An integral representation of the geometric mean
Now weareinapositionto establishanintegralrepresentationofthe geometric
mean G (a+z).
n
Theorem 3.1. Let 0 < a ≤ a for 1 ≤ k ≤ n−1 and a+z = (a +z,a +
k k+1 1 2
z,...,a +z) for z ∈ C\(−∞,−a ]. Then the principal branch of the geometric
n 1
mean G (a+z) has the integral representation
n
1 n−1 ℓπ aℓ+1 n 1/n dt
G (a+z)=A (a)+z− sin (a −t) . (3.1)
n n π Xℓ=1 n Zaℓ (cid:12)(cid:12)kY=1 k (cid:12)(cid:12) t+z
(cid:12) (cid:12)
Proof. By standard arguments, it is not difficult to(cid:12) see that (cid:12)
(cid:12) (cid:12)
lim [zh (z)]=0 and h (z)=h (z). (3.2)
n n n
z→0+
For any but fixed point z ∈ C\(−∞,0], choose 0 < ε < 1 and r > 0 such that
0<ε<|z|<r, and consider the positively oriented contour C(ε,r) in C\(−∞,0]
consisting of the half circle z = εeiθ for θ ∈ −π,π and the half lines z = x±iε
2 2
for x ≤ 0 until they cut the circle |z| = r, which close the contour at the points
(cid:2) (cid:3)
−r(ε)±iε,where 0<r(ε)→r as ε→0. By the famous Cauchy’sintegralformula
4 F.QI,X.-J.ZHANG,ANDW.-H.LI
in the theory of complex functions, we have
1 h (w)
h (z)= n dw
n 2πi w−z
IC(ε,r)
1 −π/2 iεeiθh εeiθ arg[−r(ε)+iε] ireiθh reiθ
= dθ+ dθ (3.3)
2πi εeiθ−z reiθ −z
(cid:20)Zπ/2 (cid:0) (cid:1) Zarg[−r(ε)−iε] (cid:0) (cid:1)
0 h (x+iε) −r(ε) h (x−iε)
+ n dx+ n dx .
x+iε−z x−iε−z
Z−r(ε) Z0 (cid:21)
By the limit in (3.2), it follows that
−π/2 iεeiθh εeiθ
n
lim dθ =0. (3.4)
ε→0+Zπ/2 εeiθ−(cid:0) z (cid:1)
By virtue of the limit (2.2) in Lemma 2.1, we deduce that
arg[−r(ε)+iε] ireiθh reiθ π ireiθh reiθ
n n
lim dθ = lim dθ
εr→→0∞+Zarg[−r(ε)−iε] reiθ −(cid:0) z (cid:1) r→∞Z−π reiθ −(cid:0) z (cid:1) (3.5)
=2A (a−a )πi,
n 1
where a−a =(0,a −a ,...,a −a ). Utilizing the second formula in (3.2) and
1 2 1 n 1
the limit (2.4) in Lemma 2.2 results in
0 h (x+iε) −r(ε) h (x−iε)
n dx+ n dx
x+iε−z x−iε−z
Z−r(ε) Z0
0 h (x+iε) h (x−iε)
= n − n dx
x+iε−z x−iε−z
Z−r(ε)(cid:20) (cid:21)
0 (x−iε−z)h (x+iε)−(x+iε−z)h (x−iε)
= n n dx
(x+iε−z)(x−iε−z)
Z−r(ε)
0 (x−z)[h (x+iε)−h (x−iε)]−iε[h (x−iε)+h (x+iε)]
= n n n n dx
(x+iε−z)(x−iε−z)
Z−r(ε)
0 (x−z)ℑh (x+iε)−εℜh (x+iε)
=2i n n dx
(x+iε−z)(x−iε−z)
Z−r(ε)
0 lim ℑh (x+iε)
→2i ε→0+ n dx
x−z
Z−r
r lim ℑh (−t+iε)
=−2i ε→0+ n dt
t+z
Z0
∞ lim ℑh (−t+iε)
→−2i ε→0+ n dt
t+z
Z0
n−1 ℓπ aℓ+1−a1 n 1/n dt
=−2i sin |a −a −t| (3.6)
n k 1 t+z
ℓ=1 Zaℓ−a1 "k=1 #
X Y
A NEW PROOF OF THE GEOMETRIC-ARITHMETIC MEAN INEQUALITY 5
asε→0+ andr →∞. Substituting equations(3.4),(3.5), and(3.6)into (3.3)and
simplifying generate
1 n−1 ℓπ aℓ+1−a1 n 1/n dt
h (z)=A (a−a )− sin |a −a −t| . (3.7)
n n 1 π n k 1 t+z
ℓ=1 Zaℓ−a1 "k=1 #
X Y
From f (z)=h (z+a )+a and (3.7), it is immediate to deduce that
n n 1 1
1 n−1 ℓπ aℓ+1−a1 n 1/n dt
f (z)=A (a−a )+a − sin |a −a −t|
n n 1 1 π n k 1 t+z+a
ℓ=1 Zaℓ−a1 "k=1 # 1
X Y
1 n−1 ℓπ aℓ+1 n 1/n dt
=A (a)− sin |a −t| ,
n π n k t+z
ℓ=1 Zaℓ "k=1 #
X Y
from which the integral representation (3.1) follows. Theorem 3.1 is proved. (cid:3)
4. A new proof of the GA mean inequality
With the aid of the integral representation (3.1) in Theorem 3.1, we can easily
deduce the GA mean inequality (1.3) as follows.
Taking z =0 in the integral representation (3.1) yields
1 n−1 ℓπ aℓ+1 n 1/ndt
G (a)=A (a)− sin |a −t| ≤A (a).
n n π n k t n
ℓ=1 Zaℓ "k=1 #
X Y
From this, it is also obvious that the equality in (1.3) is valid if and only if a =
1
a =···=a . The proof of the GA mean inequality (1.3) is complete.
2 n
References
[1] E.F.Beckenbach andR.Bellman,Inequalities,Springer,Berlin,1983.
[2] P. S. Bullen, Handbook of Means and Their Inequalities, Mathematics and its Applications
(Dordrecht) 560,KluwerAcademicPublishers,Dordrecht,2003.
[3] G. H. Hardy, J. E. Littlewood, and G. P´olya, Inequalities, 2nd ed., Cambridge University
Press,Cambridge,1952.
[4] J.-C. Kuang, Cha´ngy`ong Bu`dˇengsh`ı (Applied Inequalities), 3rd ed., Sh¯and¯ong K¯exu´e J`ıshu`
Chu¯bˇan Sh`e (Shandong Science and Technology Press), Ji’nan City, Shandong Province,
China,2004.(Chinese)
[5] D.S.Mitrinovi´c,Analytic Inequalities,Springer,NewYork-Heidelberg-Berlin,1970.
[6] D.S.Mitrinovi´candP.M.Vasi´c,Sredine,MatematiˇckaBiblioteka40,Beograd,1969.
(Qi)DepartmentofMathematics,SchoolofScience,TianjinPolytechnicUniversity,
TianjinCity, 300387,China
E-mail address: [email protected], [email protected], [email protected]
URL:http://qifeng618.wordpress.com
(Zhang)DepartmentofMathematics,SchoolofScience,TianjinPolytechnicUniver-
sity,Tianjin City, 300387,China
E-mail address: [email protected]
(Li)DepartmentofMathematics,SchoolofScience,TianjinPolytechnicUniversity,
TianjinCity, 300387,China
E-mail address: [email protected]