ANOTHER NEW PROOF METHOD OF ANALYTIC INEQUALITY

This paper gives another new proof method of analytic inequality involving n variables.. As its applications, we give proof of some known-well inequalities and prove five new analytic in

XIAO-MING ZHANG, BO-YAN XI, AND YU-MING CHU

Abstract This paper gives another new proof method of analytic inequality involving n variables.

As its applications, we give proof of some known-well inequalities and prove five new analytic

inequalities.

1 monotonicity on special variables Throughout the paper R denotes the set of real numbers and R+ denotes the set of strictly positive real numbers Let n ≥ 2, n ∈ N The arithmetic mean A(x) and the power mean

Mr(x) of order r with respect to the positive real numbers x1, x2, · · · , xn are defined respectively

as A(x) = 1nPn

i=1xi, Mr(x) = n1Pn

i=1xr i

1/r

for r 6= 0, and M0(x) = (Qn

i=1xi)1/n

In paper [4], the author puts up a new proof method of analytic inequality In this section, we shall provide another new proof method of analytic inequality involving n variables

Lemma 1.1 Let interval I = [m, M ] ⊂ R, Ddef.= {(x1, x2) |m ≤ x2 ≤ x1 ≤ M } ⊂ I2 and function

f : I2→ R have continuous partial derivative Then ∂f/∂x1 ≥ (≤) ∂f /∂x2 hold in D, if and only

if f (a, b) ≥ (≤) f (a − l, b + l) hold for all a, b ∈ I and l with b < b + l ≤ a − l < a

Proof Without the losing of generality, we only prove the case ∂f /∂x1 ≥ ∂f /∂x2

For all x1, x2 ∈ D and l ∈ R+with m ≤ x2< x2+l ≤ x1−l < x1≤ M , we have f (x1− l, x2+ l)−

f (x1, x2) ≤ 0 Then it exists ξl∈ (0, l) such that

l



−∂f (x1− ξl, x2+ ξl)

∂x1

+∂f (x1− ξl, x2+ ξl)

∂x2



≤ 0,

−∂f (x1− ξl, x2+ ξl)

∂x1

+∂f (x1− ξl, x2+ ξl)

∂x2

≤ 0

Let l → 0+, we get

∂f (x1, x2)

∂x1

≥ ∂f (x1, x2)

∂x2

According to continuity of partial derivative, we know

∂f (x1, x1)

∂x1

≥ ∂f (x1, x1)

∂x2

hold also

Date: September 13, 2009.

2000 Mathematics Subject Classification Primary 26A48, 26B35, 26D20,

Key words and phrases monotone, maximum, minimum, inequality.

This paper was typeset using AMS-L A TEX.

1

On the other hand, assumes ∂f /∂x1≥ ∂f /∂x2 hold in D For all a, b ∈ I and l with b < b + l ≤

a − l < a, it exists ξl∈ (0, l) such that

f (a, b) − f (a − l, b + l) = − (f (a − l, b + l) − f (a, b))

= −l



−∂f (a − ξl, b + ξl)

∂f (a − ξl, b + ξl)

∂x2



= l ∂f (a − ξl, b + ξl)

∂x1

−∂f (a − ξl, b + ξl)

∂x2



≥ 0

Theorem 1.1 (Compressed independent variables theorem) Suppose that D ⊂ Rn be a symmetric with respect to permutations and convex set, and it have a nonempty interior set D0, function f : D → R and its partial derivatives be continue, and



x ∈ D|xi = max

1≤j≤n{xj}



− {x ∈ D|x1= x2 = · · · = xn} ,



x ∈ D|xi = min

1≤j≤n{xj}



− {x ∈ D|x1 = x2 = · · · = xn} ,

i = 1, 2, · · · , n For all i, j = 1, 2, · · · , n, i 6= j,

∂f /∂xi > (<) ∂f /∂xj

in D^i∩D_j Then

(1.3) f (a1, a2, · · · , an) ≥ (≤) f (A (a) , A (a) , · · · , A (a))

for all a = (a1, a2, · · · , an) ∈ D, equality hold if only if a1 = a2= · · · = an

Proof If n = 2, Let l → |a1− a2|/2 in Lemma 1.1, we complete the proof of theorem We suppose

n ≥ 3 Without the losing of generality, we only prove the case ∂f /∂xi > ∂f /∂xj with i 6= j

If a1 = a2 = · · · = an, the inequality (1.3) hold obviously If max

1≤j≤n{aj} 6= min

1≤j≤n{aj}, we let

a1= max

1≤j≤n{aj} and an= min

1≤j≤n{aj}

(1) If a1 > max

2≤j≤n{aj}, an< min

1≤j≤n−1{aj}, because ∂f /∂xi > ∂f /∂xj hold inD^1∩D_n, According

to Lemma 1.1, exist a(1)1 , a(1)n such that l = a1− a(1)1 = a(1)n − an> 0 and a(1)1 = ai0 = max

2≤j≤n−1{aj} (let a(1)1 = a2 briefly ), or a(1)n = aj 0 = min

2≤j≤n−1{aj} (let a(1)n = an−1 briefly ), we have

f (a1, a2, a3, · · · , an) ≥ f



a(1)1 , a2, a3, · · · , a(1)n

 Simply, we denote a(1)i = ai, 2 ≤ i ≤ n − 1 Consequently,

f (a1, a2, a3, · · · , an) ≥ f



a(1)1 , a(1)2 , a(1)3 , · · · , a(1)n



If a(1)1 = a(1)2 = · · · = a(1)n , implies Theorem 1.1 hold Otherwise, for a(1)1 = a(1)2 > a(1)n , owing to

∂f (x)

∂x1

x=a(1),a(1),a(1),··· ,a(1)

> ∂f (x)

∂xn

x=a(1),a(1),a(1),··· ,a(1)

,

and the continuity of partial derivatives, it exists ε > 0 such that

∂f (x)

∂x1

x=s,a(1)2 ,a(1)3 ,··· ,t

> ∂f (x)

∂xn

x=s,a(1)2 ,a(1)3 ,··· ,t

,

where s ∈

h

a(1)1 − ε, a(1)1 it ∈

h

a(1)n , a(1)n + ε

i Denote a(2)1 = a(1)1 − ε, a(2)n = a(1)n + ε, a(2)i = a(1)i (2 ≤

i ≤ n − 1) By Lemma 1.1, we get



a(1)1 , a(1)2 , a(1)3 , · · · , a(1)n



≥ fa(2)1 , a(2)2 , a(2)3 , · · · , a(2)n

 , and a(2)2 = max

1≤i≤n

n

a(2)i o For a(1)1 > a(1)n−1 = a(1)n , after a similar argument, we get inequality (1.4) and a(2)n−1= min

1≤i≤n

n

a(2)i

o Repeated the above steps, we getna(i)1 , a(i)2 , · · · , a(i)n o(i = 1, 2, · · ·),Pn

j=1a(i)j are constant,na(i)j o (i = 1, 2, · · ·) are monotone increasing (decreasing) sequences if aj ≥ (≤) A (a) , j = 1, 2, 3, · · · , n, and

fa(1)1 , a(1)2 , a(1)3 , · · · , a(1)n ≥ fa(i)1 , a(i)2 , a(i)3 , · · · , a(i)n 

If exists i ∈ N, a(i)1 = a(i)2 = · · · = a(i)n , we complete the proof of theorem Otherwise, let α = inf

i∈N

n

max

n

a(i)1 , a(i)2 , · · · , a(i)n

oo Without the losing of generality, we suppose max

n

a(ij )

1 , a(ij )

2 , · · · , a(ij )

n

o

= a(ij )

and

lim

j→+∞



a(i1j), a(i2j), · · · , a(inj)



= (α, b2, b3, · · · , bn) , where {ij}+∞j=1 is a subsequence of N Because of the continuity of function f , it have

f (a1, a2, a3, · · · , an) ≥ f (α, b2, b3, · · · , bn)

If α 6= min {b2, b3, · · · , bn}, we can repeat the above arguments, this contradicts with the definition

of α Then α = b2 = b3 = · · · = bn.Owing to α +Pn

i=2bi =Pn

i=1ai, we have α = b2 = b3 = · · · =

bn= A(a), the proof of Theorem 1.1 is completed

(2) For the case a1 = max

2≤j≤n{aj}, or an= min

1≤j≤n−1{aj}, it have been proved in (1) 

In particular, according to Theorem 1.1 the following corollary hold

Corollary 1.1 Suppose that D ⊂ Rn is a symmetric with respect to permutations and convex set, and it has a nonempty interior set D0, function f : D → R is symmetric, f and its partial derivatives is continue Let

^

D1 =



x ∈ D|x1 = max

1≤j≤n{xj}



− {x ∈ D|x1= x2 = · · · = xn} ,

_

D2 =



x ∈ D|x2 = min

1≤j≤n{xj}



− {x ∈ D|x1= x2 = · · · = xn} ,

If ∂f /∂x1 > (<) ∂f /∂x2 hold in D∗, then

f (a1, a2, · · · , an) ≥ (≤) f (A (a) , A (a) , · · · , A (xa)) for all a = (a1, a2, · · · , an) ∈ D, equality hold if only if a1 = a2 = · · · = an

2 Unifying Proof of Some Well-known Inequality Take advantage of Theorem 1.1 and Corollary 1.1, we can prove some well-known inequality, for example, Power Mean Inequality, Holder-Inequality, Minkowski-Inequality In this section, we only prove an example

Proposition 2.1 (Holder-Inequality) Let (x1, x2, · · · , xn), (y1, y2, · · · , yn) ∈ Rn+, p, q > 1, and 1/p + 1/q = 1 Then

Xn k=1xpk1/pXn

k=1yqk1/q ≥Xn

k=1xkyk Proof Let (a1, a2, · · · , an) ∈ Rn+ and function

f : b →Xn

k=1ak1/pXn

k=1akbk1/q−Xn

k=1akb1/qk , b ∈ Rn+

We have

∂f

∂bi = 1/q ·

Xn k=1ak

1/pXn

k=1akbk

1/q−1

ai− 1/q · aib1/q−1i ,

∂f

∂bi

− ∂f

∂bj

= 1 q

 Pn k=1ak

Pn k=1bkak

1/p

(ai− aj) − 1

q



aib−1/pi − ajb−1/pj  Let b ∈D^i∩D_j (see (1.1) and (1.2)) (1)If ai≥ aj,

∂f

∂bi

− ∂f

∂bj

≥ 1 q

 Pn k=1ak

biPnk=1ak

1/p

(ai− aj) −1

q



aib−1/pi − ajb−1/pj 

= 1

qaj



b−1/pj − b−1/pi > 0

(2)If ai ≤ aj,

∂f

∂bi

− ∂f

∂bj

≥ 1 q

 Pn k=1ak

bjPnk=1ak

1/p

(ai− aj) −1

q



aib−1/pi − ajb−1/pj 

= 1

qai



b−1/pj − b−1/pi > 0

According to 1.1, we get

f (b) ≥ f (A (b) , A (b) , · · · , A (b)) that is

k=1ak

1/pXn

k=1akbk

1/q

−Xn

k=1akb1/qk ≥ 0

Let ak= xpk, , bk= yqkxp

3 Five new inequalities

Let n ≥ 3, a = (a1, a2, · · · , an) ∈ Rn+,Qk

1≤i 1 <···<i k ≤n

1 k

Pk j=1ai j

!1 ,

n k

!

is the third symmetric mean of a (see [2])

Theorem 3.1 Let 2 ≤ k ≤ n − 1, p = (k − 1)/(n − 1) Then

n(a) ≥ [A (a)]p[M0(a)]1−p,

p = (k − 1)/(n − 1) is the best constant

Proof Suppose

f (a) =hYn

i=1aii−

(n−k)·



 n k





n(n−1)

1≤i 1 <···<i k ≤n

1 k

Xk j=1aij, a = (a1, a2, · · · , an) ∈ Rn+ Then

∂f

∂a1

= − (n − k) ·

 n k



n (n − 1) a1

hYn i=1ai

i−

(n−k)·



 n k





n(n−1)

1≤i 1 <···<ik≤n

1 k

Xk j=1ai j,

∂f

∂a1 −

∂f

∂a2 = −

(n − k) ·

 n k



n (n − 1)

hYn i=1ai

i−

(n−k)·



 n k





n(n−1)

1≤i 1 <···<i k ≤n

1 k

Xk j=1ai j

 1

a1 −

1

a2



= (n − k) ·

 n k



n (n − 1) a1a2

hYn i=1aii−

(n−k)·



 n k





n(n−1)

1≤i 1 <···<i k ≤n

1 k

Xk j=1aij(a1− a2)

−hYn

i=1aii

−

(n−k)·



 n k





n(n−1)

1≤i 1 <···<i k ≤n

1 k

Xk j=1aij

3≤i 1 <···i k−1 ≤n

a1− a2



a1+Pk−1

j=1ai j

 

a2+Pk−1

j=1ai j

 (3.2)

If a ∈ D∗ (see (1.5)),

a1+ (k − 1) a2 > a1, (n − k) ·

 n k



n (n − 1) a1a2 >



n − 2

k − 1



ka2(a1+ (k − 1) a2), (n − k) ·

 n k



n (n − 1) a1a2 >

X

3≤i 1 <···i k−1 ≤n

1 (a1+ (k − 1) a2) ka2,

(3.3)

(n − k) ·

 n k



n (n − 1) a1a2 >

X

3≤i 1 <···i k−1 ≤n

1



a1+Pk−1

j=1aij a2+Pk−1

j=1aij Combining inequality (3.2) to inequality (3.3), we have ∂f /∂a1− ∂f /∂a2 > 0 Owing to Corollary 1.1, we get

f (a1, a2, · · · , an) ≥ (≤) f (A (a) , A (a) , · · · , A (a))

for all a = (a1, a2, · · · , an) ∈ Rn+ It implies

hYn

i=1ai

i−

(n−k)·



 n k





n(n−1)

1≤i 1 <···<ik≤n

k−1Xk

j=1ai j ≥ [A (a)]

(k−1)·



 n k





Inequality (3.1) is proved

Let a1 = a2 = · · · = an−1= 1, an= x in inequality (3.1), it lead to

 x + k − 1 k

 n − 1

k − 1

!,

n k

!

≥ x + n − 1

n

p

n

√

x
1−p

p ≤

k

nlnx+k−1k −1

nln x

ln (x + n − 1) − ln n√n

x. Let x → +∞ in above inequality,

p ≤ lim

x→+∞

k

n·x+k−11 −nx1

1 x+n−1 − 1

nx

= lim

x→+∞

kx x+k−1 − 1

nx x+n−1− 1 =

k − 1

n − 1.

Theorem 3.2 Suppose n ≥ 3, a = (a1, a2, · · · , an) ∈ Rn+, β > 0 > α If β + α > 0, let λ = n(β−α)−2α

If β + α ≤ 0, let λ = n1 Then

(3.4) [Mα(a)]1−λ· [Mβ(a)]λ ≤ M0(a)

Proof Let f (x) = nβ1 ln (Qn

i=1xi) −1−λα lnn1 Pn

i=1xα/βi , x ∈ Rn+ Then

∂f (x)

∂xj

nβxj

− 1 − λ β

xα/β−1j

Pn i=1xα/βi , j = 1, 2,

∂x1

−∂f (x)

∂x2

= x2− x1 nβx1x2

−1 − λ β

xα/β−11 − xα/β−12

Pn i=1xα/βi . Case 1 :α + β > 0 Let

g (t) = β + α

β − αt

β−α− tβ+ t−α−β + α

β − α, t ∈ (1, +∞) Then

tα+1g0(t) = (β + α) tβ− βtβ+α− α,

tα+1g0(t)0 = (β + α) βtβ+α−1 t−α− 1
> 0

Therefore tα+1g0(t) is monotone increasing function in (1, +∞) Meanwhile

lim

t→1+tα+1g0(t) = lim

t→1+

h (β + α) tβ− βtβ+α− αi= 0

Thence, tα+1g0(t) > 0, g0(t) > 0 In addition to lim

t→1+g (t) = 0, we have g (t) > 0 and

β + α

β − αt

β−α− tβ+ t−α− β + α

β − α > 0

β + α

β − αt

β−



1 + 2α

β − α



tα− tα+β+ 1 > 0,

β + α

β − αt

β−



n − 1 + 2α

β − α



tα− tα+β+ (n − 1) > 0,

(3.6) (1 − nλ) tβ− (n − 1 − nλ) tα− tα+β+ (n − 1) > 0,

α−β

tα+ (n − 1) >

tβ− 1

ntβ

We assume that x ∈ D∗ (see (1.5)), let t = (x1/x2)1/β in above inequality It has

(1 − λ) x

α/β−1

2 − xα/β−11

xα/β1 + (n − 1) xα/β2

> x1− x2

nx1x2 ,

β

xα/β−12 − xα/β−11

Pn i=1xα/βi >

x1− x2 nβx1x2 Combining inequality (3.5) to inequality (3.8), we have ∂f (x)/∂x1 − ∂f (v)/∂x2 > 0 According

to Corollary 1.1, we get

f (x1, x2, · · · , xn) ≥ f (A (x) , A (x) , · · · , A (x)) , 1

nβln

Yn i=1xi−1 − λ

α ln

 1 n

Xn i=1xα/βi



≥ λ

βln

 1 n

Xn i=1xi

 Let ai = x1/βi , i = 1, 2, · · · , n, it get

[Mα(a)]1−λ· [Mβ(a)]λ≤ M0(a) Case 2: α + β < 0 Let t > 1, because α < 0, α + β < 0, it has

(n − 1) > (n − 2) tα+ tα+β

We find that inequality (3.7) hold.The rest of the proofs are similar, so we shall omit them 

A similar argument lead to Theorem 3.3

Theorem 3.3 Suppose n ≥ 3, a = (a1, a2, · · · , an) ∈ Rn+, β > 0 > α If β + α > 0, let θ = n−1n

If β + α ≤ 0, let θ = 1 − n(β−α)2β Then

(3.9) M0(a) ≤ [Mα(a)]1−θ· [Mβ(a)]θ

Remark 3.1 Inequality 3.4 3.9 improve the well-known Sierpinski-Inequality

[M−1(a)]n−1n [A (a)]n1 ≤ M0(a) ≤ [M−1(a)]n1 [A (a)]n−1n For n ≥ 2, a = (a1, a2, · · · , an) ∈ Rn+, paper [1] introduces an inequality

n − 1

n A (a) +

1

nM−1(a) ≥ M0(a)

We can improve the inequality by Theorem 3.4 and Theorem 3.5

Theorem 3.4 Suppose p = n2 n2+ 4n − 4
, then

(3.10) pA (a) + (1 − p) M−1(a) ≥ M0(a)

Proof Firstly, Let p > n2

n2+ 4n − 4
,

f (b) = p/n ·Xn

i=1ebi+ (1 − p) n ·Xn

i=1e−bi

−1

, b = (b1, b2, · · · , bn) ∈ Rn Then

∂f

∂b1 =

p

ne

b 1+ (1 − p) n

(Pn i=1e−bi)2e

−b 1,

∂f

∂b1

− ∂f

∂b2

= p n



eb1 − eb2



− (1 − p) n

(Pn i=1e−b i)2



e−b2 − e−b1



If b1 = max

1≤i≤n{bi} > b2 = min

1≤i≤n{bi}, t = eb 1 −b2 > 1 We have

∂f

∂b1 −

∂f

∂b2 ≥

p n



eb1− eb2− (1 − p) n

((n − 1) e−b 1+ e−b 2)2



e−b2 − e−b1

b 1 − eb 2

n ((n − 1) eb 2 + eb 1)2



p(n − 1) eb2+ eb12

− (1 − p) n2eb1eb2



3b 2(t − 1)

n ((n − 1) eb 2 + eb 1)2

h

p (n − 1 + t)2− n2t + pn2ti

> e

3b 2(t − 1)

n ((n − 1) eb 2 + eb 1)2



n2

n2+ 4n − 4(n − 1 + t)

2− n2t + n

2

n2+ 4n − 4n

2t



3b 2(t − 1) (t − n + 1)2 (n2+ 4n − 4) ((n − 1) eb 2 + eb 1)2 ≥ 0.

According to Corollary 1.1, we get

f (b) ≥ f (A (b) , A (b) , · · · , A (b)) , p

n

Xn i=1ebi+ (1 − p)Pnn

i=1e−bi ≥ eA(b) = n

r

Yn i=1eb i Let ebi = ai in above inequality, we know inequality (3.10) hold Because of continuity, if p =

n2

Theorem 3.5 Suppose p =1 − n −√5n2− 6n + 1.2n, then

n − 1

n A (a) +

1

nMp(a) ≥ M0(a) Proof Let

f (a) = n

r

Yn i=1a1/pi − (n − 1)n2·Xn

i=1a1/pi , a ∈ Rn+ Then

∂f

∂a1 =

1 npa1

n

r

Yn i=1a1/pi − n − 1

n2p a

1/p−1

∂f

∂a1

− ∂f

∂a2

= −a1− a2 npa1a2

Yn i=1a1/npi −n − 1

n2p



a1/p−11 − a1/p−12 

If a1 = max

1≤i≤n{ai} > a2 = min

1≤i≤n{ai} > 0, a1/a2= t > 1 Owing to p < 0,−a1 −a 2

npa 1 a 2 > 0, we have

∂f

∂a1 −

∂f

∂a2 ≤ −

a1− a2 npa1a2a

1/(np)

1 a(n−1)/(np)2 −n − 1

n2p



a1/p−11 − a1/p−12 

= a

1/p−1 1

n2p



−nt − 1

t t

1−(n−1)/(np)− (n − 1)1 − t1−1/p

 (3.11)

Let g (t) = −nt1−(n−1)/(np)+ nt−(n−1)/(np)+ (n − 1) t1−1/p− (n − 1) , t > 1.

g0(t) =



−n +n − 1

p



t−(n−1)/(np)−n − 1

p t

−1−(n−1)/(np)+ (n − 1)



1 −1 p



t−1/p,

t1+(n−1)/(np)g0(t) =



−n + n − 1

p



t − n − 1

p + (n − 1)



1 −1 p



t1−1/(np)



t1+(n−1)/(np)g0(t)

0

=



−n + n − 1

p

 + (n − 1)



1 −1 p

 

1 − 1 np



t−1/(np)

>



−n + n − 1

p

 + (n − 1)



1 −1 p

 

1 − 1 np



= 0

Thus t1+(n−1)/(np)g0(t) is a monotone increasing function Meanwhile,

lim

t→1+t1+(n−1)/(np)g0(t) = lim

t→1+



−n +n − 1

p



t − n − 1

p + (n − 1)



1 −1 p



t1−np1



= −1 −n − 1

p ≥ 0.

Therefore t1+(n−1)/(np)g0(t) > 0, g0(t) > 0, g (t) is monotone increasing function Meanwhile, lim

t→1+g (t) = 0, then g (t) > 0 By (3.11), we know ∂f /∂a1− ∂f /∂a2 < 0 According to Corollary 1.1,

f (a) ≤ f (A (a) , A (a) , · · · , A (a)) ,

n

r

Yn i=1a1/pi −n − 1

n2 ·Xn

i=1a1/pi ≤ 1

n· A

1/p(a) Finally, let ai → api(i = 1, 2, · · · , n) in the above inequality, we know Theorem 3.5 hold  Remark 3.2 More applications of Theorem 1.1 and Corollary 1.1 will appear in book [5] and http://old.irgoc.org/Article/ShowArticle.asp?ArticleID=391

Acknowledgments This work was supported by the NSF of China Central Radio and TV University under Grant No GEQ1633 and Foundation of the Educational Committee of Zhejiang Province under Grant Y200804124

References

[1] Alzer, H., Sierpinski’s inequality, J Belgian Math Soc B, 41 (1989), 139-144.

[2] J.-J Wen, Hardy mean and their inequalities, J of Math., 27, 4 (2007), 447–450(in chinese).

[3] J.-J Wen, W.-L Wang, The optimizations for the inequalities of power means Joural of In-equalities and Applications, 2006 (2006), Article ID 46782, 25 page [ONLINE]Available online at http://www.hindawi.com/journals/jia/volume-2006/regular.91.html

[4] X.-M Zhang, A new proof method of analytic inequality Research Report Collection, 12, 1(2009), Art 2 [ON-LINE] Available online at http://www.staff.vu.edu.au/RGMIA/v12n1.asp

[5] X.-M Zhang, Y.-M Chu, New discussion to analytic inequality, HarBin: HarBin Institute of Technology Press,

2009 (in chinese)

[6] X.-M Zhang, Optimization of Schur-Convex Functions, Mathematical Inequalities and Applications, 1, 3 (1998), 319-330 [ONLINE] Available online at http://www.mia-journal.com/miasup.asp

(X.-M Zhang) Zhejiang Broadcast and TV University Haining College, Haining City, Zhejiang Province, 314400, P R China

E-mail address: zjzxm79@126.com

(B.-Y Xi) Department of Mathematics of Inner Mongolia University for the Nationalities, Tongliao, Inner Mongolia, 028000, P R China

E-mail address: baoyintu68@sohu.com

(Y.-M Chu) Department of Mathematics, Huzhou Teachers College, Huzhou 313000, P.R.China E-mail address: chuyuming@hutc.zj.cn

2 Unifying Proof of Some Well-known Inequality Take advantage of Theorem 1.1 and Corollary 1.1, we can prove some well-known inequality, for example, Power Mean Inequality, Holder -Inequality, ... http://www.hindawi.com/journals/jia/volume-2006/regular.91.html

[4] X.-M Zhang, A new proof method of analytic inequality Research Report Collection, 12, 1(2009), Art [ON-LINE] Available online... Y.-M Chu, New discussion to analytic inequality, HarBin: HarBin Institute of Technology Press,

2009 (in chinese)

[6] X.-M Zhang, Optimization of Schur-Convex