Seitz inequality, Convex function, Difference, Monotonicity, Exponential convex

http://jipam.vu.edu.au/

Volume 5, Issue 2, Article 39, 2004

ON CERTAIN INEQUALITIES RELATED TO THE SEITZ INEQUALITY

LIANG-CHENG WANG AND JIA-GUI LUO DEPARTMENT OFMATHEMATICS

DAXIANTEACHER’SCOLLEGE

DAZHOU635000, SICHUANPROVINCE

THEPEOPLE’SREPUBLIC OFCHINA. wangliangcheng@163.com DEPARTMENT OFMATHEMATICS

ZHONGSHANUNIVERSITY

GUANGZHOU510275 PEOPLE’SREPUBLIC OFCHINA.

Received 18 August, 2003; accepted 11 April, 2004 Communicated by F. Qi

ABSTRACT. In this paper, we investigate the monotonicity of difference results from the G. Seitz inequality. An application is given, with some resulting inequalities.

Key words and phrases: G. Seitz inequality, Convex function, Difference, Monotonicity, Exponential convex.

2000 Mathematics Subject Classification. 26D15.

1. INTRODUCTION

For a given positive integer n ≥ 2, let X = (x₁, x₂, . . . , x_n), Y = (y₁, y₂, . . . , y_n), U = (u₁, u₂, . . . , u_n) and Z = (z₁, z₂, . . . , z_n)be known sequences of real numbers, and let t_i >

0 (i = 1,2, . . . , n), T_j = Pj

i=1t_i (j = 1,2, . . . , n)anda_ij (i, j = 1,2, . . . , n) be known real numbers. Define the functionsA,J,C,W andGby

A(n)=⁴

n

X

i=1

x_i−n

n

Y

i=1

x_i

!¹_n

(x_i >0, i= 1,2, . . . , n),

J(n)=⁴

n

X

i=1

tif(vi)−Tnf 1 T_n

n

X

i=1

tivi

! ,

ISSN (electronic): 1443-5756

c 2004 Victoria University. All rights reserved.

This author is partially supported by the Key Research Foundation of the Daxian Teacher’s College under Grant 2003–81.

112-03

wheref is convex function on the intervalIandv_i ∈I(i= 1,2, . . . , n), C(n)=⁴

" _n X

i=1

x²_i

! _n X

i=1

y_i²

!#¹₂

−

n

X

i=1

x_iy_i,

W(n)=⁴ T_n

n

X

i=1

t_ix_iz_i−

n

X

i=1

t_ix_i

! _n X

i=1

t_iz_i

! ,

and

G(n)=⁴

n

X

i,j=1

a_ijx_iz_j

! _n X

i,j=1

a_ijy_iu_j

!

−

n

X

i,j=1

a_ijx_iu_j

! _n X

i,j=1

a_ijy_iz_j

! .

Rade investigated the monotonicity of difference forA−G mean inequality, and obtained the following inequality [2]

(1.1) A(n)≥A(n−1).

P. M. Vasi´c and J. E. Peˇcari´c generalized inequality (1.1) to convex functions, and obtained the following inequality [4, 6]

(1.2) J(n)≥J(n−1).

Recently the first author and Xu Zhang studied inequality (1.2) in depth, and obtained some inequalities. L.-C. Wang also obtained some applications, one of them is the following inequal- ity [8]

(1.3) C(n)≥C(n−1).

Inequality (1.3) resulted from the Cauchy inequality (1.4)

n

X

i=1

x²_i

! _n X

i=1

y_i²

!

≥

n

X

i=1

x_iy_i

!2

.

In [7], L.-C. Wang proved the following inequality

(1.5) W(n)≥W(n−1),

withXandZ both increasing or both decreasing. If one ofX orZ is increasing and the other decreasing, then the inequality (1.5) reverses.

Inequality (1.5) resulted from the following Chebyshev inequality

(1.6) T_n

n

X

i=1

t_ix_iz_i ≥

n

X

i=1

t_ix_i

! _n X

i=1

t_iz_i

! ,

withXandZ both increasing or both decreasing. If one ofX orZ is increasing and the other decreasing, then the inequality (1.6) reverses.

Assume thati, j, r, s∈Nsuch that1≤i < j ≤nand1≤r < s≤n, we have (1.7)

x_i x_j y_i y_j

z_r z_s u_r u_s

≥0

and (1.8)

a_ir a_is a_jr a_js

≥0.

When both (1.7) and (1.8) are true, the following inequality by G. Seitz [1] holds:

(1.9)

n

P

i,j=1

a_ijx_iz_j

n

P

i,j=1

aijxiuj

≥

n

P

i,j=1

a_ijy_iz_j

n

P

i,j=1

aijyiuj

.

If

(1.10) X =Z, Y =U and aij =

( 1 i=j

0 i6=j (i, j = 1,2, . . . , n),

then inequality (1.9) changes into (1.4). If

(1.11) Y =U = (1,1, . . . ,1) and a_ij =

( t_i i=j

0 i6=j (i, j = 1,2, . . . , n), then inequality (1.9) changes into (1.6).

In this paper, we investigate inequality (1.9) in depth, obtaining the following main result.

Theorem 1.1. If both inequalities (1.7) and (1.8) are true, then we have

(1.12) G(n)≥G(n−1).

Remark 1.2. If we put (1.10) and (1.11) into (1.12), then (1.12) becomes (1.3) and (1.5), re- spectively. Hence, (1.12) is an extension of (1.3) and (1.5).

2. PROOF OFTHEOREM1.1 Using

a_ijx_iz_j

a_iny_iu_n

=

a_ijy_iz_j

a_inx_iu_n

(i, j = 1,2, . . . , n−1),

a_ijy_iu_j

a_inx_iz_n

=

a_ijx_iu_j

a_iny_iz_n

(i, j = 1,2, . . . , n−1), and (1.7) – (1.8), we have

n−1

X

i,j=1

a_ijx_iz_j

n−1

X

i=1

a_iny_iu_n−

n−1

X

i,j=1

a_ijy_iz_j

n−1

X

i=1

a_inx_iu_n (2.1)

+

n−1

X

i,j=1

a_ijy_iu_j

n−1

X

i=1

a_inx_iz_n−

n−1

X

i,j=1

a_ijx_iu_j

n−1

X

i=1

a_iny_iz_n

=

n−1

X

i=1 n−1

X

j=1

a_ijx_iz_j

n−1

X

k=1

a_kny_ku_n−

n−1

X

i=1 n−1

X

j=1

a_ijy_iz_j

n−1

X

k=1

a_knx_ku_n

+

n−1

X

i=1 n−1

X

j=1

a_ijy_iu_j

n−1

X

k=1

a_knx_kz_n−

n−1

X

i=1 n−1

X

j=1

a_ijx_iu_j

n−1

X

k=1

a_kny_kz_n

=

n−1

X

i=1 n−1

X

j=1 n−1

X

k=1,k6=i

a_ija_kn

x_iz_jy_ku_n+x_kz_ny_iu_j−y_iz_jx_ku_n−x_iu_jy_kz_n

=

n−1

X

j=1 n−2

X

i=1 n−1

X

k=2,i<k

+

n−1

X

i=2 n−2

X

k=1,i>k

!

a_ija_kn

x_iz_jy_ku_n+x_kz_ny_iu_j −y_iz_jx_ku_n−x_iu_jy_kz_n

=

n−1

X

j=1 n−2

X

i=1 n−1

X

k=2,i<k

a_ija_kn

x_iz_jy_ku_n+x_kz_ny_iu_j−y_iz_jx_ku_n−x_iu_jy_kz_n

+

n−1

X

j=1 n−1

X

k=2 n−2

X

i=1,k>i

a_kja_in

x_kz_jy_iu_n+x_iz_ny_ku_j−y_kz_jx_iu_n−x_ku_jy_iz_n

=

n−1

X

j=1

X

1≤i<k<n

a_ija_kn−a_kja_in

x_iz_jy_ku_n+x_kz_ny_iu_j−y_iz_jx_ku_n−x_iu_jy_kz_n

=

n−1

X

j=1

X

1≤i<k<n

a_ij a_in akj akn

x_i x_k yi yk

z_j z_n uj un

≥0.

Using

a_ijx_iz_j

a_njy_nu_j

=

a_ijx_iu_j

a_njy_nz_j

(i, j = 1,2, . . . , n−1),

a_ijy_iu_j

a_njx_nz_j

=

a_ijy_iz_j

a_njx_nu_j

(i, j = 1,2, . . . , n−1), (1.7) – (1.8) and the same method as in the proof of (2.1), we obtain

n−1

X

i,j=1

a_ijx_iz_j

n−1

X

j=1

a_njy_nu_j −

n−1

X

i,j=1

a_ijx_iu_j

n−1

X

j=1

a_njy_nz_j (2.2)

+

n−1

X

i,j=1

a_ijy_iu_j

n−1

X

j=1

a_njx_nz_j −

n−1

X

i,j=1

a_ijy_iz_j

n−1

X

j=1

a_njx_nu_j

=

n−1

X

i=1 n−1

X

j=1 n−1

X

k=1,k6=j

a_ija_nk

x_iz_jy_nu_k+y_iu_jx_nz_k−x_iu_jy_nz_k−y_iz_jx_nu_k

=

n−1

X

i=1

X

1≤j<k<n

a_ij a_ik a_nj a_nk

x_i x_n y_i y_n

z_j z_k u_j u_k

≥0.

Using

a_iny_iu_n

a_jnx_jz_n

=

a_iny_iz_n

a_jnx_ju_n

(i, j = 1,2, . . . , n) and

a_niy_nu_i

a_njx_nz_j

=

a_niy_nz_i

a_njx_nu_j

(i, j = 1,2, . . . , n−1),

we obtain (2.3)

n

X

i=1

a_iny_iu_n

n

X

j=1

a_jnx_jz_n−

n

X

i=1

a_iny_iz_n

n

X

j=1

a_jnx_ju_n= 0

and (2.4)

n−1

X

i=1

a_niy_nu_i

n−1

X

j=1

a_njx_nz_j −

n−1

X

i=1

a_niy_nz_i

n−1

X

j=1

a_njx_nu_j = 0,

respectively.

Using

a_nny_nu_n

a_njx_nz_j

=

a_njy_nz_j

a_nnx_nu_n

(j = 1,2, . . . , n−1),

anjynuj

annxnzn

=

annynzn

anjxnuj

(j = 1,2, . . . , n−1), and (1.7) – (1.8), we have

n

X

i=1

a_iny_iu_n

n−1

X

j=1

a_njx_nz_j−

n−1

X

j=1

a_njy_nz_j

n

X

i=1

a_inx_iu_n

! (2.5)

+

n−1

X

j=1

a_njy_nu_j

n

X

i=1

a_inx_iz_n−

n

X

i=1

a_iny_iz_n

n−1

X

j=1

a_njx_nu_j

!

+

n−1

X

i,j=1

a_ija_nnx_iz_jy_nu_n−

n−1

X

i,j=1

a_ija_nny_iz_jx_nu_n

!

+

n−1

X

i,j=1

a_ija_nny_iu_jx_nz_n−

n−1

X

i,j=1

a_ija_nnx_iu_jy_nz_n

!

=

n−1

X

i=1

a_iny_iu_n

n−1

X

j=1

a_njx_nz_j −

n−1

X

j=1

a_njy_nz_j

n−1

X

i=1

a_inx_iu_n

!

+

n−1

X

j=1

a_njy_nu_j

n−1

X

i=1

a_inx_iz_n−

n−1

X

i=1

a_iny_iz_n

n−1

X

j=1

a_njx_nu_j

!

+

n−1

X

i=1 n−1

X

j=1

a_ija_nnx_iz_jy_nu_n−

n−1

X

i=1 n−1

X

j=1

a_ija_nny_iz_jx_nu_n

!

+

n−1

X

i=1 n−1

X

j=1

a_ija_nny_iu_jx_nz_n−

n−1

X

i=1 n−1

X

j=1

a_ija_nnx_iu_jy_nz_n

!

=

n−1

X

i=1 n−1

X

j=1

a_ina_nj

y_iu_nx_nz_j+x_iz_ny_nu_j−x_iu_ny_nz_j−y_iz_nx_nu_j

+

n−1

X

i=1 n−1

X

j=1

a_ija_nn

x_iz_jy_nu_n+x_nz_ny_iu_j −y_iz_jx_nu_n−x_iu_jy_nz_n

=

n−1

X

i=1 n−1

X

j=1

a_ija_nn−a_ina_nj

x_iz_jy_nu_n+x_nz_ny_iu_j −y_iz_jx_nu_n−x_iu_jy_nz_n

=

n−1

X

i=1 n−1

X

j=1

a_ij a_in a_nj a_nn

x_i x_n y_i y_n

z_j z_n u_j u_n

≥0.

By (2.1)-(2.5) and definition ofG(n), we have

G(n)−G(n−1) =

n−1

X

i,j=1

a_ijx_iz_j+

n

X

i=1

a_inx_iz_n+

n−1

X

j=1

a_njx_nz_j

!

×

n−1

X

i,j=1

aijyiuj +

n

X

i=1

ainyiun+

n−1

X

j=1

anjynuj

!

−

n−1

X

i,j=1

a_ijx_iu_j+

n

X

i=1

a_inx_iu_n+

n−1

X

j=1

a_njx_nu_j

!

×

n−1

X

i,j=1

aijyizj +

n

X

i=1

ainyizn+

n−1

X

j=1

anjynzj

!

−

n−1

X

i,j=1

a_ijx_iz_j

n−1

X

i,j=1

a_ijy_iu_j −

n−1

X

i,j=1

a_ijx_iu_j

n−1

X

i,j=1

a_ijy_iz_j

!

=

n−1

X

i,j=1

a_ijx_iz_j

n−1

X

i=1

a_iny_iu_n−

n−1

X

i,j=1

a_ijy_iz_j

n−1

X

i=1

a_inx_iu_n

!

+

n−1

X

i,j=1

a_ija_nnx_iz_jy_nu_n−

n−1

X

i,j=1

a_ija_nny_iz_jx_nu_n

!

+

n−1

X

i,j=1

a_ijy_iu_j

n−1

X

i=1

a_inx_iz_n−

n−1

X

i,j=1

a_ijx_iu_j

n−1

X

i=1

a_iny_iz_n

!

+

n−1

X

i,j=1

a_ija_nny_iu_jx_nz_n−

n−1

X

i,j=1

a_ija_nnx_iu_jy_nz_n

!

+

n−1

X

i,j=1

a_ijx_iz_j

n−1

X

j=1

a_njy_nu_j−

n−1

X

i,j=1

a_ijx_iu_j

n−1

X

j=1

a_njy_nz_j

!

+

n−1

X

i,j=1

a_ijy_iu_j

n−1

X

j=1

a_njx_nz_j−

n−1

X

i,j=1

a_ijy_iz_j

n−1

X

j=1

a_njx_nu_j

!

+

n

X

i=1

a_iny_iu_n

n

X

j=1

a_jnx_jz_n−

n

X

i=1

a_iny_iz_n

n

X

j=1

a_jnx_ju_n

!

+

n−1

X

i=1

a_niy_nu_i

n−1

X

j=1

a_njx_nz_j −

n−1

X

i=1

a_niy_nz_i

n−1

X

j=1

a_njx_nu_j

!

+

n

X

i=1

a_iny_iu_n

n−1

X

j=1

a_njx_nz_j −

n−1

X

j=1

a_njy_nz_j

n

X

i=1

a_inx_iu_n

!

+

n−1

X

j=1

a_njy_nu_j

n

X

i=1

a_inx_iz_n−

n

X

i=1

a_iny_iz_n

n−1

X

j=1

a_njx_nu_j

!

≥0,

i.e., inequality (1.12) is true. This completes the proof of theorem . 3. APPLICATIONS

LetEbe a convex subset of an arbitrary real linear spaceK, and letf :E 7→(0,+∞). f is an exponential convex function onE, if and only if

(3.1) f

tu+ (1−t)v

≤f^t(u)f^1−t(v)

for anyu, v ∈ E and anyt ∈ [0,1]. f is an exponential concave function onE, if and only if the inequality (3.1) reverses (see [4]).

For anyu, v ∈ E(u 6= v)and α_ki, β_ki ∈ [0,1], we letx_ki = α_kiu+ (1−α_ki)v andy_ki = β_kiu+ (1−β_ki)v (k= 1,2;i= 1,2, . . . , n;n >2). Define a functionLby

L(n) =

n

X

i,j=1

a_ijf(x_1i)f(x_2j)

n

X

i,j=1

a_ijf(y_1i)f(y_2j)−

n

X

i,j=1

a_ijf(x_1i)f(y_2j)

n

X

i,j=1

a_ijf(y_1i)f(x_2j).

Proposition 3.1. Let f be an exponential convex (or concave) function on E and inequality (1.8) be true. Fork = 1,2and every pair of positive integersiandj such that1≤i < j ≤n, if

(3.2) α_ki ≤β_ki ≤α_kj and α_kj−α_ki =β_kj−β_ki, then we have

(3.3) L(n)≥L(n−1).

Proof. (1) Supposef is an exponential convex function onE. For k = 1,2 and1 ≤ i <

j ≤n, from (3.2), we haveβ_kj =α_kj+β_ki−α_ki ≥α_kj. Case 1. When α_ki < β_ki ≤ α_kj < β_kj, we take t = ^β_β^ki−αki

kj−αki, then1−t = _β^βkj−βki

kj−αki. Hence, we have

(3.4) ty_kj+ (1−t)x_ki =β_kiu+ (1−β_ki)v =y_ki. From (3.1) and (3.4), we have

(3.5) f(y_ki)≤f^t(y_kj)f^1−t(x_ki).

From (3.2), we get the other form oftand1−t: t= ^β_β^kj−αkj

kj−α_ki and1−t= ^α_β^kj−αki

kj−α_ki. Then we have

(3.6) (1−t)y_kj +tx_ki =α_kju+ (1−α_kj)v =x_kj. From (3.1) and (3.6), we have

(3.7) f(x_kj)≤f^1−t(y_kj)f^t(x_ki).

From (3.5) and (3.7), we obtain (3.8)

f(x_ki) f(x_kj) f(y_ki) f(y_kj)

≥0.

Case 2. When α_ki = β_ki (or α_kj = β_kj), by (3.2), then we haveα_kj = β_kj (orα_ki = β_ki). Hence, the equality of (3.8) holds.

For any1≤i < j ≤nand any1≤r < s≤n, by (3.8), we obtain (3.9)

f(x_1i) f(x_1j) f(y_1i) f(y_1j)

f(x_2r) f(x_2s) f(y_2r) f(y_2s)

≥0.

(2) Let f be an exponential concave function on E. Then (3.5), (3.7) and (3.8) reverse.

Hence, (3.9) is still valid.

Replacingx_i,y_i,z_i andu_i in Theorem 1.1 withf(x_1i),f(y_1i),f(x_2i)andf(y_2i)(i= 1,2, . . . , n), respectively, we obtain (3.3). This completes the proof of Proposition 3.1.

Remark 3.2. In Proposition 3.1, whenEis a real intervalI, we only need

xki ≤yki ≤xkj and xkj−xki =ykj −yki,

wherek = 1,2,i, jare every pair of positive integers such that1≤i < j ≤n,x_ki, x_kj, y_ki, y_kj ∈ I.

In order to verify Proposition 3.4, the following lemma is necessary.

Lemma 3.3. Letc, d: [a, b]7→ R(b > a)be the monotonic functions, both increasing or both decreasing. Furthermore, letq: [a, b]7→(0,+∞)be an integrable function. Then

(3.10)

Z b a

q(x)c(x)dx Z b

a

q(x)d(x)dx≤ Z b

a

q(x)dx Z b

a

q(x)c(x)d(x)dx.

If one of the functions ofcordis increasing and the other decreasing, then the inequality (3.10) reverses. (see [2, 3]).

Let p : [a, b] 7→ (0,+∞) be continuous, g : [a, b] 7→ (1,+∞)be monotonic continuous.

Define a functionM by M(n) =

n

X

i,j

a_ijh^(k+i)(x)h^(m+j)(x)

n

X

i,j

a_ijh^(l+i)(x)h^(w+j)(x)

−

n

X

i,j

a_ijh^(k+i)(x)h^(w+j)(x)

n

X

i,j

a_ijh^(l+i)(x)h^(m+j)(x),

wherek, l, m, w ∈N,i, j = 1,2, . . . , nand

(3.11) h:R7→R⁺, h(x) =

Z b a

p(t) g(t)x

dt (see [5]).

Proposition 3.4. Let the inequalities in (1.8) hold. Ifk < l,m < w ork > l, m > w, then we have

(3.12) M(n)≥M(n−1).

Proof. For (3.11), by continuity ofpandg,we may change the order of derivation and integra- tion. By direct computation, we get

(3.13) h⁽ⁿ⁾(x) =

Z b a

p(t) (g(t))^x(lng(t))ⁿdt.

For every pair of integersi, jsuch that1≤i < j ≤n, whenk < l, replaceq,canddin Lemma 3.3 byp(t) (g(t))^x(lng(t))^k+i,(lng(t))^j−iand(lng(t))^l−k, respectively. Using (3.13), we get (3.14) h^(k+i)(x)h^(l+j)(x)≥h^(k+j)(x)h^(l+i)(x).

By (3.14), we have (3.15)

h^(k+i)(x) h^(k+j)(x) h^(l+i)(x) h^(l+j)(x)

≥0.

Similarly we obtain (3.16)

h^(m+r)(x) h^(m+s)(x) h^(w+r)(x) h^(w+s)(x)

≥0, wherer,sare pair of integer such that1≤r < s≤nandm < w.

Replacingx_i,y_i,z_iandu_iin Theorem 1.1 byh^(k+i)(x),h^(l+i)(x),h^(m+i)(x)andh^(w+i)(x)(i= 1,2, . . . , n), respectively, we obtain (3.12).

By Lemma 3.3, whenk > landm > w, both (3.15) and (3.16) reverse. Hence, the product on the left for both (3.15) and (3.16) is still nonnegative, hence, by Theorem 1.1, (3.12) is also satisfied.

This completes the proof of Proposition 3.4.

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