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volume 5, issue 3, article 77, 2004.

Received 20 January, 2004;

accepted 05 July, 2004.

Communicated by:F. Qi

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Journal of Inequalities in Pure and Applied Mathematics

GENERALIZATIONS OF A CLASS OF INEQUALITIES FOR PRODUCTS

SHAN-HE WU AND HUAN-NAN SHI

Department of Mathematics Longyan College

Longyan City, Fujian 364012, China.

EMail:wushanhe@yahoo.com.cn Department of Electronic Information Vocational-Technical Teacher’s College Beijing Union University, Beijing 100011, China.

EMail:shihuannan@yahoo.com.cn

2000c Victoria University ISSN (electronic): 1443-5756 019-04

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Generalizations of a Class of Inequalities for Products Shan-He Wu and Huan-Nan Shi

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J. Ineq. Pure and Appl. Math. 5(3) Art. 77, 2004

http://jipam.vu.edu.au

Abstract

In this paper, a class of inequalities for products of positive numbers are generalized.

2000 Mathematics Subject Classification:Primary 26D15.

Key words: Inequality, Product, Arithmetic-geometric mean inequality, Jensen’s in- equality.

The authors would like to express heartily many thanks to the anonymous referees and to the Editor, Professor Dr. F. Qi, for their making great efforts to improve this paper in language and mathematical expressions and typesetting.

1. Introduction and Main Results

In 1987, H.-Sh. Huang [2] proved the following algebraic inequality for products:

(1.1)

n

Y

i=1

1 x_i +xi

≥

n+ 1 n

n

,

wherex₁,x₂,. . .,x_nare positive real numbers withPn

i=1x_i = 1.

In 2002, X.-Y. Yang [4] considered an analogous form of inequality (1.1) and posed an interesting open problem as follows.

Open Problem. Assumex₁,x₂,. . .,x_nare positive real numbers withPn

i=1x_i = 1forn ≥3. Then

(1.2)

n

Y

i=1

1 x_i −x_i

≥

n− 1 n

n

.

In [1], Ch.-H. Dai and B.-H. Liu gave an affirmative answer to the above open problem.

In this article, by using the arithmetic-geometric mean inequality, inequalities (1.1) and (1.2) are refined and generalized as follows.

Theorem 1.1. Let x₁, x₂, . . ., x_n be positive real numbers with Pn

i=1x_i = k

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andk≤n, wherekandnare natural numbers. Then we have form∈N

n

Y

i=1

1

x^m_i +x^m_i

≥ n^m

k^m + k^m n^m

n n

Y

i=1

nx_i k

!^m(k

2m−n2m) k2m+n2m

(1.3)

≥ n^m

k^m + k^m n^m

n

.

Theorem 1.2. Let x₁, x₂, . . ., x_n be positive real numbers with Pn

i=1x_i = k fork ≤1andn ≥3. Then form∈Nwe have

n

Y

i=1

1

x^m_i −x^m_i

≥ n^m

k^m − k^m n^m

n n

Y

i=1

nxi

k

!^m_n⁻^m₃ (1.4)

≥ n^m

k^m − k^m n^m

n

.

Remark 1.1. Choosingm= 1andk = 1in Theorem1.1and Theorem1.2, we can obtain inequalities (1.1) and (1.2) respectively.

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2. Lemmas

To prove Theorem1.1and Theorem1.2, we will use following lemmas.

Lemma 2.1. Letx1,x2,. . .,xnbe positive real numbers withPn

i=1xi = 1and n ≥3. Then

(2.1)

n

Y

i=1

1 x_i −x_i

≥

n− 1 n

n" _n Y

i=1

(nx_i)

#¹_n⁻¹₃ .

Proof. From the conditions of Lemma2.1and by using the arithmetic-geometric mean inequality, we have for1≤p, q ≤nandp6=q

(1−x_p)(1−x_q) = 1−x_p−x_q+x_px_q (2.2)

= X

k6=p,q

x_k+x_px_q

= X

k6=p,q





 x_k

n +· · ·+x_k n

| {z }

n





+xpxq

≥[n(n−2) + 1]

"

Y

k6=p,q

x_k n

n

x_px_q

#_n(n−2)+1¹

= (n−1)²

"

1 n

n(n−2)

Y

k6=p,q

x_k

!n

x_px_q

#_(n−1)2¹

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= (n−1)² 1

n

ⁿ⁽ⁿ⁻²⁾_(n−1)2 ⁿ Y

k=1

x_i

! ⁿ

(n−1)2

(x_px_q)¹⁻ⁿ¹

then (2.3)

n

Y

i=1

(1−x_i)≥(n−1)ⁿ 1

n

^n2(n−2)_2(n−1)2 ⁿ Y

i=1

x_i

!ⁿ

2−2n+2 2(n−1)2

.

By the arithmetic-geometric mean inequality, we obtain

n

Y

i=1

(1 +x_i) =

n

Y

i=1





 1

n +· · ·+ 1 n

| {z }

n

+x_i







≥

n

Y

i=1

(

(n+ 1) 1

n

nx_i _n+1¹ )

= (n+ 1)ⁿ 1

n ⁿ

2 n+1 n

Y

i=1

xi

!_n+1¹ . (2.4)

Utilizing (2.3) and (2.4) yields

n

Y

i=1

1 x_i −xi

(2.5)

=

" _n Y

i=1

(1−x_i)

# " _n Y

i=1

(1 +x_i)

# _n Y

i=1

1 x_i

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≥(n−1)ⁿ(n+ 1)ⁿ 1

n

^n2(n−2)_2(n−1)2+_n+1ⁿ² n

Y

i=1

x_i

!ⁿ

2−2n+2

2(n−1)2 +_n+1¹ −1

=

n− 1 n

n" _n Y

i=1

(nx_i)

#⁻ⁿ³⁺³ⁿ

2−2n+2 2(n+1)(n−1)2

.

From the arithmetic-geometric mean inequality andPn

i=1x_i = 1forn ≥3, we have

(2.6) 0<

n

Y

i=1

(nx_i)≤

n

X

i=1

x_i

!n

= 1.

Sincen≥3, it follows that

−n³+ 3n²−2n+ 2 2(n+ 1)(n−1)² = 1

n −1

3 − n(n−3) (n²+ 2n+ 8) + 10n+ 6 6n(n+ 1)(n−1)²

< 1 n −1

3

≤0.

Therefore, by the monotonicity of the exponential function, we obtain

(2.7)

" _n Y

i=1

(nx_i)

#⁻ⁿ³⁺³ⁿ

2−2n+2 2(n+1)(n−1)2

≥

" _n Y

i=1

(nx_i)

#¹_n−¹₃

.

Combining inequalities (2.5) and (2.7) leads to inequality (2.1).

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Lemma 2.2. Letx₁,x₂,. . .,x_nbe positive real numbers withPn

i=1x_i = 1for n ≥3andma natural number. Then

(2.8)

n

Y

i=1

1

x^m_i −x^m_i

≥

n^m− 1 n^m

n" _n Y

i=1

(nx_i)

#^m_n⁻^m₃ . Proof. Using the arithmetic-geometric mean inequality, we obtain

m−1

X

j=0

x^2j_i =

m−2

X

j=0





 x^2j_i

n^2(m−j−1) +· · ·+ x^2j_i n^2(m−j−1)

| {z }

n^2(m−j−1)







+x^2m−2_i

≥

"_m−1 X

j=0

n^2j

#

x^2(m−1)_i

m−2

Y

j=0

x^2j_i n^2(m−j−1)

!n^2(m−j−1)



Pm−1 j=0 n^2j

= n^2m−1

n^2(m−1)(n²−1)(nxi)

Pm−1

j=0 [2(m−j−1)]n2j Pm−1

j=0 n2j

. (2.9)

Hence 1

x^m_i −x^m_i = 1

xi

−x_i

x^1−m_i

m−1

X

j=0

x^2j_i

≥ 1

x_i −xi

x^1−m_i n^2m−1

n^2(m−1)(n²−1)(nxi)

Pm−1

j=0[2(m−j−1)]n2j Pm−1

j=0 n2j

= 1

x_i −x_i

n^2m−1

n^m−1(n²−1)(nx_i)

(m−1)Pm−1

j=0 n2j−Pm−1 j=1 2jn2j Pm−1

j=0 n2j

, (2.10)

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and then (2.11)

n

Y

i=1

1

x^m_i −x^m_i

≥n^n(1−m)

n^2m−1 n²−1

n" _n Y

i=1

1 x_i −x_i

# " _n Y

i=1

(nx_i)

#

(m−1)Pm−1

j=0 n2j−Pm−1 j=1 2jn2j Pm−1

j=0 n2j

.

In the following, we prove that forn≥3 (2.12) (m−1)Pm−1

j=0 n^2j −Pm−1 j=1 2jn^2j Pm−1

j=0 n^2j ≤(m−1)

1 n − 1

3

.

Form= 1, the equality in (2.12) holds. Form ≥2, we have (m−1)Pm−1

j=0 n^2j−Pm−1 j=1 2jn^2j Pm−1

j=0 n^2j −(m−1)

1 n − 1

3 (2.13)

= (m−1) ⁴₃ − _n¹ Pm−1

j=0 n^2j−Pm−1 j=1 2jn^2j Pm−1

j=0 n^2j

= (m−1) ⁴₃ − _n¹ Pm−2

j=0 n^2j−Pm−2 j=1 2jn^2j Pm−2

j=0 n^2j

− (m−1) _n¹ +²₃

n^2(m−1) Pm−2

j=0 n^2j

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=

(m−1)

hn^2(m−1)−1 n²−1

4 3 − _n¹

− ¹_n+ ²₃

n^2(m−1) i

−Pm−2 j=1 2jn^2j Pm−2

j=0 n^2j

< (m−1)₁

8 4 3 − ¹_n

n^2(m−1)− _n¹ +²₃

n^2(m−1)

−Pm−2 j=1 2jn^2j Pm−2

j=0 n^2j

= (m−1) −_8n⁹ − ¹₂

n^2(m−1)−Pm−2 j=1 2jn^2j Pm−2

j=0 n^2j

<0.

Hence inequality (2.12) holds.

Considering inequality (2.6) and the monotonicity of the exponential function and combining inequality (2.11) with (2.12) reveals

(2.14)

n

Y

i=1

1

x^m_i −x^m_i

≥n^n(1−m)

n^2m−1 n²−1

n" _n Y

i=1

1 x_i −x_i

# " _n Y

i=1

(nx_i)

#(m−1)(n¹−¹₃) . Substituting inequality (2.1) into (2.14) produces

n

Y

i=1

1

x^m_i −x^m_i

≥n^n(1−m)

n^2m−1 n²−1

n n− 1

n n

(2.15)

×

" _n Y

i=1

(nx_i)

#_n¹⁻¹₃ " _n Y

i=1

(nx_i)

#^(m−1)(¹_n⁻¹₃)

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=

n^m− 1 n^m

n" _n Y

i=1

(nxi)

#m(¹_n⁻¹₃) .

The proof is complete.

Lemma 2.3. Letx₁,x₂,. . .,x_nbe positive real numbers withPn

i=1x_i =k ≤1 forn≥3. Then for any natural numberm, we have

(2.16)

n

Y

i=1

1

x^m_i −x^m_i

≥

n^m− 1 n^m

−n n^m k^m − k^m

n^m n n

Y

i=1

k^m x^m_i − x^m_i

k^m

. Proof. It is easy to see that

(2.17)

n

Y

i=1

1

x^m_i −x^m_i ⁿ

Y

i=1

k^m x^m_i − x^m_i

k^m −1

=k^nm

n

Y

i=1

1−x^2m_i k^2m−x^2m_i .

Define

(2.18) f(x) = ln 1−x^2m

k^2m−x^2m

forx∈(0, k),m≥1andk≤1. Direct calculation shows that (2.19) f⁰(x) = 2m(1−k^2m)x^2m−1

(1−x^2m)(k^2m−x^2m),

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f⁰⁰(x) = 2mx^2(m−1)(1−k^2m)

(1−x^2m)²(k^2m−x^2m)²[(2m−1)(1−x^2m)(k^2m−x^2m) (2.20)

+ 2mx^2m(k^2m−x^2m+ 1−x^2m)]

≥0.

This means thatf is convex in the interval(0, k). Using Jensen’s inequality [3], we obtain

(2.21) 1

n

X

i=1

ln 1−x^2m_i

k^2m−x^2m_i ≥ln 1−[¹_nPn

i=1x_i]^2m k^2m−₁

n

Pn

i=1x_i2m

for any0< x_i < k≤1andi∈N. UsingPn

i=1x_i =kin (2.21), it follows that

(2.22)

n

Y

i=1

1−x^2m_i

k^2m−x^2m_i ≥ 1− ^k_n^2m2m

k^2m− ^k_n^2m2m

!n

,

therefore

(2.23) k^nm

n

Y

i=1

1−x^2m_i k^2m−x^2m_i ≥

n^m− 1 n^m

n^m n

.

Substituting (2.17) into (2.23) leads to (2.16). The proof is complete.

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3. Proofs of Theorems

Proof of Theorem1.1. Using the arithmetic-geometric mean inequality, we ob- tain

1

x^m_i +x^m_i = 1

n^2mx^m_i +· · ·+ 1 n^2mx^m_i

| {z }

n^2m

+ x^m_i

k^2m +· · ·+ x^m_i k^2m

| {z }

k^2m

≥(n^2m+k^2m)

"

1 n^2mx^m_i

n^2m x^m_i k^2m

k^2m# ¹

k2m+n2m

= (n^2m+k^2m)

k^−2mk^2mn^−2mn^2mx^mk_i ^2m^−mn^2m ¹

k2m+n2m

, (3.1)

therefore (3.2)

n

Y

i=1

1

x^m_i +x^m_i

≥ n^2m+k^2mn

k^−2mk^2mn^−2mn^2m_k_2m_+nⁿ _2m Yⁿ

i=1

x_i

!^m(k

2m−n2m) k2m+n2m

,

that is

(3.3)

n

Y

i=1

1

x^m_i +x^m_i

≥ n^m

k^m + k^m n^m

n n

Y

i=1

nxi

k

!^m(k

2m−n2m) k2m+n2m

.

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FromPn

i=1x_i =k and the arithmetic-geometric mean inequality, it follows that

(3.4)

n

Y

i=1

nx_i k ≤

n

X

i=1

x_i k

!n

= 1,

and then, consideringk≤n, we have

(3.5)

n^m k^m + k^m

n^m

n n

Y

i=1

nx_i k

!^m(k

2m−n2m) k2m+n2m

≥ n^m

k^m +k^m n^m

n

.

Inequality (1.3) is then deduced by combining (3.3) and (3.5). This completes the proof of Theorem1.1.

Proof of Theorem1.2. ApplyingPn i=1

xi

k = 1to Lemma2.2, we have (3.6)

n

Y

i=1

k^m x^m_i − x^m_i

k^m

≥

n^m− 1 n^m

n n

Y

i=1

nx_i k

!^m_n−^m₃

.

Substituting inequality (3.6) into Lemma2.3gives

n

Y

i=1

1

x^m_i −x^m_i

≥

n^m− 1 n^m

n^m n n

Y

i=1

k^m x^m_i − x^m_i

k^m

≥ n^m

k^m − k^m n^m

n n

Y

i=1

nx_i k

!^m_n⁻^m₃ . (3.7)

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Since

(3.8) 0<

n

Y

i=1

nx_i k ≤

n

X

i=1

x_i k

!n

= 1

and ^m_n −^m₃ ≤0, we have

(3.9)

n^m k^m − k^m

n^m

n n

Y

i=1

nx_i k

!^m_n⁻^m₃

≥ n^m

k^m − k^m n^m

n

.

Combining (3.7) and (3.9), we immediately obtain inequality (1.4). This completes the proof of Theorem1.2.

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References

[1] Ch.-H. DAI AND B.-H. LIU, The proof of a conjecture, Shùxué T¯ongxùn, 23 (2002), 27–28. (Chinese)

[2] H.-Sh. HUANG, On inequality Qn i=1

1

xi +x_i

≥ n+_n¹n

, Shùxué T¯ongxùn, 5 (1987), 5–7. (Chinese)

[3] D.S. MITRINOVI ´C, J.E. PE ˇCARI ´C AND A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, 1993.

[4] X.-Y. YANG, The generalization of an inequality, Shùxué T¯ongxùn, 19 (2002), 29–30. (Chinese)