Key words and phrases: Inequality, Product, Arithmetic-geometric mean inequality, Jensen’s inequality

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

Volume 5, Issue 3, Article 77, 2004

GENERALIZATIONS OF A CLASS OF INEQUALITIES FOR PRODUCTS

SHAN-HE WU AND HUAN-NAN SHI DEPARTMENT OFMATHEMATICS

LONGYANCOLLEGE

LONGYANCITY, FUJIAN364012, CHINA. wushanhe@yahoo.com.cn DEPARTMENT OFELECTRONICINFORMATION

VOCATIONAL-TECHNICALTEACHER’SCOLLEGE

BEIJINGUNIONUNIVERSITY, BEIJING100011, CHINA. shihuannan@yahoo.com.cn

Received 20 January, 2004; accepted 05 July, 2004 Communicated by F. Qi

ABSTRACT. In this paper, a class of inequalities for products of positive numbers are general- ized.

Key words and phrases: Inequality, Product, Arithmetic-geometric mean inequality, Jensen’s inequality.

2000 Mathematics Subject Classification. Primary 26D15.

1. INTRODUCTION ANDMAIN RESULTS

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

(1.1)

n

Y

i=1

1 x_i +x_i

≥

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 inter- esting 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

.

ISSN (electronic): 1443-5756 c

2004 Victoria University. All rights reserved.

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.

019-04

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

i=1x_i =kandk ≤n, where kandnare natural numbers. Then we have form ∈N

(1.3)

n

Y

i=1

1

x^m_i +x^m_i

≥ n^m

k^m + k^m n^m

n n

Y

i=1

nx_i k

!^m(k

2m−n2m) k2m+n2m

≥ n^m

k^m + k^m n^m

n

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

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

(1.4)

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

≥ n^m

k^m − k^m n^m

n

.

Remark 1.3. Choosing m = 1 and k = 1 in Theorem 1.1 and Theorem 1.2, we can obtain inequalities (1.1) and (1.2) respectively.

2. LEMMAS

To prove Theorem 1.1 and Theorem 1.2, we will use following lemmas.

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

i=1x_i = 1andn≥3. Then (2.1)

n

Y

i=1

1 x_i −x_i

≥

n− 1 n

n" _n Y

i=1

(nx_i)

#_n¹−¹₃

.

Proof. From the conditions of Lemma 2.1 and by using the arithmetic-geometric mean inequal- ity, we have for1≤p, q ≤nandp6=q

(1−x_p)(1−x_q) = 1−x_p−x_q+x_px_q

= X

k6=p,q

x_k+x_px_q

= X

k6=p,q





 x_k

n +· · ·+ x_k n

| {z }

n





+x_px_q

≥[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

= (n−1)² 1

n

n(n−2)

(n−1)2 n

Y

k=1

x_i

!_(n−1)2ⁿ

(x_px_q)¹⁻ⁿ¹ , (2.2)

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+1}¹ )

= (n+ 1)ⁿ 1

n ⁿ

2 n+1 n

Y

i=1

x_i

!_n+1¹ . (2.4)

Utilizing (2.3) and (2.4) yields

n

Y

i=1

1 x_i −x_i

=

" _n Y

i=1

(1−x_i)

# " _n Y

i=1

(1 +x_i)

# _n Y

i=1

1 x_i (2.5)

≥(n−1)ⁿ(n+ 1)ⁿ 1

n

^n2(n−2)_2(n−1)2+_n+1ⁿ² n

Y

i=1

xi

!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=1xi = 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)

#_n¹−¹₃

.

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

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

i=1xi = 1forn≥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)(nx_i)

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

x_i −x_i

x^1−m_i

m−1

X

j=0

x^2j_i

≥ 1

x_i −x_i

x^1−m_i n^2m−1

n^2(m−1)(n²−1)(nx_i)

Pm−1

j=0 [2(m−j−1)]n2j Pm−1

j=0 n2j

= 1

xi

−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)

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−(m−1) _n¹ + ²₃

n^2(m−1) Pm−2

j=0 n^2j

=

(m−1)h

n^2(m−1)−1 n²−1

4 3 −_n¹

− _n¹ + ²₃

n^2(m−1)i

−Pm−2 j=1 2jn^2j Pm−2

j=0 n^2j

< (m−1)₁

8 4 3 − _n¹

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 (2.15)

≥n^n(1−m)

n^2m−1 n²−1

n n− 1

n

n" _n Y

i=1

(nx_i)

#¹_n−¹₃ " _n Y

i=1

(nx_i)

#(m−1)(n¹−¹₃)

=

n^m− 1 n^m

n" _n Y

i=1

(nx_i)

#m(n¹−¹₃) .

The proof is complete.

Lemma 2.3. Let x₁, x₂, . . ., x_n be positive real numbers with Pn

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),

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

n

X

i=1

ln 1−x^2m_i

k^2m−x^2m_i ≥ln 1−[_n¹Pn

i=1x_i]^2m k^2m−₁

n

Pn

i=1x_i2m

for any0< xi < k≤1andi∈N. UsingPn

i=1xi =kin (2.21), it follows that (2.22)

n

Y

i=1

1−x^2m_i

k^2m−x^2m_i ≥ 1−_n^k2m2m

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 n^m k^m − k^m

n^m n

.

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

3. PROOFS OF THEOREMS

Proof of Theorem 1.1. Using the arithmetic-geometric mean inequality, we obtain 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^−2mk2mn^−2mn2mx^mk_i ^2m−mn2m_k2m+n¹ _2m , (3.1)

therefore (3.2)

n

Y

i=1

1

x^m_i +x^m_i

≥ n^2m+k^2mn

k^−2mk2mn^{−2mn2m n}

k2m+n2m

n

Y

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

nx_i k

!^m(k

2m−n2m) k2m+n2m

. FromPn

i=1x_i =kand 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

Theorem 1.1.

Proof of Theorem 1.2. ApplyingPn i=1

xi

k = 1to Lemma 2.2, 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 Lemma 2.3 gives

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

≥ n^m

k^m − k^m n^m

n n

Y

i=1

nxi

k

!^m_n−^m₃

. (3.7)

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 Theorem 1.2.

REFERENCES

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

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

1 xi +xi

≥ n+_n¹n

, Shùxué T¯ongxùn, 5 (1987), 5–7.

(Chinese)

[3] D.S. MITRINOVI ´C, J.E. PE ˇCARI ´C ANDA.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)