Journal of Inequalities and Applications
Volume 2010 (2010), Article ID 698012, 13 pages
doi:10.1155/2010/698012
Research Article

A New Method to Study Analytic Inequalities

Xiao-Ming Zhang and Yu-Ming Chu

Department of Mathematics, Huzhou Teachers College, Huzhou 313000, China

Received 16 October 2009; Accepted 24 December 2009

Academic Editor: Kunquan Lan

Copyright © 2010 Xiao-Ming Zhang and Yu-Ming Chu. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

We present a new method to study analytic inequalities involving 𝑛 variables. Regarding its applications, we proved some well-known inequalities and improved Carleman's inequality.

1. Monotonicity Theorems

Throughout this paper, we denote ℝ the set of real numbers and ℝ + the set of strictly positive real numbers, 𝑛 ∈ ℕ , 𝑛 ≥ 2 .

In this section, we present the main results of this paper.

Theorem 1.1. Suppose that 𝑎 , 𝑏 ∈ ℝ with 𝑎 < 𝑏 and 𝑐 ∈ [ 𝑎 , 𝑏 ] , 𝑓 ∶ [ 𝑎 , 𝑏 ] 𝑛 → ℝ has continuous partial derivatives and If 𝜕 𝑓 ( 𝐱 ) / 𝜕 𝑥 𝑚 > 0 for all 𝐱 ∈ 𝐷 𝑚 ( 𝑚 = 1 , 2 , … , 𝑛 − 1 ) , then for all 𝑦 𝑚 ∈ [ 𝑐 , 𝑏 ] ( 𝑚 = 1 , 2 , … , 𝑛 − 1 ) .

Proof. Without loss of generality, since we assume that 𝑛 = 3 and 𝑦 1 > 𝑦 2 > 𝑐 .
For 𝑥 1 ∈ [ 𝑦 2 , 𝑦 1 ] , we clearly see that ( 𝑥 1 , 𝑦 2 , 𝑐 ) ∈ 𝐷 1 , then
From the continuity of the partial derivatives of 𝑓 and we know that there exists 𝜀 > 0 such that 𝑦 2 − 𝜀 ≥ 𝑐 and for any 𝑥 1 ∈ [ 𝑦 2 − 𝜀 , 𝑦 2 ] . Hence, since 𝑓 ( ⋅ , 𝑦 2 , 𝑐 ) ∶ 𝑥 1 ∈ [ 𝑦 2 − 𝜀 , 𝑦 1 ] → 𝑓 ( 𝑥 1 , 𝑦 2 , 𝑐 ) is strictly monotone increasing, then we have Next, for 𝑥 2 ∈ [ 𝑦 2 − 𝜀 , 𝑦 2 ] , then ( 𝑦 2 − 𝜀 , 𝑥 2 , 𝑐 ) ∈ 𝐷 2 and Hence, we get If 𝑦 2 − 𝜀 = 𝑐 , then Theorem 1.1 is true. Otherwise, we repeat the above process and we clearly see that the first and second variables in 𝑓 are decreasing and no less than 𝑐 . Let 𝑠 , 𝑡 be their limit values, respectively, then 𝑓 ( 𝑦 1 , 𝑦 2 , 𝑐 ) > 𝑓 ( 𝑠 , 𝑡 , 𝑐 ) and 𝑠 , 𝑡 ≥ 𝑐 . If 𝑠 = 𝑐 , 𝑡 = 𝑐 , then Theorem 1.1 is also true; otherwise, we repeat the above process again and denote 𝑝 and 𝑞 the greatest lower bounds for the first and the second variables , respectively. We clearly see that 𝑝 = 𝑞 = 𝑐 ; therefore, 𝑓 ( 𝑦 1 , 𝑦 2 , 𝑐 ) > 𝑓 ( 𝑐 , 𝑐 , 𝑐 ) and Theorem 1.1 is true.

Similarly, we have the following theorem.

Theorem 1.2. Suppose that 𝑎 , 𝑏 ∈ ℝ with 𝑎 < 𝑏 and 𝑐 ∈ [ 𝑎 , 𝑏 ] , 𝑓 ∶ [ 𝑎 , 𝑏 ] 𝑛 → ℝ has continuous partial derivatives and If 𝜕 𝑓 ( 𝐱 ) / 𝜕 𝑥 𝑚 < 0 for all 𝐱 ∈ 𝐸 𝑚 ( 𝑚 = 1 , 2 , … , 𝑛 − 1 ) , then for all 𝑦 𝑚 ∈ [ 𝑎 , 𝑐 ] ( 𝑚 = 1 , 2 , … , 𝑛 − 1 ) .

It follows from Theorems 1.1 and 1.2 that we get the following Corollaries 1.3–1.6.

Corollary 1.3. Suppose that 𝑎 , 𝑏 ∈ ℝ with 𝑎 < 𝑏 , 𝑓 ∶ [ 𝑎 , 𝑏 ] 𝑛 → ℝ has continuous partial derivatives and If 𝜕 𝑓 ( 𝐱 ) / 𝜕 𝑥 𝑚 > 0 for all 𝐱 ∈ 𝐷 𝑚 and 𝑚 = 1 , 2 , … , 𝑛 , then for all 𝑥 𝑚 ∈ [ 𝑎 , 𝑏 ] ( 𝑚 = 1 , 2 , … , 𝑛 ) with 𝑥 m i n = m i n 1 ≤ 𝑘 ≤ 𝑛 { 𝑥 𝑘 } .

Corollary 1.4. Suppose 𝑎 , 𝑏 ∈ ℝ with 𝑎 < 𝑏 , then and 𝑓 ∶ [ 𝑎 , 𝑏 ] 𝑛 → ℝ is symmetric with continuous partial derivatives. If 𝜕 𝑓 ( 𝐱 ) / 𝜕 𝑥 1 > 0 for all 𝐱 = ( 𝑥 1 , 𝑥 2 , … , 𝑥 𝑛 ) ∈ 𝐷 1 , then where 𝑥 m i n = m i n 1 ≤ 𝑘 ≤ 𝑛 { 𝑥 𝑘 } . Equality holds if and only if 𝑥 1 = 𝑥 2 = ⋯ = 𝑥 𝑛 .

Corollary 1.5. Suppose 𝑎 , 𝑏 ∈ ℝ with 𝑎 < 𝑏 , 𝑓 ∶ [ 𝑎 , 𝑏 ] 𝑛 → ℝ has continuous partial derivatives and If 𝜕 𝑓 ( 𝐱 ) / 𝜕 𝑥 𝑚 < 0 for all 𝐱 ∈ 𝐸 𝑚 and 𝑚 = 1 , 2 , … , 𝑛 , then where 𝑥 m a x = m a x 1 ≤ 𝑘 ≤ 𝑛 { 𝑥 𝑘 } . Equality holds if and only if 𝑥 1 = 𝑥 2 = ⋯ = 𝑥 𝑛 .

Corollary 1.6. Suppose 𝑎 , 𝑏 ∈ ℝ with 𝑎 < 𝑏 , then and 𝑓 ∶ [ 𝑎 , 𝑏 ] 𝑛 → ℝ is symmetric with continuous partial derivatives . If 𝜕 𝑓 ( 𝐱 ) / 𝜕 𝑥 𝑛 < 0 for all 𝐱 = ( 𝑥 1 , 𝑥 2 , … , 𝑥 𝑛 ) ∈ 𝐸 𝑛 , then where 𝑥 m a x = m a x 1 ≤ 𝑘 ≤ 𝑛 { 𝑥 𝑘 } . Equality holds if and only if 𝑥 1 = 𝑥 2 = ⋯ = 𝑥 𝑛 .

2. Unifying Proof of Some Well-Known Inequality

In this section, we denote 𝐚 = ( 𝑎 1 , 𝑎 2 , … , 𝑎 𝑛 ) , 𝑎 m i n = m i n 1 ≤ 𝑘 ≤ 𝑛 { 𝑎 𝑘 } , 𝑎 m a x = m a x 1 ≤ 𝑘 ≤ 𝑛 { 𝑎 𝑘 } , and

Proposition 2.1 (Power Mean Inequality). If the power mean 𝑀 𝑟 ( 𝐚 ) of order 𝑟 is defined by 𝑀 𝑟 ∑ ( 𝐚 ) = ( ( 1 / 𝑛 ) 𝑛 𝑖 = 1 𝑎 𝑟 𝑖 ) 1 / 𝑟 for 𝑟 ≠ 0 and 𝑀 0 ∏ ( 𝐚 ) = 𝑛 𝑖 = 1 𝑎 𝑖 1 / 𝑛 , then 𝑀 𝑟 ( 𝐚 ) ≥ 𝑀 𝑠 ( 𝐚 ) for 𝑟 > 𝑠 ; equality holds if and only if 𝑎 1 = 𝑎 2 = ⋯ = 𝑎 𝑛 .

Proof. It is well known that 𝑀 𝑟 ( 𝐚 ) is symmetric with respect to 𝑎 1 , 𝑎 2 , … , 𝑎 𝑛 and 𝑟 ↦ 𝑀 𝑟 ( 𝐚 ) is continuous. Without loss of generality, we assume that 𝑟 , 𝑠 ≠ 0 . Then If 𝐚 ∈ 𝐷 1 , then 𝜕 𝑓 ( 𝐚 ) / 𝜕 𝑎 1 > 0 . It follows from Corollary 1.4 that we get Equality holds if and only if 𝑎 1 = 𝑎 2 = ⋯ = 𝑎 𝑛 .

Proposition 2.2 (Holder Inequality). Suppose that ( 𝑥 1 , 𝑥 2 , … , 𝑥 𝑛 ) , ( 𝑦 1 , 𝑦 2 , … , 𝑦 𝑛 ) ∈ ℝ 𝑛 + ( 𝑝 , 𝑞 > 1 ) . If 1 / 𝑝 + 1 / 𝑞 = 1 , then

Proof. Let 𝐛 = ( 𝑏 1 , 𝑏 2 , … , 𝑏 𝑛 ) ∈ ℝ 𝑛 + and If 𝐚 ∈ 𝐷 1 , then Similarly, if 𝐚 ∈ 𝐷 𝑚 ( 𝑚 = 2 , 3 , … , 𝑛 ) , then 𝜕 𝑓 ( 𝐚 ) / 𝜕 𝑎 𝑚 > 0 . From Theorem 1.1, we get Therefore, Proposition 2.2 follows from 𝑎 𝑘 = 𝑦 𝑞 𝑘 / 𝑥 𝑝 𝑘 and 𝑏 𝑘 = 𝑥 𝑝 𝑘 .

Proposition 2.3 (Minkowski Inequality). Suppose that ( 𝑥 1 , 𝑥 2 , … , 𝑥 𝑛 ) , ( 𝑦 1 , 𝑦 2 , … , 𝑦 𝑛 ) ∈ ℝ 𝑛 + . If 𝑝 > 1 , then

Proof. Let 𝐛 = ( 𝑏 1 , 𝑏 2 , … , 𝑏 𝑛 ) ∈ ℝ 𝑛 + and If 𝐚 ∈ 𝐷 1 , then Similarly, If 𝐚 ∈ 𝐷 𝑚 ( 𝑚 = 2 , 3 , … , 𝑛 ) , then 𝜕 𝑓 ( 𝐚 ) / 𝜕 𝑎 𝑚 > 0 . It follows from Theorem 1.1 that we get Therefore, Proposition 2.3 follows from 𝑎 𝑘 = 𝑦 𝑝 𝑘 / 𝑥 𝑝 𝑘 and 𝑏 𝑘 = 𝑥 𝑝 𝑘 .

3. A Brief Proof for Hardy's Inequality

If 𝑎 𝑛 ≥ 0 ( 𝑛 ∈ ℕ , 𝑛 ≥ 1 ) with ∑ ∞ 𝑛 = 1 𝑎 𝑝 𝑛 < + ∞ , then the well-known Hardy's inequality (see [1,Theorem 3 2 6 ]) is

In this section, we establish the following result involving Hardy's inequality.

Theorem 3.1. Let 𝑛 ∈ ℕ , 𝑛 ≥ 1 , and 𝑎 𝑘 ≥ 0 ( 𝑘 ∈ ℕ , 𝑘 ≥ 1 ) . If then

Proof. Let 𝑏 𝑘 = ( 𝑘 − 1 / 2 ) 1 / 𝑝 𝑎 𝑘 , then inequality (3.3) is equivalent to and 𝐵 𝑛 = m i n 1 ≤ 𝑘 ≤ 𝑛 { 𝑏 𝑘 } . Let If 𝐛 ∈ 𝐷 𝑚 ( 𝑚 = 1 , 2 , … , 𝑛 ) , then Making use of the well-known Hadamard's inequality of convex functions, we get Then Theorem 1.1 leads to and we clearly see that inequalities (3.4) and (3.3) are true.

Corollary 3.2. Let 𝑛 ∈ ℕ , 𝑛 ≥ 1 , and 𝑎 𝑘 ≥ 0 ( 𝑘 ∈ ℕ , 𝑘 ≥ 1 ) . If then

Proof. From inequality (3.3), we clearly see that

Remark 3.3. If 𝑛 → + ∞ , then inequality (3.1) follows from inequality (3.10).

4. A Refinement of Carleman's Inequality

If 𝑎 𝑛 ≥ 0 ( 𝑛 ∈ ℕ , 𝑛 ≥ 1 ) with ∑ 0 < ∞ 𝑛 = 1 𝑎 𝑛 < ∞ , then the well-known Carleman's inequality is

with the best possible constant factor 𝑒 (see [2]).

Recently, Yang and Debnath [3] gave a strengthened version of (4.1) as follows:

Some other strengthened versions of (4.1) were given in [4–9]. In this section, we give a refinement for Carleman's inequality (see Corollary 4.4).

Lemma 4.1. If 𝑚 ∈ ℕ and 𝑚 ≥ 1 , then

Proof. Let 𝜓 ( 𝑚 ) > 𝜓 ( 𝑚 + 1 ) , then inequality 1 − 2 𝑚 + 2 + 3 𝑚 + 7 2 𝑚 > 3 𝑚 + 1 0 𝑚 + 1 𝑒 ( 𝑚 ! ) 1 / 𝑚 . ( 4 . 5 ) is equivalent to inequality
If 𝑚 ≥ 1 7 , then simple computation leads to inequality (4.5).
If √ 2 𝜋 𝑚 ≥ 𝑒 7 / 3 , then it is not difficult to verify that √ 2 𝜋 𝑚 ≥ 𝑒 ( 2 1 𝑚 2 + 7 1 𝑚 + 7 0 ) / ( 9 𝑚 2 + 3 9 𝑚 + 5 0 ) . ( 4 . 6 ) and
If 𝑒 > ( 1 + 1 / 𝑥 ) 𝑥 , then 𝑒 > 1 + 2 1 𝑚 2 + 7 1 𝑚 + 7 0 9 𝑚 2 𝑚 + 3 9 𝑚 + 5 0 ( 9 𝑚 2 + 3 9 𝑚 + 5 0 ) 𝑚 / ( 2 1 𝑚 2 + 7 1 𝑚 + 7 0 ) . ( 4 . 7 ) ; this implies that From inequalities (4.6) and (4.7), we get From the well-known Stirling Formula 0 < 𝜃 𝑚 < 1 ( √ 𝑚 ! > 𝑚 2 𝜋 𝑚 𝑒 𝑚 . ( 4 . 9 ) ), we get Therefore, inequality (4.5) follows from inequalities (4.8) and (4.9).
From the monotonicity of sequence l i m 𝑚 → + ∞ 𝜓 ( 𝑚 ) = 0 and 𝜓 ( 𝑚 ) > 0 , we get √ 2 𝜋 ( 𝑚 + 1 ) > 𝑒 2 / 3 , √ 2 𝜋 ( 𝑚 + 1 ) > 𝑒 ( 2 𝑚 + 2 ) / ( 3 𝑚 + 8 ) , √ 2 2 𝜋 ( 𝑚 + 1 ) > 1 + 3 𝑚 + 8 ( 3 𝑚 + 8 ) / 2 ⋅ ( 2 𝑚 + 2 ) / ( 3 𝑚 + 8 ) , ( 2 𝜋 ( 𝑚 + 1 ) ) 1 / ( 2 𝑚 + 2 ) > 3 𝑚 + 1 0 , 𝑒 2 3 𝑚 + 8 1 − 1 3 𝑚 + 1 0 > 𝑒 𝑚 + 1 ( 𝑚 + 1 ) ( 2 𝜋 ( 𝑚 + 1 ) ) 1 / ( 2 𝑚 + 2 ) . ( 4 . 1 0 ) ; therefore, inequality (4.3) is proved.
Meanwhile, we have
Therefore, inequality (4.4) follows from inequalities (4.10) and

Theorem 4.2. Let 𝑛 ≥ 1 , 𝑎 𝑘 > 0 ( 𝑘 = 1 , 2 , … , 𝑛 ) , and 𝐵 𝑛 = m i n 1 ≤ 𝑘 ≤ 𝑛 { 𝑘 𝑎 𝑘 } . If 𝑒 𝑛 𝑘 = 1 2 1 − 𝑎 3 𝑘 + 7 𝑘 − 𝑛 𝑘 = 1 𝑘 𝑗 = 1 𝑎 𝑗 1 / 𝑘 ≥ 𝐵 𝑛 𝑒 𝑛 𝑘 = 1 2 1 − 1 3 𝑘 + 7 𝑘 − 𝑛 𝑘 = 1 1 ( 𝑘 ! ) 1 / 𝑘 . ( 4 . 1 2 ) , then

Proof. Let 𝑘 = 1 , 2 , … , 𝑛 , 𝐛 = ( 𝑏 1 , 𝑏 2 , … , 𝑏 𝑛 ) , and 𝐷 𝑚 = 𝐛 ∣ 𝑏 𝑚 = m a x 1 ≤ 𝑘 ≤ 𝑛 𝑏 𝑘 > m i n 1 ≤ 𝑘 ≤ 𝑛 𝑏 𝑘 > 0 , 𝑚 = 1 , 2 , … , 𝑛 , 𝑓 ∶ 𝐛 ∈ ℝ 𝑛 + ⟶ 𝑒 𝑛 𝑘 = 1 2 1 − 𝑏 3 𝑘 + 7 𝑘 𝑘 − 𝑛 𝑘 = 1 1 𝑘 ! 𝑘 𝑗 = 1 𝑏 𝑗 1 / 𝑘 , 𝐛 ∈ ℝ 𝑛 + . ( 4 . 1 3 ) , Then inequality (4.12) is equivalent to the following inequality: where 𝐛 ∈ 𝐷 𝑚 ( 𝑚 = 1 , 2 , … , 𝑛 ) .
If 𝜕 𝑓 ( 𝐛 ) 𝜕 𝑏 𝑚 2 = 𝑒 1 − 1 3 𝑚 + 7 𝑚 − 𝑛 𝑘 = 𝑚 1 𝑘 𝑏 𝑚 1 𝑘 ! 𝑘 𝑗 = 1 𝑏 𝑗 1 / 𝑘 2 > 𝑒 1 − 1 3 𝑚 + 7 𝑚 − 𝑛 𝑘 = 𝑚 1 𝑘 ( 𝑘 ! ) 1 / 𝑘 2 > 𝑒 1 − 1 3 𝑚 + 7 𝑚 − ∞ 𝑘 = 𝑚 1 𝑘 ( 𝑘 ! ) 1 / 𝑘 . ( 4 . 1 5 ) , then
From inequality (4.3) and 𝑓 𝑏 1 , 𝑏 2 , … , 𝑏 𝑛 𝐵 ≥ 𝑓 𝑛 , 𝐵 𝑛 , … , 𝐵 𝑛 . ( 4 . 1 6 ) together with Theorem 1.1, we clearly see that Therefore, inequality (4.14) is proved.

Corollary 4.3. Let 𝑛 ≥ 1 , 𝑎 𝑘 > 0 ( 𝑘 = 1 , 2 , … , 𝑛 ) , and 𝐵 𝑛 = m i n 1 ≤ 𝑘 ≤ 𝑛 { 𝑘 𝑎 𝑘 } . If 𝑒 𝑛 𝑘 = 1 2 1 − 𝑎 3 𝑘 + 7 𝑘 − 𝑛 𝑘 = 1 𝑘 𝑗 = 1 𝑎 𝑗 1 / 𝑘 ≥ 𝐵 𝑛 4 5 . 𝑒 − 1 ( 4 . 1 7 ) , then

Proof. Let 𝑚 = 1 , 2 , … , 𝑛 ( { 𝑇 ( 𝑚 ) } 𝑛 𝑚 = 1 ), then inequality (4.4) implies that 𝑒 𝑛 𝑘 = 1 2 1 − 𝑎 3 𝑘 + 7 𝑘 − 𝑛 𝑘 = 1 𝑘 𝑗 = 1 𝑎 𝑗 1 / k ≥ 𝐵 𝑛 𝑇 ( 𝑛 ) ≥ 𝐵 𝑛 𝑇 ( 1 ) = 𝐵 𝑛 4 5 . 𝑒 − 1 ( 4 . 1 8 ) is a strictly increasing sequence. Then from inequality (4.12) we get

Let 𝑎 𝑛 ≥ 0 ( 𝑛 ∈ ℕ , 𝑛 ≥ 1 ) ; thus, we know that Corollary 4.4 is true.

Corollary 4.4. If ∑ 0 < ∞ 𝑛 = 1 𝑎 𝑛 < ∞ with ∞ 𝑛 = 1 𝑛 𝑗 = 1 𝑎 𝑗 1 / 𝑛 ≤ 𝑒 ∞ 𝑛 = 1 2 1 − 𝑎 3 𝑛 + 7 𝑛 . ( 4 . 1 9 ) , then

Remark 4.5. Many other applications for Theorem 1.1 appeared in [10].

Acknowledgments

The authors wish to thank the anonymous referees for their very careful reading of the manuscript and fruitful comments and suggestions. This work was partly supported by the National Nature Science Foundation of China under Grant no. 60850005, the Nature Science Foundation of Zhejiang Province under Grant no. Y607128, the Nature Science Foundation of China Central Radio & TV University under Grant no. GEQ1633, and the Nature Science Foundation of Zhejiang Broadcast & TV University under Grant no. XKT-07G19.

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