A New Method to Study Analytic Inequalities

Throughout this paper, we denote the set of real numbers and the set of strictly positive real numbers, , .

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

Theorem 1.1.

Suppose that with and , has continuous partial derivatives and

(1.1)

If for all , then

(1.2)

for all .

Proof.

Without loss of generality, since we assume that and .

For , we clearly see that , then

(1.3)

From the continuity of the partial derivatives of and

(1.4)

we know that there exists such that and

(1.5)

for any . Hence, since is strictly monotone increasing, then we have

(1.6)

Next, for , then and

(1.7)

Hence, we get

(1.8)

If , then Theorem 1.1 is true. Otherwise, we repeat the above process and we clearly see that the first and second variables in are decreasing and no less than . Let be their limit values, respectively, then and . If , then Theorem 1.1 is also true; otherwise, we repeat the above process again and denote and the greatest lower bounds for the first and the second variables , respectively. We clearly see that ; therefore, and Theorem 1.1 is true.

Similarly, we have the following theorem.

Theorem 1.2.

Suppose that with and , has continuous partial derivatives and

(1.9)

If for all , then

(1.10)

for all .

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

Corollary 1.3.

Suppose that with , has continuous partial derivatives and

(1.11)

If for all and , then

(1.12)

for all with .

Corollary 1.4.

Suppose with , then

(1.13)

and is symmetric with continuous partial derivatives. If for all , then

(1.14)

where . Equality holds if and only if .

Corollary 1.5.

Suppose with , has continuous partial derivatives and

(1.15)

If for all and , then

(1.16)

where . Equality holds if and only if .

Corollary 1.6.

Suppose with , then

(1.17)

and is symmetric with continuous partial derivatives . If for all , then

(1.18)

where . Equality holds if and only if .

In this section, we denote , , and

(2.1)

Proposition 2.1 (Power Mean Inequality).

If the power mean of order is defined by for and , then for ; equality holds if and only if .

Proof.

It is well known that is symmetric with respect to and is continuous. Without loss of generality, we assume that . Then

(2.2)

If , then . It follows from Corollary 1.4 that we get

(2.3)

Equality holds if and only if .

Proposition 2.2 (Holder Inequality).

Suppose that , . If , then

(2.4)

Proof.

Let and

(2.5)

If , then

(2.6)

Similarly, if , then . From Theorem 1.1, we get

(2.7)

Therefore, Proposition 2.2 follows from and .

Proposition 2.3 (Minkowski Inequality).

Suppose that , . If , then

(2.8)

Proof.

Let and

(2.9)

If , then

(2.10)

Similarly, If , then . It follows from Theorem 1.1 that we get

(2.11)

Therefore, Proposition 2.3 follows from and .

If with , then the well-known Hardy's inequality (see [1,Theorem ]) is

(3.1)

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

Theorem 3.1.

Let , , and . If

(3.2)

then

(3.3)

Proof.

Let , then inequality (3.3) is equivalent to

(3.4)

and . Let

(3.5)

If , then

(3.6)

Making use of the well-known Hadamard's inequality of convex functions, we get

(3.7)

Then Theorem 1.1 leads to

(3.8)

and we clearly see that inequalities (3.4) and (3.3) are true.

Corollary 3.2.

Let , , and . If

(3.9)

then

(3.10)

Proof.

From inequality (3.3), we clearly see that

(3.11)

Remark 3.3.

If , then inequality (3.1) follows from inequality (3.10).

If with , then the well-known Carleman's inequality is

(4.1)

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

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

(4.2)

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 and , then

(4.3)

(4.4)

Proof.

Let , then inequality is equivalent to inequality

(4.5)

If , then simple computation leads to inequality (4.5).

If , then it is not difficult to verify that and

(4.6)

If , then ; this implies that

(4.7)

From inequalities (4.6) and (4.7), we get

(4.8)

From the well-known Stirling Formula ( ), we get

(4.9)

Therefore, inequality (4.5) follows from inequalities (4.8) and (4.9).

From the monotonicity of sequence and , we get ; therefore, inequality (4.3) is proved.

Meanwhile, we have

(4.10)

Therefore, inequality (4.4) follows from inequalities (4.10) and

(4.11)

Theorem 4.2.

Let , , and . If , then

(4.12)

Proof.

Let , , and ,

(4.13)

Then inequality (4.12) is equivalent to the following inequality:

(4.14)

where .

If , then

(4.15)

From inequality (4.3) and together with Theorem 1.1, we clearly see that

(4.16)

Therefore, inequality (4.14) is proved.

Corollary 4.3.

Let , , and . If , then

(4.17)

Proof.

Let ( ), then inequality (4.4) implies that is a strictly increasing sequence. Then from inequality (4.12) we get

(4.18)

Let ; thus, we know that Corollary 4.4 is true.

Corollary 4.4.

If with , then

(4.19)

Remark 4.5.

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