Some Equivalent Forms of Bernoulli’s Inequality: A Survey

Abstract

The main purpose of this paper is to link some known inequalities which are equivalent to Bernoulli’s inequality.

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Li, Y. and Yeh, C. (2013) Some Equivalent Forms of Bernoulli’s Inequality: A Survey. Applied Mathematics, 4, 1070-1093. doi: 10.4236/am.2013.47146.

1. Introduction

Based on the theory of inequalities, many classical inequalities not only promote the development of the inequality theory, but also lead to many applications in pure mathematics and in applied mathematics. Bernoulli’s inequality is one of the most distinguished inequalities. In this paper, a new proof of Bernoulli’s inequality via the dense concept is given. Some strengthened forms of Bernoulli’s inequality are established. Moreover, some equivalent relations between this inequality and other known inequalities are tentatively linked. The organization of this paper is as follows:

In Section 2, a new proof of Bernoulli’s inequality by means of the concept of density is raised. In Section 3, some strengthened forms of Bernoulli’s inequality are establised. In Section 4, we link some known inequalities which are equivalent to Bernoulli’s inequality. In Section 5, we collect some variants of Young’s inequality which are equivalent to Bernoulli’s inequality. For related results, we refer to [1-35].

2. Preliminaries

In order to complete these tasks, we need the definition and some basic results of the convex function as follows:

Definition 2.1

Let be a function, where I is an interval of R.

1) Suppose that P and Q are any two points on the graph of, if the chord can not below the arc PQ of the graph of f, then we say that f is a convex function on I. That is, for any two point and any,

(1)

then f is a convex function on I. We say that f is called concave on I if is convex on I.

If, for any two points with and any,

then we say that is a strictly convex function on I.

2) I is said to be midpoint convex or J-convex on I if for any two points,

(2)

It is well-known fact that every convex function on an interval is continuous; if f is mid-point convex and continuous on an interval I, then it is convex on I. The following Jensen’s inequality can be shown by the mathematical induction directly.

Lemma 2.2 (Jensen’s inequality, [3], page 31) Let be a convex function on I. Then for any with and for any ,

(3)

If f is strictly convex, then (3) is strictly unless the are all identically.

Lemma 2.3 Let be a function. Then the following statements are equivalent:

1) f is strictly convex on I2) For any two distinct points and any satisfying

3) For any two distinct points and any satisfying

Proof 1) Þ 2). Let be distinct and let be arbitrary. If, then

and x is between y and z. It follows from the strict convexity of f on I that

Hence 2) holds.

2) Þ 3). Let be distinct and let be arbitrary. If, then

and both y and z are distinct. By the assumption of (b), we have

It follows from that 3) holds.

3) Þ 1). Let be distinct and let be arbitrary. If, then

and. It follows from the assumption of 3) that

This prove 1) holds. Thus the proof is complete.

Next, we will prove Bernoulli’s inequality by means of the concept of density without differentiation or integration.

Lemma 2.4

(4)

The equality is obvious for case x = 0 or for case or 1.

Proof Let

Claim 1: E is dense in.

It suffices to show that E satisfies the following three properties.

1).

2) If, then.

3) If, then and.

Let be arbitrary with. Then

and. Thus

So, This proves 1) and hence E is nonempty.

If, then

This proves 2).

Next, if are such that, then for every with,

This proves the first part of 3). On the other hand, it follows from that

and

Therefore,

(5)

Thus, we complete the proof of 3). Since 1)-3) imply that for and

Therefore E must be dense in.

Finally, if is arbitrary and with, then for every,

This proves

Similarly, if is arbitrary and with, then, for every,

This proves

Therefore, for every, we have

It follows from (5) again that (4) holds. This completes the proof.

Corollary 2.5 The following statements are equivalent:

1) is strictly convex on

2) for all x, y > 0 with and for all

3) Young’s inequality holds, that is,

where X, Y > 0 with and with

4) (4) holds.

Proof

The equality of Young’s inequality is clear for case with. This completes the proof.

Next, we prove some equivalent results which are related to:

Lemma 2.6 For any, the following statements are equivalent:

1) is strictly convex on;

2) is strictly convex on;

3) is strictly convex on.

Proof Clearly, 1) Þ 2) and 3).

Now, we prove 3) Þ 1) and 2) Þ 1). Let be with and let be arbitrary. Since

we have

Thus, if t is small such that, we obtain that 3) implies 1). Similarly, if is enough large so that, we obtain that 2) implies 1). This completes the proof.

Lemma 2.7 Let, , satisfying cixi, i = 1, 2, ··· be all positive or all negative. If, for all with, then

Proof This lemma is true for by assumption. Suppose that this lemma holds for. Let

. If, then, clearly, the conclusion holds. Now, we assume. Since,

are all positive or all negative, we see that,. Therefore,

This completes our proof.

3. Variants of Bernoulli’s Inequality

In this section, we establish some variants of Bernoulli’s Inequality.

Since is strictly concave and strictly increasing on, its inverse function is strictly convex and strictly increasing. Using Lemma 2.7, we have the following Theorem 3.1 The following inequalities are equivalent:

is strictly convex on

, where and, that is, , where and

, where, 0 < yi, satisfy

, where and, that is, , where and

, where, , satisfy

, where and, that is, , where and

, where, ,

, where and, that is, , where and

, where, ,

, where and, that is, , where and

, where , ,

, where and, that is, , where and

, where, ,

, where and y > 0;

, where and

, where and

, where and

, where and

, where and

, where and y > 0;

, where and

, where and

, where and y < 0;

, where and;

, where and.

Proof Let, where, then, , is a strictly convex function, and.

Let. WLOG, we assume y1 > y2 > 0 and. Then

Now, we assume holds for . Set. We have for and with. Let. It follows from above argument and the induction assumption that

This proves. is obvious.

Moreover, it follows from Lemma 2.7 that, , and.

By, and,

holds

holds

, where and

, where and

, where and

, where and

holds.

holds

, where and

, where and

, where and

holds.

holds

, where and

, where and x > 0

, where and

holds.

It follows from that

holds

, where and

, where and

, where and

holds.

holds

, where and

, where and

, where and

holds.

holds

, where and

, where and x > 0

holds.

holds

, where and

, where and

, where and

holds.

It follows from and, where, that holds

, where and

, where, and

holds.

holds

, where and

, where and

, where and

holds.

holds

, where and

, where and

, where and

, where and

holds.

holds

, where and

, where and

, where and x > 0

holds.

holds

, where and

, where and

, where, −1 < x < 0

holds.

holds

, where and

, where and

, where and

holds.

holds

, where and

, where and

, where and x > 0

holds.

holds

, where and

, where and

, where and

holds.

holds

, where and

, where and

, where and

holds.

This prove our Theorem.

By Theorem 3.1, we have the following Corollary 3.2 Let be a constant. If and, then the following three inequalities are equivalent:

1)2)3)

Proof Clearly, it follows from Theorem 3.1 that 1) holds and hold;

2) holds and hold;

3) holds and hold.

4. Main Results

Now, we can state and prove some inequalities which are equivalent to each other in the following Theorem 4.1 Let,

, and, where n is a positive integer. Then the following some statements are equivalent:

, where and

, where and

and

, where, and

, where and, ,

, where and

, where and or

and

, where and,

, where and

, where and or

and

, where and,

, where and

where, and

, where and

, where and

, where

is convex on

, where

, where

, where and, hence, where and. Thus, is an increasing function of q, where and

, where q < p < 0 and hence, where p > q > 0 and y < q;

, where or and

, where

, where

if with (Hölder’s inequality);

(Cauchy’s inequality);

, where

, where,

Here In particular,

, (Schlömich’s inequality);

for (Minkowski’ inequality)

for (Minkowski’ inequality)

, hence where.

In general, (AGM inequality)

Shanon’s inequality:

(see [7]) where, , , , , see the following figure:

where

, which is equivalent to

, where are positive integers, and, that is, , where is a rational number;

, where and

, where

,

is (strictly) increasing on

is (strictly) increasing on

is (strictly) decreasing on

, , where

, where, it has following some variants:

, where

, where,

, where,

or, where,

, where,

, where,

, where,

, where,

, where,

, where or,

, where, ,

, where, ,

, where,

, where,

, where and

, where, that is, , where

, where

, where or that is, ,

, where that is, and

, where

, where

, where, that is, (Jacobsthal’s inequality);

where and or.

Proof Taking in Corollary 3.2, we see that, and are equivalent. Similarly, replacing x by in Corollary 2.2, we get that, and are equivalent. Hence, it follows from Theorem 3.1 that, , , and are equivalent. If, then, clearly,

, , , ,.

with i = 1, 2, 3 follows by taking y = x + 1.

: We see that iff x > −1. Hence

Similarly, we can prove.

follows from and in Theorem 3.1.

and follows from Theorem 3.1 too.

: Let, satisfy. By, we see that. Thus, it follows from that

Hence holds.

by replacing yi and ci by, respectively.

Similarly, we can prove.

: Let, , where, then. It follows from that

Hence,

This completes the proof of.

, see Hardy etc. ([8], Theorem 9, 11 and 16).

: It follows from and that

Thus, (see Maligrands [18] or Rooin [28]). Hence,

Therefore, holds.

: Taking and in, we see that

Hence

Dividing both sides by, we get.

is clear.

: We show by mathematical induction on n. If, then is obvious by.

Suppose holds for with.

Set n = m + 1. If, it is easy to see that each by the assumption, and hence. Therefore holds. Assume. Since, we have

Thus holds.

: Taking and in, we see that holds.

: Clearly,. Let, such that. If, then it follows from that

(5)

By and, ,. Clearly,. By (5) and,

Thus, is proved.

: Let. If, then. By,

This completes the proof of.

: Without loss of generality, we my assume that x, y > 0 and. Since is strictly increasing,

By the definition of the convex function, .

: Taking and in, we see that holds.

: Let. Then, by,

Let. Then, by,

: Let. Then, by,

Hence

and so

: If, then, by,

Replacing by,

Thus, is proved.

Similarly, we can prove.

: Let, then, by,

Replacing by,

Hence

and

f, then,. If, then,. Hence,

It follows by taking that holds.

Similarly, we can prove,.

: Let. Replacing and

by and in, respectively, for, we obtain thus, we complete the proof.

follows by taking p = q = 2 in.

: Let and

for. Then, it follows from that

Thus, is midconvex on, and hence is convex on. Hence, for any,

which implies

Letting in the both sides of the above inequality,

This shows that holds, see Li and Shaw [15].

: Let, ai > 0, qi ≥ 0, and, , where. Thus. By,

(6)

It follows from (6) and

that

Hence, holds.

: Let. Then, by,

where and are defined as above. Hence,

Thus,

This completes the proof of.

: Taking and replacing ai by in, for we obtain.

follows by taking in for.

: Taking and replacing by in, for thus we complete the proof.

: see p. 55 of Mitrinovic [19]. Similarly, we can prove.

follows by taking in.

follows by taking in.

follows by taking in.

follows by taking in

follows by taking and in

, (with) and (with) are clear.

: Replacing by in, we proved. Similarly, we can prove.

: Taking, , , , , and, we see easily that both and are equivalent.

: Without loss of generality, we may assume that,. Thus, by,

Hence,

This completes the proof of.

: Taking the natural logarithm in the both sides of, we get. Conversely, deleting the natural logarithm of the both sides of, we get.

: Taking in, we get.

: Taking in, we see that holds.

: Taking in, we get.

: Taking, in and using the following figure, we get.

is clear.

: Taking and in, we get.

(see [33]): It follows from that

follows by taking and in.

: Let. Then, by,

Thus,. Clearly, if, then the above inequality holds too.

: Clearly, if, then holds.

Suppose that holds for. Thus, for, if, then. Hence. This and complete the proof of.

: If, then. Therefore, by,

Thus,

Thus, holds.

Let. Then,. Thus,

holds, that is, is a strictly increasing function on.

, where and

,

holds by Theorem 1.

holds, that is, is a strictly increasing function on.

, where and

, ,

, ,

holds by Theorem 3.1.

holds, that is, is a strictly increasing function on.

is a strictly increasing function on.

is a strictly increasing function on.

holds.

Thus, , and are equivalent.

: For all, since and are equivalent,

In particular, for all x > 0, and

approaches to 0 as. Thus, by, and, we get, see [2].

: By the first inequality of,

Hence,

By the second inequality of,

Hence,

It follows from for that, for each

If or, then, clearly, holds. This completes the proof of.

: By,

Hence, , , that is, the first inequality of holds. Next, by,

Hence, if, then

where Thus, the second inequality of holds.

is clear.

: Taking in of,

where. Summing this n inequalities, we get holds.

:

see Bullen ([3], p. 117) or Kuang ([14], p. 33).

follows by taking.

follows by taking.

: By,. Thus. Replacing by, we completes the proof, see Cloud and Dranchman ([6], p. 32).

is clear.

: By,

Hence,. Thus, holds, see [3] and [29].

, , and are equivalent, we can also refer to [12].

: Without loss of generality, we may assume that By,

Hence

Thus, holds.

: For any x, a > 0, it follows from that. Taking, we get.

see Hardy etc. ([8], pp. 40-41) or Wang, Su, Wang [33].

: Taking in,

Hence,.

: Taking in,

.

Hence,

This completes the proof of, see Bullen ([3], p. 98) or Kuang ([14], p. 33).

follows by taking ck = 1 for k = 1, 2, ···, n.

: By,.

Hence, is midpoint convex on. Since is continuous on, is a convex function on. Thus, holds.

We can also prove by using the mathematical induction.

Thus, our proof is complete.

5. Other Equivalent Forms of Bernoulli’s Inequality

In this section, we shall collect some variants of Young’ inequality which is equivalent to the Bernoulli’s inequality.

Theorem 5.1 Let be positive numbers for

If, where the real numbers p, q satisfy, , then the following some inequalities are equivalent:

and there exists exactly one of is positive, the other are negative;

satisfying and;

satisfying and;

Proof Clearly, are variant of, respectively. Hence and are equivalent.

, is clear.

: Replacing by in, respectively, we get, where.

: Let, , , where. Then , by,

Hence

Thus,

This completes our proof.

: For all, satisfying and, by,

This prove the proof of.

follows by taking, in.

: Let, satisfy. Then, for,

Thus, holds.

follows by replacing by in, respectively. Conversely, is clear.

Similarly, we can prove,.

and can be proved similarly.

follows by using the mathematical induction.

If, then and.

,: Replacing a, b by in, respectively, we get.

Similarly, , where follows by replacing a, b by, in, respectively.

: Replacing a, b by in, respectively, , where. Similarly, we can prove, where.

: Replacing a, b by in, respectively, we get.

: Replacing a, b by in, respectively, we get.

: Replacing a, b by in, respectively, we get, Similarly, we can prove , where.

: Replacing by in, respectively, we get, where.

: It follows from Theorem 3.1 that, and are equivalent. Without loss of generality, we may assume that in. Replacing by in, respectively, where . It follows from and that

holds.

: Replacing by in, we get, where.

: Replacing by in, respectively, we get, where.

: Replacing by in, we get, where. Similarly, we can prove, where.

follows by replacing by in, respectively.

follows by taking by in, where.

follows by replacing by, respectively.

: Since, therefore, by, we get.

: Taking in, we get.

: Letting in, And then, by taking, we get This completes the proof of.

is a variant of.

and, see Sun [31].

follows by taking, where.

: By Theorem 3.1, and are equivalent. It suffices to show. We assume and. Replacing by in, respectively, and then, replacing by and taking, it follows from that holds.

: It follows from and that

Replacing by, we complete our proof.

Similarly, we can prove.

and are clear.

Remark 5.2 For inequality, we refer to Isumino and Tominaga [13]. For inequality, we refer to [3].

For inequality and, we refer to Sun [31].

For inequality and, we refer to Kuang

[14]. For inequality, , we refer to [34].

For inequality, we refer to Sun [31].

Remark 5.3 There are many variants of Hölder’s inequality, Schlömich’s inequality, AGM inequality, Minkowski’s inequality, and so on, we omit the detail.

NOTES

Conflicts of Interest

The authors declare no conflicts of interest.

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