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
.
and
, see also Sun [31].
: 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