On some inequalities

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Tiberiu Popoviciu

(original), Inequalities, paper

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Tiberiu Popoviciu (Institutul de Calcul)

Tiberiu Popoviciu

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T. Popoviciu, Asupra unor inegalităţi, Gaz. Mat. Fiz. Ser. A 11 (64) (1959) no. 8, pp. 451-461 (in Romanian)

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Gazeta Matematica

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1959-e-Popoviciu-GM-On-some-inequalities

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TIB. POPOVICIU, university professor, corresponding member of the RPR Academy

and

The well-known Cauchy-Buniakowski inequality
(1)

{(\sum A_{and} b_{and})}^{2} ≦ (\sum A_{and}^{2}) (\sum b_{and}^{2})

is checked for any strings of real numbers
(2)

A_{1}, A_{2}, \dots, A_{n}; b_{1}, b_{2}, \dots, b_{n} .

I. Aczél proved [1] the inequality
(3)

{(A_{0} b_{0} - \sum A_{and} b_{and})}^{2} ≧ (A_{0}^{2} - \sum A_{and}^{2}) (b_{0}^{2} - \sum b_{and}^{2}),

which is checked regardless of the strings of real numbers
(4)

A_{0}, A_{1}, \dots, A_{n}; b_{0}, b_{1}, \dots, b_{n},

so that we have
(5)

A_{0}^{2} > \sum A_{and}^{2} (or b_{0}^{2} > \sum b_{and}^{2}) .

In these formulas, as well as in the following ones, we denote, for brevity, with

\sum

, the meaning

\sum_{and = 1}^{n}, n

being a given natural number.
2. Several proofs of inequality (1) are known. One of these proofs is based on the observation that the quadratic form

\sum {(A_{and} x - b_{and} y)}^{2}

is nonnegative and, in general, defined. It follows that in (1) the equality holds if and only if the series (2) are proportional. Two series, with the same number of terms, such as, for example, series (2) (respectively series (4)), are said to be proportional (or linearly dependent), if there are two numbers

λ, μ

, not both zero, so that we have

λ A_{and} = μ b_{and}, and = 1, 2, \dots, n

(respectively

and = 0, 1, \dots, n

).
I. Aczél gives an analogous proof for inequality (3), noting that the quadratic form

\begin{matrix} {(A_{0} x - b_{0} y)}^{2} - \sum {(A_{and} x - b_{and} y)}^{2} = \\ = (A_{0}^{2} - \sum A_{and}^{2}) x^{2} - 2 (A_{0} b_{0} - \sum A_{and} b_{and}) x y + (b_{0}^{2} - \sum b_{and}^{2}) y^{2} \end{matrix}

is indefinite and deduces that in (3) equality occurs if and only if the sequences (4) are proportional.

If

n > 1

, the conclusion relative to the case of equality in formula (3) is valid only under hypothesis (5). If

A_{0}^{2} = \sum A_{and}^{2}

(or

b_{0}^{2} = \sum b_{and}^{2}

), inequality (3) holds (obviously), but for equality it is not then necessary that the strings (4) be proportional. To see this, it is sufficient to take, for example, for the strings (4) the strings

(n = 2) 5, 3, 4; 8, 4, 7

.
3. A second, very simple proof of inequality (1) is based on the equality

\sum_{and < j}^{1, 2, \dots, n} {(A_{and} b_{j} - A_{j} b_{and})}^{2} = (\sum A_{and}^{2}) (\sum b_{and}^{2}) - {(\sum A_{and} b_{and})}^{2} .

An analogous equality allows us to deduce inequality (3) from inequality (1).

If

\sum A_{and}^{2} > 0

, we have

\begin{aligned} {(A_{0} b_{0} - \sum A_{and} b_{and})}^{2} - (A_{0}^{2} - \sum A_{and}^{2}) (b_{0}^{2} - \sum b_{and}^{2}) = \\ = \frac{1}{\sum A_{and}^{2}} {{(b_{0} \sum A_{and}^{2} - A_{0} \sum A_{and} b_{and})}^{2} + \\ + (A_{0}^{2} - \sum A_{and}^{2}) [(\sum A_{and}^{2}) (\sum b_{and}^{2}) - {(\sum A_{and} b_{and})}^{2}]} \end{aligned}

from which, taking into account condition (5) and inequality (1), it follows that we have (3). For equality to occur in (3), it is necessary and sufficient that we have equality in (1) and that

b_{0} \sum A_{and}^{2} = A_{0} \sum A_{and} b_{and}

. From the proportionality of the series (2) we then immediately deduce the proportionality of the series (4).
4. It is known that the inequality (1) is generalized by Hölder's inequality

\begin{matrix} (6) & \sum A_{and} b_{and} ≦ {(\sum b_{and}^{p})}^{\frac{1}{p}} {(\sum b_{and}^{q})}^{\frac{1}{q}} \end{matrix}

where

A_{and}, b_{and} ≧ 0, and = 1, 2, \dots, n

, and

p

,

q

are two positive numbers (

> 1

) and conjugates, that is

\frac{1}{p} + \frac{1}{q} = 1

.

The equality in (3) holds if and only if the strings
(7)

A_{1}^{p}, A_{2}^{p}, \dots, A_{n}^{p}; b_{1}^{q}, b_{2}^{q}, \dots, b_{n}^{q}

are proportional.
To prove inequality (6), we start from the inequality between the generalized arithmetic and geometric means,

\begin{matrix} (8) & ξ^{\frac{R}{R + S}} η^{\frac{S}{R + S}} ≦ \frac{R ξ + S η}{R + S} \end{matrix}

which is checked for

ξ, η ≧ 0, R, S > 0

and in which equality occurs if and only if

ξ = η

.

If we assume that

\sum A_{and}^{p} > 0, \sum b_{and}^{q} > 0

and if we put

R = \frac{1}{p}, S = \frac{1}{q}, ξ = \frac{A_{and}^{p}}{\sum A_{and}^{p}}, η = \frac{b_{and}^{q}}{\sum b_{and}^{q}}

inequality (8) becomes

\begin{matrix} (9) & \frac{A_{and} b_{and}}{{(\sum A_{and}^{p})}^{\frac{1}{p}} {(\sum b_{and}^{q})}^{\frac{1}{q}}} ≦ \frac{A_{and}^{p}}{p \sum A_{and}^{p}} + \frac{b_{and}^{q}}{q \sum b_{and}^{q}} \end{matrix}

Doing here

and = 1, 2, \dots, n

and adding the inequalities thus obtained member by member, we have, taking into account also

\frac{1}{p} + \frac{1}{q} = 1

,

\frac{\sum A_{and} b_{and}}{{(\sum A_{and}^{p})}^{\frac{1}{p}} {(\sum b_{and}^{q})}^{\frac{1}{q}}} ≦ 1,

which is equivalent to inequality (6).
The equality condition follows from the equality condition of relation (8).
In the previous proof we assumed that the numbers

A_{1}, A_{2}, \dots, A_{n}

, on the one hand and the numbers

b_{1}, b_{2}, \dots, b_{n}

, on the other hand, are not all zero. It is seen that the previous results are also valid if

A_{1} = A_{2} = \dots = A_{n} = 0

or if

b_{1} = b_{2} = \dots = b_{n} = 0

For the equality case, we rely on the fact that any sequence is proportional to the sequence that has all terms equal to 0.

Note. Inequality (8) can be proven in many ways. For example, it is easily deduced if we observe that, assuming

η

given, the function of

ξ, \frac{R ξ + S η}{R + S} - ξ^{\frac{R}{R + S}} η^{\frac{S}{R + S}}

whose derivative is equal to

\frac{R}{R + S} [1 - {(\frac{η}{ξ})}^{\frac{S}{R + S}}]

, reaches its absolute minimum for and only for

ξ = η

.
5. We have the inequality

\begin{matrix} (10) & A_{0} b_{0} - \sum A_{and} b_{and} ≧ {(A_{0}^{p} - A_{and}^{p})}^{\frac{1}{p}} {(b_{0}^{q} - \sum b_{and}^{q})}^{\frac{1}{q}} \end{matrix}

which is checked if

p, q (> 1)

are two conjugate numbers and

a_{i}, b_{i} ≧ 0, i = 1, 2, \dots, n, a_{0}^{p} > \sum a_{i}^{p}, b_{0}^{q} > \sum b_{i}^{q}

, the equality in (10) holds if and only if the strings

\begin{matrix} (11) & a_{0}^{p}, a_{1}^{p}, \dots, a_{n}^{p}; b_{0}^{q}, b_{1}^{q}, \dots, b_{n}^{q} \end{matrix}

are proportional.
To prove inequality (10), we will first show that it is sufficient to deal with the case

n = 1

Indeed, either

n ≧ 1

and let's put

A = {(\sum a_{i}^{p})}^{\frac{1}{p}}, B = {(\sum b_{i}^{q})}^{\frac{1}{q}}

. Then

a_{0} > A ≧ 0, b_{0} > B ≧ 0

and based on Hölder's inequality,

\begin{matrix} (12) & \sum a_{i} b_{i} ≦ A B \end{matrix}

Assuming that inequality (10) is true for

n = 1

, we can write

\begin{matrix} (13) & a_{0} b_{0} - A B ≧ {(a_{0}^{p} - A^{p})}^{\frac{1}{p}} {(b_{0}^{q} - B^{q})}^{\frac{1}{q}} \end{matrix}

From (12), (13) the inequality (10) immediately follows (for

n

It remains to prove inequality (13) where

a_{0}, b_{0}, A, B

are real numbers such that

a_{0} > A ≧ 0, b_{0} > B ≧ 0

.

Based on inequality (6) (where

n = 2, a_{1} = A, a_{2} = {(a_{0}^{p} - A^{p})}^{\frac{1}{p}}

,

b_{0} = B, b_{1} = {(b_{0}^{q} - B^{q})}^{\frac{1}{q}})

, we have

\begin{matrix} (14) & A B + {(a_{0}^{p} - A^{p})}^{\frac{1}{p}} {(b_{0}^{q} - B^{q})}^{\frac{1}{q}} ≦ a_{0} b_{0} \end{matrix}

which is equivalent to inequality (13).
In order to have equality in (10), it is necessary and sufficient that we have equality in (12) and (13). We immediately find the proportionality of the series (11) as a necessary and sufficient condition for this equality.
6. Minkowski's inequality
(15)

{[Σ {(a_{i} + b_{i})}^{p}]}^{\frac{1}{p}} ≦ {(\sum a_{i}^{p})}^{\frac{1}{p}} + {(\sum b_{i}^{p})}^{\frac{1}{p}}

which is checked for

a_{i}, b_{i} ≧ 0, i = 1, 2, \dots, n, p > 1

, immediately follows from inequality (6). It is sufficient to add the inequalities term by term

\begin{matrix} (16) & {\begin{cases} \sum a_{i} {(a_{i} + b_{i})}^{p - 1} ≦ {(\sum a_{i}^{p})}^{\frac{1}{p}} {(\sum {(a_{i} + b_{i})}^{q (p - 1)})}^{\frac{1}{q}} \\ \sum b_{i} {(a_{i} + b_{i})}^{p - 1} ≦ {(\sum b_{i}^{p})}^{\frac{1}{p}} {(\sum {(a_{i} + b_{i})}^{q (p - 1)})}^{\frac{1}{q}} \end{cases} \end{matrix}

where

\frac{1}{p} + \frac{1}{q} = 1

and we obtain an inequality equivalent to (15).
In order to have the equality in (15) it is necessary and sufficient that we have the equality in both formulas (16). For the first equality, it is necessary and sufficient that the strings
and therefore the strings

a_{1}^{p}, a_{2}^{p}, \dots, a_{n}^{p}; {(a_{1} + b_{1})}^{p}, {(a_{2} + b_{2})}^{p}, \dots, {(a_{n} + b_{n})}^{p}

a_{1}, a_{2}, \dots, a_{n}; a_{1} + b_{1}, a_{2} + b_{2}, \dots, a_{n} + b_{n}

to be proportional, and for the second equality as the strings

b_{1}, b_{2}, \dots, b_{n}; a_{1} + b_{1}, a_{2} + b_{2}, \dots, a_{n} + b_{n}

to be proportional.
It follows that a necessary and sufficient condition for (15) to have equality is that the series (2) (with non-negative terms) are proportional.

We have based ourselves here on the following facts:

1^{\circ}

the proportionality coefficients (see no. 2) can be chosen non-negative if

a_{i}, b_{i} ≧ 0, 2^{\circ}

the string with all zero terms is proportional to any string (observation already made),

3^{\circ}

proportionality is transitive for sequences that do not have all zero terms.
7. R. Bellmann's inequality [2]

\begin{matrix} (17) & {[{(a_{0} + b_{0})}^{p} - \sum {(a_{i} + b_{i})}^{p}]}^{\frac{1}{p}} ≧ {(a_{0}^{p} - \sum a_{i}^{p})}^{\frac{1}{p}} + {(b_{0}^{p} - \sum b_{i}^{p})}^{\frac{1}{p}} \end{matrix}

which is checked for

a_{i}, b_{i} ≧ 0, i = 0, 1, \dots, n, p > 1, a_{0}^{p} > \sum a_{i}^{p}

,

b_{0}^{p} > \sum b_{i}^{p}

, is deduced from (15) exactly as inequality (10) is deduced from (6).

we

A = {(\sum a_{i}^{p})}^{\frac{1}{p}}, B = {(\sum b_{i}^{p})}^{\frac{1}{p}}

. Then

a_{0}^{p} > A^{p}, b_{0}^{p} > B^{p}

and based on Minkowski's inequality, we have

\begin{matrix} (18) & {[Σ {(a_{i} + b_{i})}^{p}]}^{\frac{1}{p}} ≦ A + B \end{matrix}

Assuming that inequality (17) is true for

n = 1

, we have

\begin{matrix} (19) & {[{(a_{0} + b_{0})}^{p} - (A + B)^{p}]}^{\frac{1}{p}} ≧ {(a_{0}^{p} - A^{p})}^{\frac{1}{p}} + {(b_{0}^{p} - B^{p})}^{\frac{1}{p}} \end{matrix}

From (18), (19) it follows (17), for

n

As for inequality (19) for

a_{0} > A ≧ 0, b_{0} > B ≧ 0

, it is equivalent to Minkowski's inequality

{(A + B)^{p} + {[{(a_{0}^{p} - A^{p})}^{\frac{1}{p}} + {(b_{0}^{p} - B^{p})}^{\frac{1}{p}}]}^{p}}^{\frac{1}{p}} ≦ a_{0} + b_{0}

We leave it to the reader to prove that the equality in (17) holds if and only if the series (4) (with non-negative terms) are proportional.

Note: In all the above formulas

x^{σ}

is the non-negative value of the power of

σ^{a}, σ > 0

his/her

x ≧ 0

.

II.

The well-known inequality of PL Chebyshev,

\begin{matrix} (20) & \frac{\sum a_{i} b_{i}}{n} \geq \frac{\sum a_{i}}{n} \cdot \frac{\sum b_{i}}{n} \end{matrix}

which is checked if the sequences (2) are monotone of the same sense, has been generalized in many ways. Thus M. Biernacki [3] puts instead of the sums

\sum a_{i}, \sum b_{i}, \sum a_{i} b_{i}

respectively the amounts

\sum ε_{i} a_{i}, \sum ε_{i} b_{i}

,

\sum_{i} ε_{i} a_{i} b_{i}

, where the terms of the string

\begin{matrix} (21) & ε_{1}, ε_{2}, \dots, ε_{n} \end{matrix}

are alternatively equal to 1 and -1 . By this substitution, inequality (20) remains true if the sequences (2) are non-increasing with non-negative terms. Inequality (20) remains true, under the same conditions, if

ε_{i}

are equal to 1 or -1, such that in the string (21) each (complete) group of consecutive terms equal to - 1 is preceded by a (complete) group of more numerous consecutive terms equal to 1.

So we have the inequality

\begin{matrix} (22) & \sum ε_{i} a_{i} b_{i} ≧ \frac{1}{n} (\sum ε_{i} a_{i}) (\sum ε_{i} b_{i}) \end{matrix}

if the string (21) verifies the above property and, in particular, the inequality

\begin{matrix} (23) & \sum (- 1)^{i - 1} a_{i} b_{i} ≧ \frac{1}{n} (\sum (- 1)^{i - 1} a_{i}) (\sum (- 1)^{i - 1} b_{i}) \end{matrix}

which are true if the sequences (2) are nonincreasing and with nonnegative terms.
9. We will give some generalizations of the previous inequalities. But first it is useful to give a shorthand notation for sequences and some notions and definitions related to them.

We will only consider strings with the same number

n

of terms. For brevity, we will denote them with a single letter

C

RANGE

c_{1}, c_{2}, \dots, c_{n}

Thus the strings (2) will be denoted by

A, B

, and the strings of variables

x_{1}, x_{2}, \dots, x_{n}; y_{1}, y_{2}, \dots, y_{n}

with

X, Y

And so on

equal

A = B

two strings

A, B

is defined by their term-by-term equality. The sum

A + B

of strings

A, B

is the sequence obtained by adding the two sequences term by term, and the product

λ A

of the string

A

by number

λ

is the string obtained by multiplying by the number

λ

each term of

A

In particular,

- A (= (- 1) A)

is the string that is deduced from

A

, changing the sign of each term. Equality and the two operations enjoy common and well-known properties (reflexivity, symmetry, transitivity, associativity, commutativity, distributivity, etc.) on which it is useless to insist here.

A string can also be interpreted as a vector whose coordinates are the terms of the string. The two operations are then vector addition and the product of a vector by a scalar.

row

\sum_{j = 1}^{k} λ^{(j)} A^{(j)}

is called a linear combination of strings (in finite number)

A^{(j)}, j = 1, 2, \dots, k

. The numbers

λ^{(j)}, j = 1, 2, \dots, k

are called the coefficients of this linear combination.

row

c_{n}, c_{n - 1}, \dots, c_{1}

formed with the terms of the string

c_{1}, c_{2}, \dots, c_{n}

considered in reverse order, is called the inversion of this string. We will denote it by

C^{*}

the inversion of the string C. We have

{(C^{*})}^{*} = C

.

Whether

\begin{matrix} (24) & F (X; Y) = F (x_{1}, x_{2}, \dots, x_{n}; y_{1}, y_{2}, \dots, y_{n}) = \sum_{i = 1}^{n} \sum_{j = 1}^{n} a_{i, j} x_{i} y_{j} \end{matrix}

real bilinear form (with coefficients $a_{i, j}$ real) in the two strings of variables $X, Y$ . The inequalities (20), (22), (23) are then equivalent to the inequality

\begin{matrix} (25) & F (A; B) ⩾ 0 \end{matrix}

where we have, respectively,

\begin{aligned} F (X; Y) = n \sum x_{i} y_{i} - (\sum x_{i}) (\sum y_{i}) \\ F (X; Y) = n \sum ε_{i} x_{i} y_{i} - (\sum ε_{i} x_{i}) (\sum ε_{i} y_{i}) \\ F (X; Y) = n \sum (- 1)^{i - 1} x_{i} y_{i} - (\sum (- 1)^{i - 1} x_{i}) (\sum (- 1)^{i - 1} y_{i}) \end{aligned}

If we want to generalize the previous inequalities, the following two problems immediately arise:

Problem 1. - Determine the real bilinear form (24), such that inequality (25) is verified if each of the sequences

A, B

has a monotonicity property of a specific meaning.

Problem 2. - Determine the real bilinear form (24), such that inequality (25) is verified if each of the series

A, B

has a monotonicity property of a determined meaning and has all terms of the same determined sign.

By the fact that one of the strings

A, B

has a monotonicity property of a certain sense, we understand that the sequence remains always non-decreasing or always non-increasing. Also, by the fact that the sequence has all terms of the same determined sign, we understand that its terms are always all non-negative or all non-positive. A sequence with all non-negative terms, resp. all non-positive terms is also called a non-negative, resp. non-positive sequence. We say that two sequences are monotonic of the same sense if they are both non-decreasing or both non-increasing, and we say that they are monotonic of the opposite sense if one is non-decreasing and the other is non-increasing. Analogous names can be used for sequences with all terms of the same determined sign.
10. It is seen that, if inequality (25) is verified under the conditions of problem 1 or under the conditions of problem 2, the analogous (somewhat contrary) inequality

\begin{matrix} (') & F (A; B) ≦ 0 \end{matrix}

is verified, under the same conditions, by the bilinear form

- F (X; Y)

and reciprocally.

In problem 1 there are four cases that can be distinguished, according to the sense of monotonicity of the sequences (2). These cases, numbered

1, 2, 3, 4

are included in the table

For example, case 2 is the case when the string

A

is non-increasing and the string

B

is non-decreasing.

The study of the four cases can be reduced to the study of case 1, noting, on the one hand, that if the string

C

is monotonous, the string -

C

is also monotonous but of opposite sense and, on the other hand, that we have

- F (- X; Y) = - F (X; - Y) = F (- X; - Y) = F (X; Y)

.

It follows that for inequality (25) to hold in case 4 and for inequality (

25^{'}

) to occur in cases 2 and 3, it is necessary and sufficient that inequality (25) occurs in case 1 .

It will therefore be sufficient to deal only with the solution of the following
Problem

1^{''}

. - Determine the real bilinear form (24) such that inequality (25) is verified regardless of the non-decreasing sequences.

A, B

.
11. Analogous considerations can be made on problem 2. We now have 16 cases numbered from 1 to 16 and which are included in the table

		nonnegative		not positive
	$A$ B	indescribable	I don't grow.	indescribable	I don't grow.
non-neg.	indescribable	1	2	3	4
non-neg.	I don't grow.	5	6	7	8
nephew	indescribable	9	10	11	12
nephew	I don't grow.	13	14	15	16

For example, case 7 is when the string

A

is nonpositive and nondecreasing and the sequence

B

is negative and non-increasing.

In problem 1, changing the signs of one or both groups of numbers

n

variables of the bilinear form

F (X; Y)

allowed us to reduce cases

2, 3, 4

corresponding to the corresponding case 1. In the case of problem 2, these transformations do not allow us to reduce, as above, all cases 2-16 to case 1, but they allow us to reduce cases 4, 13, 16 to case 1, the cases

3, 14, 15

in case 2, the cases

8, 9, 12

to case 5 and cases 7, 10, 11 to case 6. This is justified by the fact that if the string

C

is non-negative, resp. non-positive, the string -

C

is nonpositive or nonnegative and reciprocal. To reduce the cases

2, 3, 6

in case 1, we observe that, if the string

C

is monotonous, its upside down

C^{*}

is also monotonous but of opposite direction to

C

If

C

is non-negative or non-positive,

C^{*}

remains non-negative resp. non-positive. It follows that, for inequality (25) to be verified in cases 2, 5, 6, it is necessary and sufficient that the same inequality be verified for the bilinear form, respectively equal to

F (X^{*}; Y), F (X; Y^{*}), F (X^{*}; Y^{*})

in case 1.

It will therefore be sufficient to deal only with the solution of the problem
Problem 2'. - Determine the real bilinear form (24), such that the inequality (25) is verified whatever the strings are

A, B

nonnegative and nondecreasing.

Of course, we can reduce, analogously, in the case of problem 1, the 4 cases to any one of them, and, in the case of problem 2, the 16 cases to any one of them.
12. It is easily proven that any linear combination with nonnegative coefficients of nonnegative sequences is a nonnegative sequence and that any linear combination with nonnegative coefficients of nondecreasing sequences is a nondecreasing sequence.

Let's note with

U^{(- 1)}

the string with all terms equal to - 1 and

U^{(j)} (j = 1, 2, \dots, n)

the string that has the last

j

terms equal to 1, the others being equal to

0 . U^{(j)}

so the string is

\underset{⏟}{0, 0, \dots, 0}, \underset{⏟}{1, 1, \dots, 1}

. The strings

U^{(- 1)}, U^{(j)}, j = 1, 2, \dots, n

are some particular non-decreasing sequences, the last ones

n

being at the same time particular nonnegative strings.

We have
Lemma 1. - Any non-decreasing sequence is a linear combination with non-negative coefficients of the sequences

U^{(- 1)}, U^{(j)}, i = 1, 2, \dots, n

.

Lemma 2. - Any nonnegative and nondecreasing sequence is a linear combination with nonnegative coefficients of the sequences

U^{(j)}, j = 1, 2, \dots, n

.

The two lemmas follow immediately from the fact that if

C

is an arbitrary string, we have

C = \frac{| c_{1} | - c_{1}}{2} U^{(- 1)} + \frac{| c_{1} | + c_{1}}{2} U^{(n)} + \sum_{j = 1}^{n - 1} (c_{j + 1} - c_{j}) U^{(j)},

from which it is seen that the non-negativity of the coefficients is equivalent to the respective condition of monotonicity or monotonicity and invariance of the signs of the terms

c_{1}, c_{2}, \dots, c_{n}

of the string

C

.
13. If

\sum_{r = 1}^{k} λ^{(r)} A^{(r)}, \sum_{s = 1}^{l} μ^{(s)} B^{(s)}

are linear combinations of the strings

A^{(r)}, r = 1, 2, \dots, k

, respectively of the strings

B^{(s)}, s = 1, 2, \dots, l

, we have

F (\sum_{r = 1}^{k} λ^{(r)} A^{(r)}; \sum_{s = 1}^{l} μ^{(s)} B^{(s)}) = \sum_{r = 1}^{k} \sum_{s = 1}^{l} λ^{(r)} μ^{(s)} F (A^{(r)}; B^{(s)})

and if we take into account lemmas 1,2, we deduce
Theorem 1. - The necessary and sufficient condition for inequality (25) to be verified regardless of the non-decreasing series

A, B

is that this inequality is always verified when each of the strings

A, B

reduces to one of the strings

U^{(- 1)}, U^{(j)}, j = 1, 2, \dots, n

.

Theorem 2. - Necessary and sufficient condition for inequality (25) to be verified for any non-negative and non-decreasing sequences

A, B

is that this inequality is always verified when each of the strings

A, B

reduces to one of the strings

U^{(j)}

,

j = 1, 2, \dots, n

.

Considering the structure of the strings

U^{(- 1)}, U^{(j)}, j = 1, 2, \dots, n

, one can easily state the conditions relative to the coefficients

a_{i}, j

of the bilinear form (24).

We thus obtain
Theorem

1^{'}

. - The necessary and sufficient condition for the inequality (25), regarding the bilinear form (24), to be verified, whatever the non-decreasing sequences are

A, B

, is to have

\begin{matrix} \sum_{t = r}^{n} \sum_{j = s}^{n} a_{t}, j ≧ 0, r = 1, 2, \dots, n, s = 2, 3, \dots, n; \sum_{i = r}^{n} \sum_{j = 1}^{n} a_{i}, j = 0 \\ r = 1, 2, \dots, n . \end{matrix}

Theorem 2'. - The necessary and sufficient condition for the inequality (25), regarding the bilinear form (24), to be verified, whatever the non-negative and non-decreasing sequences

A, B

, is to have

\sum_{i = r}^{n} \sum_{j = s}^{n} a_{i}, j ≧ 0, r = 1, 2, \dots, n, s = 1, 2, \dots, n

As a first application, let's determine the sequence of numbers

\begin{matrix} (26) & λ_{1}, λ_{2}, \dots, λ_{n}, λ_{1} \neq 0, λ_{n} \neq 0, \end{matrix}

so that we have the inequality

\begin{matrix} (27) & (\sum λ_{i}) (\sum λ_{i} a_{i} b_{i}) ≧ (\sum λ_{i} a_{i}) (\sum λ_{i} b_{i}) \end{matrix}

whatever the non-decreasing strings are

A, B

Hypothesis

λ_{1} \neq 0, λ_{n} \neq 0

and even the assumption that all terms of the sequence (26) are different from zero does not restrict the generality of the problem (the opposite case returns to the modification of

n

).

In this case

F (X; Y) = (\sum λ_{i}) (\sum λ_{i} x_{i} y_{i}) - (\sum λ_{i} x_{i}) (\sum λ_{i} y_{i})

and if at least one of the strings

X, Y

is equal to

U^{(- 1)}

or with

U^{(n)}

, we have

F (X; Y) = 0

A simple calculation allows us to write

F (U^{(r)}; U^{(s)}) = (\sum_{j = 1}^{n - s^{'}} λ_{j}) (\sum_{j = n - r^{'} + 1}^{n} λ_{j})

where

r^{'} = min (r, s), s^{'} = max (r, s)

Applying the theorem

1^{'}

, it immediately follows that:
The necessary and sufficient condition for inequality (27) to be verified, whatever the non-decreasing sequences are

A, B

it's like numbers

\sum_{j = 1}^{r} λ_{j}, r = 1, 2, \dots, n - 1, \sum_{j = s}^{n} λ_{j}, s = 2, 3, \dots, n

be all of the same sign.
It is easy to see that this condition is equivalent to the fact that:

The numbers

λ_{1}, λ_{1} + λ_{2}, \dots, λ_{1} + λ_{2} + \dots + λ_{n - 1}

are (in a broad sense) between 0 and

Σ λ_{i}

The condition
is checked in particular for

λ_{1} = λ_{2} = \dots = λ_{n}

and we find PL Chebyshev's inequality. The condition is also verified if

n

is odd and

λ_{i} = (- 1)^{i - 1}, i = 1, 2, \dots, n

.

It follows that the inequality

\begin{matrix} (28) & \sum (- 1)^{i - 1} a_{i} b_{i} ≧ (\sum (- 1)^{i - 1} a_{i}) (\sum (- 1)^{i - 1} b_{i}) \end{matrix}

is checked whether

n

is odd and if the strings

A, B

are non-decreasing. Based on what has been said, the inequality is also verified if the strings

A, B

are non-increasing. This inequality is more precise (if

n

is odd) than inequality (23) by the lack of the second term of the factor

\frac{1}{n}

. Moreover, the second member of the inequality, under the conditions of the problem, is non-negative.

As a second example, let us determine the string (21) such that we have

\begin{matrix} (29) & \sum ε_{i} a_{i} b_{i} ≧ (\sum ε_{i} a_{i}) (\sum ε_{i} b_{i}) \end{matrix}

whatever the strings are

A, B

non-negative and non-increasing.

In this case we will apply the theorem

2^{'}

bilinear form

F (X; Y) = \sum ε_{i} x_{n - i + 1} y_{n - i + 1} - (\sum ε_{i} x_{n - i + 1}) (\sum ε_{i} y_{n - i + 1}) .

we

F (U^{(r)}; U^{(s)}) = (\sum_{i = 1}^{r^{'}} ε_{i}) (1 - \sum_{i = 1}^{s^{'}} ε_{i})

, where

r^{'} = min (r, s), s^{'} = max (r, s)

。
It immediately follows that:
The necessary and sufficient condition for inequality (29) to be verified regardless of the strings

A, B

non-negative and non-increasing is like the numbers

\sum_{i = 1}^{r} ε_{i}, r = 1, 2, \dots, n

to be (in a broad sense) between 0 and 1.
This condition is met, in particular, if

ε_{i} = (- 1)^{i - 1}

,

i = 1, 2, \dots, n

. Inequality (28) is therefore verified whatever the strings are

A, B

non-negative and non-increasing (regardless of whether

n

is even or odd). This inequality is more precise than (23).

BIBLIOGRAPHY

J. Aczél "Nokotorîe obscie metodî v teorii functionalnìh uravnenii adnoí peremenoi. Nowie primenenia functionalnîh uravnenii" Uspehi Mat. Nauk, XI, $3 (69)$ , 3-68 (1956).
M. Biernacki "Sur des inegalités remplies par des expressions dont les termes ont des signes alternés" Ann. Univ. Marie Curie-Sklodowska. A. 7, 89-99 (1954).
R. Bellmann,,On an inequality concerning an indefinite form" The Amer. Math. Monthly, 63, 108-109 (1956).

О НЕКОТОРЫХ НЕРАВЕНСТВАХ

In the first part, a new proof of inequality (3) is given. Ачеля [1] and неравенства, corresponding to неравенствам Голдер and минковского are considered. In the second part, some generalizations of the Chebyshev inequality (20) are considered. In particular, necessary and sufficient conditions are given so that inequalities (27) and (29) were fulfilled for non-decreasing, respectively non-negative and non-increasing sequences.