Inequalities play a prominent role in mathematics. At school we mainly deal with numerical inequalities, with the definition of which we will begin this article. And then we will list and justify properties of numerical inequalities, on which all principles of working with inequalities are based.
Let us immediately note that many properties of numerical inequalities are similar. Therefore, we will present the material according to the same scheme: we formulate a property, give its justification and examples, after which we move on to the next property.
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Numerical inequalities: definition, examples
When we introduced the concept of inequality, we noticed that inequalities are often defined by the way they are written. So we called inequalities meaningful algebraic expressions containing the signs not equal to ≠, less<, больше >, less than or equal to ≤ or greater than or equal to ≥. Based on the above definition, it is convenient to give a definition of a numerical inequality:
The meeting with numerical inequalities occurs in mathematics lessons in the first grade, immediately after getting acquainted with the first natural numbers from 1 to 9, and becoming familiar with the comparison operation. True, there they are simply called inequalities, omitting the definition of “numerical”. For clarity, it wouldn’t hurt to give a couple of examples of the simplest numerical inequalities from that stage of their study: 1<2 , 5+2>3 .
And further from natural numbers, knowledge extends to other types of numbers (integer, rational, real numbers), the rules for their comparison are studied, and this significantly expands the variety of types of numerical inequalities: −5>−72, 3>−0.275 (7−5, 6) , .
Properties of numerical inequalities
In practice, working with inequalities allows a number of properties of numerical inequalities. They follow from the concept of inequality we introduced. In relation to numbers, this concept is given by the following statement, which can be considered a definition of the relations “less than” and “more than” on a set of numbers (it is often called the difference definition of inequality):
Definition.
- number a is greater than b if and only if the difference a−b is a positive number;
- the number a is less than the number b if and only if the difference a−b is a negative number;
- the number a is equal to the number b if and only if the difference a−b is zero.
This definition can be reworked into the definition of the relations “less than or equal to” and “greater than or equal to.” Here is his wording:
Definition.
- number a is greater than or equal to b if and only if a−b is a non-negative number;
- a is less than or equal to b if and only if a−b is a non-positive number.
We will use these definitions when proving the properties of numerical inequalities, to a review of which we proceed.
Basic properties
We begin the review with three main properties of inequalities. Why are they basic? Because they are a reflection of the properties of inequalities in the most general sense, and not only in relation to numerical inequalities.
Numerical inequalities written using signs< и >, characteristic:
As for numerical inequalities written using the weak inequality signs ≤ and ≥, they have the property of reflexivity (and not anti-reflexivity), since the inequalities a≤a and a≥a include the case of equality a=a. They are also characterized by antisymmetry and transitivity.
So, numerical inequalities written using the signs ≤ and ≥ have the following properties:
- reflexivity a≥a and a≤a are true inequalities;
- antisymmetry, if a≤b, then b≥a, and if a≥b, then b≤a.
- transitivity, if a≤b and b≤c, then a≤c, and also, if a≥b and b≥c, then a≥c.
Their proof is very similar to those already given, so we will not dwell on them, but move on to other important properties of numerical inequalities.
Other important properties of numerical inequalities
Let us supplement the basic properties of numerical inequalities with a series of results that are of great practical importance. Methods for estimating the values of expressions are based on them; principles are based on them solutions to inequalities and so on. Therefore, it is advisable to understand them well.
In this section, we will formulate the properties of inequalities only for one sign of strict inequality, but it is worth keeping in mind that similar properties will be valid for the opposite sign, as well as for signs of non-strict inequalities. Let's explain this with an example. Below we formulate and prove the following property of inequalities: if a
For convenience, we will present the properties of numerical inequalities in the form of a list, while we will give the corresponding statement, write it formally using letters, give a proof, and then show examples of use. And at the end of the article we will summarize all the properties of numerical inequalities in a table. Go!
Consequence. Termwise multiplication of identical true inequalities of the form a
Adding (or subtracting) any number to both sides of a true numerical inequality produces a true numerical inequality. In other words, if the numbers a and b are such that a
To prove it, let’s make up the difference between the left and right sides of the last numerical inequality, and show that it is negative under the condition a (a+c)−(b+c)=a+c−b−c=a−b. Since by condition a
We do not dwell on the proof of this property of numerical inequalities for subtracting a number c, since on the set of real numbers subtraction can be replaced by adding −c.
For example, if you add the number 15 to both sides of the correct numerical inequality 7>3, you get the correct numerical inequality 7+15>3+15, which is the same thing, 22>18.
If both sides of a valid numerical inequality are multiplied (or divided) by the same positive number c, you get a valid numerical inequality. If both sides of the inequality are multiplied (or divided) by a negative number c, and the sign of the inequality is reversed, then the inequality will be true. In literal form: if the numbers a and b satisfy the inequality a b·c.
Proof. Let's start with the case when c>0. Let's make up the difference between the left and right sides of the numerical inequality being proved: a·c−b·c=(a−b)·c . Since by condition a 0 , then the product (a−b)·c will be a negative number as the product of a negative number a−b and a positive number c (which follows from ). Therefore, a·c−b·c<0
, откуда a·c We do not dwell on the proof of the considered property for dividing both sides of a true numerical inequality by the same number c, since division can always be replaced by multiplication by 1/c. Let's show an example of using the analyzed property on specific numbers. For example, you can have both sides of the correct numerical inequality 4<6
умножить на положительное число 0,5
, что дает верное числовое неравенство −4·0,5<6·0,5
, откуда −2<3
. А если обе части верного числового неравенства −8≤12
разделить на отрицательное число −4
, и изменить знак неравенства ≤ на противоположный ≥, то получится верное числовое неравенство −8:(−4)≥12:(−4)
, откуда 2≥−3
. From the just discussed property of multiplying both sides of a numerical equality by a number, two practically valuable results follow. So we formulate them in the form of consequences. All the properties discussed above in this paragraph are united by the fact that first a correct numerical inequality is given, and from it, through some manipulations with the parts of the inequality and the sign, another correct numerical inequality is obtained. Now we will present a block of properties in which not one, but several correct numerical inequalities are initially given, and a new result is obtained from their joint use after adding or multiplying their parts. If the numbers a, b, c and d satisfy the inequalities a
Let us prove that (a+c)−(b+d) is a negative number, this will prove that a+c By induction, this property extends to term-by-term addition of three, four, and, in general, any finite number of numerical inequalities. So, if for the numbers a 1, a 2, …, a n and b 1, b 2, …, b n the following inequalities are true: a 1 a 1 +a 2 +…+a n . For example, we are given three correct numerical inequalities of the same sign −5<−2
, −1<12
и 3<4
. Рассмотренное свойство числовых неравенств позволяет нам констатировать, что неравенство −5+(−1)+3<−2+12+4
– тоже верное. You can multiply numerical inequalities of the same sign term by term, both sides of which are represented by positive numbers. In particular, for two inequalities a
To prove it, you can multiply both sides of the inequality a
This property is also true for the multiplication of any finite number of true numerical inequalities with positive parts. That is, if a 1, a 2, ..., a n and b 1, b 2, ..., b n are positive numbers, and a 1 a 1 a 2…a n . Separately, it is worth noting that if the notation for numerical inequalities contains non-positive numbers, then their term-by-term multiplication can lead to incorrect numerical inequalities. For example, numerical inequalities 1<3
и −5<−4
– верные и одного знака, почленное умножение этих неравенств дает 1·(−5)<3·(−4)
, что то же самое, −5<−12
, а это неверное неравенство.
At the end of the article, as promised, we will collect all the studied properties in table of properties of numerical inequalities:
Bibliography.
- Moro M.I.. Mathematics. Textbook for 1 class. beginning school In 2 hours. Part 1. (First half of the year) / M. I. Moro, S. I. Volkova, S. V. Stepanova. - 6th ed. - M.: Education, 2006. - 112 p.: ill.+Add. (2 separate l. ill.). - ISBN 5-09-014951-8.
- Mathematics: textbook for 5th grade. general education institutions / N. Ya. Vilenkin, V. I. Zhokhov, A. S. Chesnokov, S. I. Shvartsburd. - 21st ed., erased. - M.: Mnemosyne, 2007. - 280 pp.: ill. ISBN 5-346-00699-0.
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The following properties are true for any numerical expressions.
Property 1. If we add the same numerical expression to both sides of a true numerical inequality, we obtain a true numerical inequality, that is, the following is true: ; .
Proof. If . Using the commutative, associative and distributive properties of the addition operation we have: .
Therefore, by definition of the relation “greater than” 
.
Property 2. If we subtract the same numerical expression from both sides of a true numerical inequality, we obtain a true numerical inequality, that is, the following is true: ;
Proof. By condition . Using the previous property, we add the numerical expression to both sides of this inequality, and we obtain: .
Using the associative property of the addition operation, we have: , therefore 
, hence .
Consequence. Any term can be transferred from one part of a numerical inequality to another with the opposite sign.
Property 3. If we add the correct numerical inequalities term by term, we obtain the correct numerical inequality, that is, true:
Proof. By property 1 we have: and, using the transitivity property of the relation “more”, we obtain: .
Property 4. True numerical inequalities of the opposite meaning can be subtracted term by term, preserving the sign of the inequality from which we are subtracting, that is: ;
Proof. By definition of true numerical inequalities 
. By property 3, if . As a consequence of property 2 of this theorem, any term can be transferred from one part of the inequality to another with the opposite sign. Hence, 
. Thus, if .
The property is proved in a similar way.
Property 5. If both sides of a correct numerical inequality are multiplied by the same numerical expression that takes a positive value, without changing the sign of the inequality, then we obtain a correct numerical inequality, that is:
Proof. From what 
. We have: 
Then 
. Using the distributive nature of the operation of multiplication relative to subtraction, we have: .
Then by definition the relation is “greater than”.
The property is proved in a similar way.
Property 6. If both parts of a correct numerical inequality are multiplied by the same numerical expression, which takes a negative value, changing the sign of the inequality to the opposite, then we obtain a correct numerical inequality, that is: ;
Property 7. If both sides of a true numerical inequality are divided by the same numerical expression that takes a positive value, without changing the sign of the inequality, then we obtain a true numerical inequality, that is:
Proof. We have: 
. By property 5, we get: . Using the associativity of the multiplication operation, we have: 
hence .
The property is proved in a similar way.
Property 8. If both parts of a correct numerical inequality are divided by the same numerical expression that takes a negative value, changing the sign of the inequality to the opposite, then we obtain a correct numerical inequality, that is: ;
We omit the proof of this property.
Property 9. If we multiply, term by term, correct numerical inequalities of the same meaning with negative parts, changing the sign of the inequality to the opposite, we obtain a correct numerical inequality, that is:
We omit the proof of this property.
Property 10. If we multiply, term by term, correct numerical inequalities of the same meaning with positive parts, without changing the sign of the inequality, we obtain a correct numerical inequality, that is:
We omit the proof of this property.
Property 11. If we divide the correct numerical inequality of the opposite meaning term by term with the positive parts, preserving the sign of the first inequality, we obtain a correct numerical inequality, that is:
;
.
We omit the proof of this property.
Example 1. Are inequalities 
And 
equivalent?
Solution. The second inequality is obtained from the first inequality by adding to both its parts the same expression, which is not defined at . This means that the number cannot be a solution to the first inequality. However, it is a solution to the second inequality. So there is a solution to the second inequality that is not a solution to the first inequality. Therefore, these inequalities are not equivalent. The second inequality is a consequence of the first inequality, since any solution to the first inequality is a solution to the second.
§ 1 A universal way to compare numbers
Let's get acquainted with the basic properties of numerical inequalities, and also consider a universal way to compare numbers.
The result of comparing numbers can be written using equality or inequality. Inequality can be strict or non-strict. For example, a>3 is a strict inequality; a≥3 is a weak inequality. The way numbers are compared depends on the type of numbers being compared. For example, if we need to compare decimal fractions, then we compare them place by digit; If you need to compare ordinary fractions with different denominators, then you need to bring them to a common denominator and compare the numerators. But there is a universal way to compare numbers. It consists of the following: find the difference between the numbers a and b; if a - b > 0, that is, a positive number, then a > b; if a - b< 0, то есть отрицательное число, то a < b; если a - b = 0, то a = b. Этот способ удобно использовать для доказательства неравенств. Например, доказать неравенство:
2b2 - 6b + 1 > 2b(b- 3)
Let's use a universal comparison method. Let's find the difference between the expressions 2b2 - 6b + 1 and 2b(b - 3);
2b2 - 6b + 1- 2b(b-3)= 2b2 - 6b + 1 - 2b2 + 6b; Let's add similar terms and get 1. Since 1 is greater than zero, a positive number, then 2b2 - 6b+1 > 2b(b-3).
§ 2 Properties of numerical inequalities
Property 1. If a> b, b > c, then a> c.
Proof. If a > b, then the difference a - b > 0, that is, a positive number. If b >c, then the difference b - c > 0 is a positive number. Let's add the positive numbers a - b and b - c, open the brackets and add similar terms, we get (a - b) + (b - c) = a - b + b - c = a - c. Since the sum of positive numbers is a positive number, then a - c is a positive number. Therefore, a > c, which is what needed to be proved.
Property 2. If a< b, c- любое число, то a + с < b+ с. Это свойство можно трактовать так: «К обеим частям верного неравенства можно прибавить одно и то же число, при этом знак неравенства не изменится».
Proof. Let's find the difference between the expressions a + c and b+ c, open the brackets and add similar terms, we get (a + c) - (b+ c) = a + c - b - c = a - b. By condition a< b, тогда разность a - b- отрицательное число. Значит, и разность (a + с) -(b+ с) отрицательна. Следовательно, a + с < b+ с, что и требовалось доказать.
Property 3. If a< b, c - положительное число, то aс < bс.
If a< b, c- отрицательное число, то aс >bc.
Proof. Let's find the difference between the expressions ac and bc, put c out of brackets, then we have ac-bc = c(a-b). But since a If we multiply a negative number a-b by a positive number c, then the product c(a-b) is negative, therefore, the difference ac-bc is negative, which means ac If a negative number a-b is multiplied by a negative number c, then the product c(a-b) will be positive, therefore, the difference ac-bc will be positive, which means ac>bc. Q.E.D. For example, a Since division can be replaced by multiplication by the reciprocal number, = n∙, the proven property can also be applied to division. Thus, the meaning of this property is as follows: “Both sides of an inequality can be multiplied or divided by the same positive number, and the sign of the inequality does not change. Both sides of the inequality can be multiplied or divided by a negative number, but it is necessary to change the sign of the inequality to the opposite sign.” Let us consider the corollary to property 3. Consequence. If a Proof. Let us divide both sides of the inequality a
reduce the fractions and get The statement has been proven. Indeed, for example, 2< 3, но Property 4. If a > b and c > d, then a + c > b+ d. Proof. Since a>b and c >d, the differences a-b and c-d are positive numbers. Then the sum of these numbers is also a positive number (a-b)+(c-d). Let's open the brackets and group (a-b)+(c-d) = a-b+ c-d= (a+c)-(b+ d). In view of this equality, the resulting expression (a+c)-(b+d) will be a positive number. Therefore, a+ c> b+ d. Inequalities of the form a>b, c >d or a< b, c< d называют неравенствами одинакового смысла, а неравенства a>b,c Property 5. If a > b, c > d, then ac> bd, where a, b, c, d are positive numbers. Proof. Since a>b and c is a positive number, then, using property 3, we get ac > bc. Since c >d and b is a positive number, then bc > bd. Therefore, by the first property ac > bd. The meaning of the proven property is as follows: “If we multiply term by term inequalities of the same meaning, whose left and right sides are positive numbers, we obtain an inequality of the same meaning.” For example, 6< a < 7, 4 < b< 5 тогда, 24 < ab < 35. Property 6. If a< b, a и b - положительные числа, то an< bn, где n- натуральное число. Proof. If we multiply n given inequalities term by term a< b, то, согласно утверждению свойства 5, получим an< bn. Прочесть доказанное утверждение можно так: «Если обе части неравенства - положительные числа, то их можно возвести в одну и ту же натуральную степень, сохранив знак неравенства». § 3 Application of properties Let's consider an example of the application of the properties we have considered. Let 33< a < 34, 3 < b< 4. Оценить сумму a + b, разность a - b, произведение a ∙ b и частное a: b. 1) Let's estimate the sum a + b. Using property 4, we get 33 + 3< a + b < 34 + 4 или 36 < a+ b <38. 2) Let's estimate the difference a - b. Since there is no subtraction property, we replace the difference a - b with the sum a + (-b). First let's estimate (- b). To do this, using property 3, both sides of inequality 3< b< 4 умножим на -1, при этом меняем знак неравенства на противоположный знак 3 ∙ (-1) >b∙ (-1) > 4 ∙ (-1). We get -4< -b< -3. Теперь можно сложить два неравенства одного знака 33< a < 34 и -4< -b< -3. Имеем 2 9< a - b <31. 3) Let's estimate the product a ∙ b. By property 5, we multiply inequalities of the same sign We learned about inequalities at school, where we use numerical inequalities. In this article we will consider the properties of numerical inequalities, from which the principles of working with them are built. The properties of inequalities are similar to the properties of numerical inequalities. The properties, its justification will be considered, and examples will be given. Yandex.RTB R-A-339285-1 When introducing the concept of inequalities, we have that their definition is made by the type of record. There are algebraic expressions that have signs ≠,< , >, ≤ , ≥ . Let's give a definition. Definition 1 Numerical inequality called an inequality in which both sides have numbers and numerical expressions. We consider numerical inequalities in school after studying natural numbers. Such comparison operations are studied step by step. The initial ones look like 1< 5 , 5 + 7 >3. After which the rules are supplemented, and the inequalities become more complicated, then we obtain inequalities of the form 5 2 3 > 5, 1 (2), ln 0. 73 - 17 2< 0 . To work with inequalities correctly, you must use the properties of numerical inequalities. They come from the concept of inequality. This concept is defined using a statement, which is designated as “more” or “less.” Definition 2 The definition is used when solving inequalities with the relations “less than or equal to,” “greater than or equal to.” We get that Definition 3 The definitions will be used to prove the properties of numerical inequalities. Let's look at 3 main inequalities. Use of signs< и >characteristic of the following properties: Definition 4 Example 1 For example, given the inequality 5< 11 имеем, что 11 >5, which means its numerical inequality − 0, 27 > − 1, 3 will be rewritten as − 1, 3< − 0 , 27 . Before moving on to the next property, note that with the help of asymmetry you can read the inequality from right to left and vice versa. In this way, numerical inequalities can be modified and swapped. Definition 5 Evidence 1 The first statement can be proven. Condition a< b и b < c означает, что a − b и b − c являются отрицательными, а разность а - с представляется в виде (a − b) + (b − c) , что является отрицательным числом, потому как имеем сумму двух отрицательных a − b и b − c . Отсюда получаем, что а - с является отрицательным числом, а значит, что a < c . Что и требовалось доказать. The second part with the transitivity property is proved in a similar way. Example 2 We consider the analyzed property using the example of inequalities − 1< 5 и 5 < 8 . Отсюда имеем, что − 1 < 8 . Аналогичным образом из неравенств 1 2 >1 8 and 1 8 > 1 32 it follows that 1 2 > 1 32. Numerical inequalities, which are written using weak inequality signs, have the property of reflexivity, because a ≤ a and a ≥ a can have the case of equality a = a. They are characterized by asymmetry and transitivity. Definition 6 Inequalities that have the signs ≤ and ≥ in their writing have the following properties: The proof is carried out in a similar way. To supplement the basic properties of inequalities, results that are of practical importance are used. The principle of the method is used to estimate the values of expressions, on which the principles of solving inequalities are based. This paragraph reveals the properties of inequalities for one sign of strict inequality. The same is done for non-strict ones. Let's look at an example, formulating the inequality if a< b и c являются любыми числами, то a + c < b + c . Справедливыми окажутся свойства: For a convenient presentation, we give the corresponding statement, which is written down and evidence is given, examples of use are shown. Definition 7 Adding or calculating a number to both sides. In other words, when a and b correspond to the inequality a< b , тогда для любого такого числа имеет смысл неравенство вида a + c < b + c . Evidence 2 To prove this, the equation must satisfy the condition a< b . Тогда (a + c) − (b + c) = a + c − b − c = a − b . Из условия a < b получим, что a − b < 0 . Значит, (a + c) − (b + c) < 0 , откуда a + c < b + c . Множество действительных числе могут быть изменены с помощью прибавления противоположного числа – с. Example 3 For example, if we increase both sides of the inequality 7 > 3 by 15, then we get that 7 + 15 > 3 + 15. This is equal to 22 > 18. Definition 8 When both sides of the inequality are multiplied or divided by the same number c, we obtain a true inequality. If you take a negative number, the sign will change to the opposite. Otherwise it looks like this: for a and b the inequality holds when a< b и c являются положительными числами, то a· c < b · c , а если v является отрицательным числом, тогда a · c >b·c. Evidence 3 When there is a case c > 0, it is necessary to construct the difference between the left and right sides of the inequality. Then we get that a · c − b · c = (a − b) · c . From condition a< b , то a − b < 0 , а c >0, then the product (a − b) · c will be negative. It follows that a · c − b · c< 0 , где a · c < b · c . Другая часть доказывается аналогичным образом. When proving, division by an integer can be replaced by multiplication by the inverse of the given one, that is, 1 c. Let's look at an example of a property on certain numbers. Example 4 Both sides of inequality 4 are allowed< 6 умножаем на положительное 0 , 5 , тогда получим неравенство вида − 4 · 0 , 5 < 6 · 0 , 5 , где − 2 < 3 . Когда обе части делим на - 4 , то необходимо изменить знак неравенства на противоположный. отсюда имеем, что неравенство примет вид − 8: (− 4) ≥ 12: (− 4) , где 2 ≥ − 3 . Now let us formulate the following two results, which are used in solving inequalities: When dividing both sides of inequality a< b разрешается на число a · b . Данное свойство используется при верном неравенстве 5 >3 2 we have that 1 5< 2 3 . При отрицательных a и b c условием, что a < b , неравенство 1 a >1 b may be incorrect. Example 5 For example, − 2< 3 , однако, - 1 2 >1 3 are an incorrect equation. All points are united by the fact that actions on parts of the inequality give the correct inequality at the output. Let's consider properties where initially there are several numerical inequalities, and its result is obtained by adding or multiplying its parts. Definition 9 When numbers a, b, c, d are valid for inequalities a< b и c < d , тогда верным считается a + c < b + d . Свойство можно формировать таким образом: почленно складывать числа частей неравенства. Proof 4 Let's prove that (a + c) − (b + d) is a negative number, then we get that a + c< b + d . Из условия имеем, что a < b и c < d . Выше доказанное свойство позволяет прибавлять к обеим частям одинаковое число. Тогда увеличим неравенство a < b на число b , при c < d , получим неравенства вида a + c < b + c и b + c < b + d . Полученное неравенство говорит о том, что ему присуще свойство транзитивности. The property is used for term-by-term addition of three, four or more numerical inequalities. The numbers a 1 , a 2 , … , a n and b 1 , b 2 , … , b n satisfy the inequalities a 1< b 1 , a 2 < b 2 , … , a n < b n , можно доказать метод математической индукции, получив a 1 + a 2 + … + a n < b 1 + b 2 + … + b n .
Example 6 For example, given three numerical inequalities of the same sign − 5< − 2 , − 1 < 12 и 3 < 4 . Свойство позволяет определять то, что − 5 + (− 1) + 3 < − 2 + 12 + 4 является верным. Definition 10 Termwise multiplication of both sides results in a positive number. When a< b и c < d , где a , b , c и d являются положительными числами, тогда неравенство вида a · c < b · d считается справедливым. Evidence 5 To prove this, we need both sides of the inequality a< b умножить на число с, а обе части c < d на b . В итоге получим, что неравенства a · c < b · c и b · c < b · d верные, откуда получим свойство транизитивности a · c < b · d . This property is considered valid for the number of numbers by which both sides of the inequality must be multiplied. Then a 1 , a 2 , … , a n And b 1, b 2, …, b n are positive numbers, where a 1< b 1 , a 2 < b 2 , … , a n < b n , то a 1 · a 2 · … · a n< b 1 · b 2 · … · b n
. Note that when writing inequalities there are non-positive numbers, then their term-by-term multiplication leads to incorrect inequalities. Example 7 For example, inequality 1< 3 и − 5 < − 4 являются верными, а почленное их умножение даст результат в виде 1 · (− 5) < 3 · (− 4) , считается, что − 5 < − 12 это является неверным неравенством. Consequence:
Termwise multiplication of inequalities a< b с положительными с a и b , причем получается a n < b n
. Let us consider the following properties of numerical inequalities. Corollary 1:
if a< b , то - a >-b. Corollary 2:
if a and b are positive numbers and a< b , то 1 a >1 b . Corollary 1:
If a< b , a
And b
are positive numbers, then a n< b n . If you notice an error in the text, please highlight it and press Ctrl+Enter The main types of inequalities are presented, including Bernoulli, Cauchy - Bunyakovsky, Minkowski, Chebyshev inequalities. The properties of inequalities and actions on them are considered. The basic methods for solving inequalities are given. Universal inequalities are satisfied for any values of the quantities included in them. The main types of universal inequalities are listed below. 1) | a b | ≤ |a| + |b| ; | a 1 a 2 ... a n | ≤ |a 1 | + |a 2 | + ... + |a n | 2)
|a| + |b| ≥ | a - b | ≥ | |a| - |b| | 3)
4)
Cauchy-Bunyakovsky inequality 5)
Minkowski's inequality, for p ≥ 1 Satisfiable inequalities are satisfied for certain values of the quantities included in them. 1)
Bernoulli's inequality: 2)
3)
Chebyshev's inequality 4)
Generalized Chebyshev inequalities Properties of inequalities are a set of rules that are satisfied when transforming them. Below are the properties of the inequalities. It is understood that the original inequalities are satisfied for values of x i (i = 1, 2, 3, 4) belonging to some predetermined interval. 1) When the order of the sides changes, the inequality sign changes to the opposite. 2) One equality is equivalent to two non-strict inequalities of different signs. 3) Transitivity property 4) The same number can be added (subtracted) to both sides of the inequality. 5) If there are two or more inequalities with the sign of the same direction, then their left and right sides can be added. 6) Both sides of the inequality can be multiplied (divided) by a positive number. 7) Both sides of the inequality can be multiplied (divided) by a negative number. In this case, the sign of inequality will change to the opposite. 8) If there are two or more inequalities with positive terms, with the sign of the same direction, then their left and right sides can be multiplied by each other. 9) Let f(x) be a monotonically increasing function. That is, for any x 1 > x 2, f(x 1) > f(x 2). Then this function can be applied to both sides of the inequality, which will not change the sign of the inequality. 10) Let f(x) be a monotonically decreasing function, That is, for any x 1 > x 2, f(x 1)< f(x 2)
.
Тогда к обеим частям неравенства можно применить эту функцию, от чего знак неравенства изменится на противоположный. The interval method is applicable if the inequality includes one variable, which we denote as x, and it has the form: The interval method is as follows. 1) Find the domain of definition of the function f(x) and mark it with intervals on the number axis. 2) Find the discontinuity points of the function f(x). For example, if this is a fraction, then we find the points at which the denominator becomes zero. We mark these points on the number axis. 3) Solve the equation 4) As a result, the number axis will be divided into intervals (segments) by points. Within each interval included in the domain of definition, we select any point and at this point we calculate the value of the function. If this value is greater than zero, then we place a “+” sign above the segment (interval). If this value is less than zero, then we put a “-” sign above the segment (interval). 5) If the inequality has the form: f(x) > 0, then select intervals with the “+” sign. The solution to the inequality is to combine these intervals, which do not include their boundaries. This method is applicable to inequalities of any complexity. It consists of applying the properties (presented above) to reduce the inequalities to a simpler form and obtain a solution. It is quite possible that this will result in not just one, but a system of inequalities. This is a universal method. It applies to any inequalities. References: Numerical inequalities: definition, examples
Properties of numerical inequalities
Basic properties
Other important properties of numerical inequalities
Properties of numerical inequalities
a ≤ a, a ≥ a are true inequalities.
If a< b и c - отрицательное число, то a · c >b·c.Formulas for basic inequalities
Formulas for universal inequalities
Equality occurs only when a 1 = a 2 = ... = a n.
Equality holds if and only if α a k = β b k for all k = 1, 2, ..., n and some α, β, |α| + |β| > 0 .Formulas of satisfiable inequalities
.
More generally:
,
where , numbers of the same sign and greater than -1
:
.
Bernoulli's Lemma:
.
See "Proofs of inequalities and Bernoulli's lemma".
for a i ≥ 0 (i = 1, 2, ..., n) .
at 0 < a 1 ≤ a 2 ≤ ... ≤ a n
And 0 < b 1 ≤ b 2 ≤ ... ≤ b n
.
At 0 < a 1 ≤ a 2 ≤ ... ≤ a n
And b 1 ≥ b 2 ≥ ... ≥ b n > 0
.
at 0 < a 1 ≤ a 2 ≤ ... ≤ a n
And 0 < b 1 ≤ b 2 ≤ ... ≤ b n
and k natural
.
At 0 < a 1 ≤ a 2 ≤ ... ≤ a n
And b 1 ≥ b 2 ≥ ... ≥ b n > 0
.
Properties of inequalities
If x 1< x 2
,
то x 2 >x 1 .
If x 1 ≤ x 2, then x 2 ≥ x 1.
If x 1 ≥ x 2, then x 2 ≤ x 1.
If x 1 > x 2 then x 2< x 1
.
If x 1 = x 2, then x 1 ≤ x 2 and x 1 ≥ x 2.
If x 1 ≤ x 2 and x 1 ≥ x 2, then x 1 = x 2.
If x 1< x 2
и x 2 < x 3
,
то x 1 < x 3
.
If x 1< x 2
и x 2 ≤ x 3
,
то x 1 < x 3
.
If x 1 ≤ x 2 and x 2< x 3
,
то x 1 < x 3
.
If x 1 ≤ x 2 and x 2 ≤ x 3, then x 1 ≤ x 3.
If x 1< x 2
,
то x 1 + A < x 2 + A
.
If x 1 ≤ x 2, then x 1 + A ≤ x 2 + A.
If x 1 ≥ x 2, then x 1 + A ≥ x 2 + A.
If x 1 > x 2, then x 1 + A > x 2 + A.
If x 1< x 2
,
x 3 < x 4
,
то x 1 + x 3 < x 2 + x 4
.
If x 1< x 2
,
x 3 ≤ x 4
,
то x 1 + x 3 < x 2 + x 4
.
If x 1 ≤ x 2 , x 3< x 4
,
то x 1 + x 3 < x 2 + x 4
.
If x 1 ≤ x 2, x 3 ≤ x 4, then x 1 + x 3 ≤ x 2 + x 4.
Similar expressions apply to the signs ≥, >.
If the original inequalities contain signs of non-strict inequalities and at least one strict inequality (but all signs have the same direction), then the addition results in a strict inequality.
If x 1< x 2
и A >0, then A x 1< A · x 2
.
If x 1 ≤ x 2 and A > 0, then A x 1 ≤ A x 2.
If x 1 ≥ x 2 and A > 0, then A x 1 ≥ A x 2.
If x 1 > x 2 and A > 0, then A · x 1 > A · x 2.
If x 1< x 2
и A < 0
,
то A · x 1 >A x 2.
If x 1 ≤ x 2 and A< 0
,
то A · x 1 ≥ A · x 2
.
If x 1 ≥ x 2 and A< 0
,
то A · x 1 ≤ A · x 2
.
If x 1 > x 2 and A< 0
,
то A · x 1 < A · x 2
.
If x 1< x 2
,
x 3 < x 4
,
x 1 , x 2 , x 3 , x 4 >0 then x 1 x 3< x 2 · x 4
.
If x 1< x 2
,
x 3 ≤ x 4
,
x 1 , x 2 , x 3 , x 4 >0 then x 1 x 3< x 2 · x 4
.
If x 1 ≤ x 2 , x 3< x 4
,
x 1 , x 2 , x 3 , x 4 >0 then x 1 x 3< x 2 · x 4
.
If x 1 ≤ x 2, x 3 ≤ x 4, x 1, x 2, x 3, x 4 > 0 then x 1 x 3 ≤ x 2 x 4.
Similar expressions apply to the signs ≥, >.
If the original inequalities contain signs of non-strict inequalities and at least one strict inequality (but all signs have the same direction), then multiplication results in a strict inequality.
If x 1< x 2
,
то f(x 1) < f(x 2)
.
If x 1 ≤ x 2 then f(x 1) ≤ f(x 2) .
If x 1 ≥ x 2 then f(x 1) ≥ f(x 2) .
If x 1 > x 2, then f(x 1) > f(x 2).
If x 1< x 2
,
то f(x 1) >f(x 2) .
If x 1 ≤ x 2 then f(x 1) ≥ f(x 2) .
If x 1 ≥ x 2 then f(x 1) ≤ f(x 2) .
If x 1 > x 2 then f(x 1)< f(x 2)
.
Methods for solving inequalities
Solving inequalities using the interval method
f(x) > 0
where f(x) is a continuous function with a finite number of discontinuity points. The inequality sign can be anything: >, ≥,<, ≤
.
f(x) = 0 .
We mark the roots of this equation on the number axis.
If the inequality has the form: f(x) ≥ 0, then to the solution we add points at which f(x) = 0. That is, some intervals may have closed boundaries (the boundary belongs to the interval). the other part may have open boundaries (the boundary does not belong to the interval).
Similarly, if the inequality has the form: f(x)< 0
,
то выбираем интервалы с знаком „-“
.
Решением неравенства будет объединение этих интервалов, в которые не входят их границы.
If the inequality has the form: f(x) ≤ 0, then to the solution we add points at which f(x) = 0.Solving inequalities using their properties
I.N. Bronstein, K.A. Semendyaev, Handbook of mathematics for engineers and college students, “Lan”, 2009.