Prove that the inequality is true. Proof and solution of inequalities

MOU Grishino - Slobodskaya secondary school

Module program

"Methods for Proving Inequalities"

within the elective course

"Behind the pages of a mathematics textbook"

for students in grades 10-11

Compiled by:

mathematic teacher

Pankova E.Yu.

Explanatory note

“Mathematics is called a tautological science: in other words, mathematicians are said to spend time proving that things are equal to themselves. This statement is highly inaccurate for two reasons. First, mathematics, despite its characteristic scientific language, is not a science; rather, it can be called art. Secondly the basic results of mathematics are more often expressed by inequalities than by equalities.”

Inequalities are used in practical work math all the time. They are used to obtain a number of interesting and important extreme properties of "symmetrical" figures: a square, a cube, an equilateral triangle, as well as to prove the convergence of iterative processes and calculate some limits. The role of inequalities is also important in various questions of natural science and technology.

Problems for proving inequalities are the most difficult and interesting of the traditional ones. Proving inequalities requires true ingenuity, the creativity that makes mathematics the exciting subject it is.

Evidence teaching plays a big role in the development of deductive-mathematical thinking and general thinking abilities of students. How to teach students to independently carry out proofs of inequalities? The answer is: only by considering many techniques and methods of evidence and applying them regularly.

The ideas used to prove inequalities are almost as diverse as the inequalities themselves. In specific situations, generic methods often lead to ugly solutions. But the non-obvious combination of several "basic" inequalities is possible only for a few schoolchildren. And, besides, nothing prevents the student in each specific case from looking for a better solution than that obtained by the general method. For this reason, proving inequalities is often relegated to the realm of art. And like all art, there are technique, the set of which is very wide and it is very difficult to master all of them, but every teacher should strive to expand the mathematical tool available in his stock.

This module is recommended for students in grades 10-11. Not all possible methods for proving inequalities are considered here (the method of changing a variable, proving inequalities using a derivative, the method of research and generalization, and the ordering technique are not affected). You can suggest considering other methods at the second stage (for example, in grade 11), if this module of the course arouses interest among students, as well as focusing on the success of mastering the first part of the course.


		Equations and inequalities with a parameter.
		Methods for proving inequalities.
		Equations and inequalities containing the unknown under the module sign.
		Systems of inequalities with two variables.

The content of the elective course

"Behind the pages of a mathematics textbook"

"Methods for Proving Inequalities"


		Introduction.
		Proof of inequalities based on the definition.
		Method mathematical induction.
		Application of classical inequalities.
		Graphic method.
		The opposite method.
		A technique for considering inequalities with respect to one of the variables.
		Amplification idea.
		Lesson - control.

Lesson 1. Introduction.

Proving inequalities is a fascinating and challenging topic in elementary mathematics. Absence unified approach to the problem of proving inequalities, leads to the search for a number of techniques suitable for proving inequalities certain types. This elective course will examine following methods proof of inequalities:

Repetition:

Carry out proofs of some properties.

Classical inequalities:

1)
(Cauchy's inequality)

History reference:

Inequality (1) is named after French mathematician August Cauchy. Number
called arithmetic mean numbers a and b;

number
called geometric mean numbers a and b. Thus, the inequality means that the arithmetic mean of two positive numbers is not less than their geometric mean.

Additionally:

Consider several mathematical sophisms with inequalities.

Mathematical sophism- an amazing statement, in the proof of which imperceptible, and sometimes quite subtle errors are hidden.

Sophisms are false results obtained with the help of reasoning that only seems to be correct, but necessarily contains one or another error.

Example:

four over twelve

Lesson 2. Proof of inequalities based on the definition.

The essence of this method is as follows: in order to establish the validity of the inequalities F(x,y,z)>S(x,y,z) make up the difference F(x,y,z)-S(x,y,z) and prove that it is positive. Using this method, one often singles out a square, a cube of a sum or difference, an incomplete square of a sum or difference. This helps to determine the sign of the difference.

Example. Prove the inequality (x+y)(x+y+2cosx)+2 2sin 2x

Proof:

Consider the difference (x+y)(x+y+2cosx)+2- 2sin 2 x =(x+y)(x+y+2cosx)+2cos 2 x=(x+y)(x+y+2cosx) + cos 2 x +cos 2 x= (x+y) 2 +2(x+y)cosx+ cos 2 x +cos 2 x=((x+y)+cosx) 2 + cos 2 x 0.

Prove the inequality:

1.ab(a+b)+bc(b+c)+ac(a+c) 6abc

4.
>2x-20

6.(a+b)(b+c)(c+a) 8abc

Lesson 3. The method of mathematical induction.

When proving inequalities that include natural numbers, one often resorts to the method of mathematical induction. The method is as follows:

1) check the truth of the theorem for n=1;

2) we assume that the theorem is true for some n=k, and based on this assumption we prove the truth of the theorem for n=k+1;

3) based on the first two steps and the principle of mathematical induction, we conclude that the theorem is true for any n.

Example.

Prove the inequality

Proof:

1) for n=2 the inequality is true:

2) Let the inequality be true for n=k i.e.
(*)

Let us prove that the inequality is true for n=k+1, i.e.
. Let us multiply both parts of the inequality (*) by
we get 3) From item 1. and item 2 we conclude that the inequality is true for any n.

Assignments for classroom and home work

Prove the inequality:

6)
.

Lesson 4 Application of classical inequalities.

The essence of this method is as follows: using a series of transformations, the required inequality is derived using some classical inequalities.

Example.

Prove the inequality:

Proof:

As a reference inequality, we use .

We reduce this inequality to next kind:

, then

But =
, then

Prove the inequality:

1)(p+2)(q+2)(p+q)16pq(for proof we use the inequality
)

2)
(for documentation, the inequality is used)

3) (a+b)(b+c)(c+a) 8abc (the inequality is used for proof)

4) (for the proof, the inequality is used).

Lesson 5 Graphic method.

Proof of inequalities graphic method is as follows: if we prove the inequality f(x)>g(x)(f(x)

1) build graphs of functions y=f(x) and y=g(x);

2) if the graph of the function y=f(x) is located above (below) the graph of the function y=g(x), then the inequality being proved is true.

Example.

Prove the inequality:

cosx
,x0

Proof:

Let us construct in one coordinate system the graphs of the functions y=cosx and

It can be seen from the graph that at x0 the graph of the function y=cosx lies above the graph of the function y= .

Assignments to work in the classroom and at home.

Prove the inequality:

4)
.

Lesson 6

The essence of this method is as follows: let it be necessary to prove the truth of the inequality F(x,y,z) S(x,y,z)(1). The opposite is assumed, i.e., that the inequality F(x,y,z) S(x,y,z) (2) is valid for at least one set of variables. Using the properties of inequalities, transformations of inequality (2) are performed. If as a result of these transformations a false inequality is obtained, then this means that the assumption about the validity of inequality (2) is false, and therefore inequality (1) is true.

Example.

Prove the inequality:

Proof:

Assume the opposite, i.e. .

Let's square both parts of the inequality, we get , whence
and beyond

. But this contradicts the Cauchy inequality. So our assumption is wrong, i.e., the inequality Assignments for work in the classroom and at home is true.

Lesson 9 Lesson - control of students' knowledge.

This lesson can be done in pairs or if large numbers class in groups. At the end of the lesson, each student should be assessed. This is the transcript for this course. It is not recommended to carry out control work on this topic. the proof of inequalities, as already mentioned in the explanatory note, belongs to the field of art. At the beginning, students are asked to determine the method of proving the proposed inequalities themselves. If the students have difficulties, then the teacher tells them the rational method, warning the group that this, of course, will affect their assessment.

Work in pairs.

Task examples.

________________________________________________________________

Prove the inequality:

1.
(method of mathematical induction)

2.
(a-priory)

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Program

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transcript

1 FGBOU VO "PETROZAVODSK STATE UNIVERSITY" FACULTY OF MATHEMATICS AND INFORMATION TECHNOLOGIES Department of Geometry and Topology Khalzenen Elizaveta Sergeevna Final qualifying work for the bachelor's degree Methods of proving inequalities Direction: "0.03.0" "Mathematics" Supervisor Professor, Dr. Ph.-m. Sciences, Platonov S.S. (head's signature) Petrozavodsk

2 Contents Introduction ... 3. Jensen's inequality Permutation inequality Karamata's inequality Solving problems for proving inequalities ... 3 References

3 Maintenance A method is a set of sequential actions aimed at solving a specific type of problem. Methods for proving inequalities in this paper are aimed at finding custom solution inequalities of a certain form. Using such methods, the solution is reduced at times. The result is the same, but the amount of work is less. aim final work was the study of three types of inequalities with the help of which many others are easily proved. These are Jensen's inequality, permutation inequality, Karamata's inequality. All these inequalities are mathematically beautiful, with the help of these inequalities it is possible to solve school inequalities. This topic is up to date. In my opinion, it could be useful for schoolchildren, including for raising the level of knowledge in the field of mathematics. Since the methods are not standard, it seems to me that for students with a mathematical bias, they would be useful and exciting. The task is to search for and solve thematic inequalities from the proposed literature. The work consists of four paragraphs. In this section, Jensen's inequality is described, its proof and auxiliary definitions are given. In paragraph 2, the permutation inequality, its particular cases, and the general permutation inequality. In Section 3, Karamata's inequality is without proof. Paragraph 4 is the main work of the final work, i.e. proofs of inequalities using Jensen's inequality, permutation inequality, and Karamata's inequality

4 . Jensen's inequality Definition. A subset of a plane is called convex if any two points given set can be connected by a segment that will lie entirely in this set. Definition 2. Let f(x) be defined on some interval. The set of all points (x,y) for which y f(x) is called an epigraph, where x belongs to the given interval. The set of points (x,y) for which y f(x) is called a subplot. Definition 3. Consider a function on some interval. A function is called convex if its epigraph is a convex set on this interval. A function is called concave if its subgraph is a convex set. Criterion of convexity (concavity) of a function. For a function y = f(x) continuously differentiable on the interval (a, b) to be convex (concave) on (a, b), it is necessary and sufficient that its derivative f increases (decreases) on the interval (a , b). Criterion 2 of the convexity (concavity) of the function. For a function y = f(x) to be twice differentiable on the interval (a, b) to be convex (concave) on (a, b), it is necessary and sufficient that f (x) 0(f (x) 0 ) at all points x (a, b) Definition 4. The center of mass of the points A(x, y) and B(x 2, y 2) is the point C(x, y) belonging to the segment AB, such that AC = m B, where m BC m B is the mass A of point B and m A is the mass of point A. In vector form, the center of mass is found as follows: the radius vector of the center of mass: where r i is the radius vector of points A and B, i =,2. In coordinates: r = m r +m 2 r 2 m +m 2 () x = m x +m 2 x 2 m +m 2, y = m y +m 2 y 2 m +m 2-4 -

5 Let С AB be the center of mass of points A and B. If U is a convex subset of the plane and points A and B belong to U, then С AB belongs to U, since С AB belongs to segment AB. Let A, A 2 A be arbitrary points on the plane with masses m, m 2, m. The center of mass C A,A 2 A of the system of points A, A 2 A is determined by induction on:) At = 2 the center of mass C A A 2 of the system of points A, A 2 is already defined. We will assume that the point C A A 2 has a mass m + m 2 2) Suppose that for the system of points A, A 2 A the center of mass A, A 2 A is already defined. Let us denote the center of mass of points A, A 2 A by B and assume that the mass of point B is equal to m B = m + m m. By definition, we set C A,A 2 A = C BA, i.e. the center of mass of the system of points A, A 2 A, A is equal to the center of mass of two points B and A. We assume that the mass of the point C A,A 2 A is equal to m B + m = m + m m. It follows from the definition of the center of mass that if all points A, A 2 A belong to a convex set U, then their center of mass also belongs to U. Lemma. Let A, A 2 A be points on the plane with masses m, m 2, m and let r i be the radius vector of the point A i, i =,. If C is the center of mass of a system of points A, A 2 A, then the radius vector r C of point C can be calculated by the formula Proof. r C = m r +m 2 r 2 + +m r m +m 2 + +m (2) We will prove formula (2) by induction on. For = 2, the formula has already been proven (see formula ()). Let us assume that formula (2) has already been proved for (). Let B be the center of mass of the system of points A, A 2 A. Then - 5 -

6 r B = m r + m 2 r m r m + m m, the mass of point B is m B = m + m m. By definition, the center of mass C of a system of points A, A 2 A coincides with the center of mass of a pair of points B and A. The radius vector of point C is calculated by the formula () r C = m Br B + m r m B + m = m r + m 2 r m r m + m m which proves formula (2) for points. In coordinates, formula (2) has the form: x C = m x + m 2 x m k x k m + m m k y C = m y + m 2 y m k y k m + m m k Jensen's theorem. Let y = f(x) be a convex function on some interval, x, x 2, x - numbers from this interval; m, m 2, m - positive numbers, satisfying the condition m + m m =. Then the Jensen inequality holds: f(m x + m 2 x m x) m f(x) + m 2 f(x 2) + + m f(x) If the function y = f(x) is concave on some interval, x, x 2, x - numbers from this interval; m, m 2, m -positive numbers that also satisfy the condition m + m m =. Then Jensen's inequality has the form: f(m x + m 2 x m x) m f(x) + m 2 f(x 2) + + m f(x)" Proof: Consider a function f(x) convex on the interval (a, b) . Consider points A, A 2, A on its graph and let A i = (x i, y i), y i = f(x i). Let us take arbitrary masses m, m 2, m for points A, A 2, A, so that m + m m =. From the fact that f(x) is a convex function it follows that - 6 -

7 that the epigraph of the function is a convex set. Therefore, the center of mass of points A, A 2, A belongs to the epigraph. Find the coordinates of the center of mass: x c = m x + m 2 x m x m + m m = m x + m 2 x m x y c = m y + m 2 y m y m + m m = m f(x) + m 2 f(x 2) + + m f(x) Since C belongs to the epigraph, then we get a p.t.d. y c f(x c) m f(x) + m 2 f(x 2) + + m f(x) f(m x + m 2 x m x) (a + a a) a a 2 a We take the logarithm of inequality (3), we obtain the equivalent inequality (3) l (a + a 2 + +a) l(a a 2 a) (4) Using the properties of logarithms, we rewrite inequality (4) in the form : l (a +a 2 + +a) l a + l a l a (5) The resulting inequality is a particular case of Jensen's inequality for the case when f(x) = l(x), m = m 2 = = m =. Note that the function y = l(x) is concave on the interval (0, +), since y =< 0, поэтому неравенство (5) есть special case inequalities x2-7 -

8 Jensen for a concave function f(x) = l(x). Since inequality (5) is valid, the equivalent inequality (3) is also valid. 2. Permutation inequality Definition. A one-to-one correspondence of a set of numbers (,2,3, ) into itself is called a permutation of elements. Let's denote the permutation by σ so that σ(), σ(2), σ(3) σ() are numbers,2,3, in a different order. Consider two sets of numbers a, a 2, a and b, b 2, b. Sets a, a 2, a and b, b 2, b are called identically ordered if for any numbers i and j from the fact that a i a j implies that b i b j. In particular, the largest number from the set a, a 2, a corresponds to the largest number from the set b, b 2, b, for the second largest number from the first set, there is the second largest number from the second set, and so on. Sets a, a 2, a and b, b 2, b are said to be reverse-ordered if, for any numbers i and j, the fact that a i a j implies that b i b j. It follows from this that the largest number from the set a, a 2, a corresponds to smallest number from the set b, b 2, b, the second largest number from the set a, a 2, a corresponds to the second smallest number from the end of the set b, b 2, b and so on. Example.) Let two collections be given, such that a a 2 a and b b 2 b, then according to the definitions given by us, these collections are equally ordered. 2) Let two sets be given, such that a a 2 a and b b 2 b, in this case the sets of numbers a, a 2, a and b, b 2, b will be in reverse order. Everywhere below a, a 2, a and b, b 2 , b - positive real numbers «Theorem. (Permutation inequality) Let there be two sets of numbers a, a 2, a and b, b 2, b. Consider the set of their possible permutations. Then the value of the expression is 8 -

9 S = a b σ + a 2 b σ2 + + a b σ () will be largest when the sets a, a 2, a and b, b 2, b are equally ordered, and smallest when a, a 2, a and b, b 2, b are in reverse order. For all other permutations, the sum S will be between the smallest and largest values. Example. According to the theorem a b + b c + c a 3, since the set a, b, c and a, b, c are in reverse order and the value a a + b b + c c = 3 will be the smallest. Proof of the theorem. Consider two sets of numbers: the first is a, a 2, a and the second is b, b 2, b. Assume that these sets are not ordered in the same way, those. There are indices i and k such that a i > a k and b k > b i. Let's swap the numbers b k and b i in the second set (such a transformation is called “sorting”). Then in the sum S the terms a i b i and a k b k will be replaced by a i b k and a k b i, and all other terms will remain unchanged. Note that a i b i + a k b k< a i b k + a k b i, так как (a i b i + a k b k) (a i b k + a k b i) = a i (b i b k) a k (b i b k) = (a i a k)(b i b k) < 0 Поэтому сумма Sувеличится. Выполняем сортировку пока это возможно. Если процесс прекратился, то это означает, что мы получили правильный порядок, а это и есть highest value. The smallest value is obtained similarly, only we sort until the sets are in reverse order. As a result, we will come to the smallest value. “Theorem 2. Consider two positive sets a, a 2, a 3 a and b, b 2, b 3 b and all its possible permutations. Then the value of the product (a i + b σ(i)) will be largest when the sets a, a 2, a 3 a and b, b 2, b 3 b are equally ordered, and smallest when they are reverse ordered

10 Theorem 3. Consider two sets a, a 2, a 3 a and b, b 2, b 3 b the elements of this set are positive. Then the value of () a i + b σ(i) will be largest when the sets a, a 2, a 3 a and b, b 2, b 3 b are equally ordered and smallest when they are reverse ordered.” Theorems 2,3 are special cases of more general theorem, which is discussed below. General permutation inequality “Theorem 4 (General permutation inequality). Let the function f be continuous and convex on some interval in R. Then for any sets of numbers a, a 2, a 3 a and b, b 2, b 3 b from the interval, the value of the expression f (a + b σ()) + f ( a 2 + b σ(2)) + f (a + b σ()) will be largest when the sets are in the same order and smallest when the sets are in reverse order. Theorem 5. Let the function f be continuous and concave on some interval in R Then: the value of the expression f (a + b σ()) + f (a 2 + b σ(2)) + f (a + b σ()) will be largest when the numbers are in reverse order and smallest when the sets a, a 2, a 3 a and b, b 2, b 3 b are equally ordered. Proof.") Consider the case = 2. Let the function f be convex and there are two sets a > a 2 and b > b 2. We need to prove that Let f(a + b) + f(a 2 + b 2) f(a + b 2) + f(a 2 + b) (2) x = a + b 2, k = a a 2, m = b b 2. Then - 0 -

11 a + b 2 = x + k, a 2 + b = x + m, a + b = x + k + m, so inequality (2) takes the form f(x + k + m) + f(x + k ) f(x + k) + f(x + m) (3) To prove the inequality, we use the figure The figure shows a graph of a convex function y = f(x) the points A(x, f(x)), C(x + k, f(x + k)), D(x + m, f(x + m)), B (x + k + m, f(x + k + m)). and on The convexity of the function f implies that the chord CD lies below the chord AB. Let K be the midpoint of CD, M3 the midpoint of AB. Note that the abscissas of the points K and M are the same, since x k = 2 ((x + k) + (x + m)) = (2x + k + m) 2 x m = 2 (x + (x + k + m) ) = (2x + k + m) 2 Therefore, the points K and M lie on the same vertical line, which implies that y m y k. - -

12 Since y m = (f(x) + f(x + k + m)) 2 y k = (f(x + k) + f(x + m)) 2 This implies inequalities (3) and (2). Q.E.D. 2) Let > 2. Suppose that the sets a, a 2, a 3 a and b, b 2, b 3 b are not ordered in the same way, i.e. there are indices i and k such that a i > a k and b i< b k. Поменяем во втором наборе числа b i и b k местами. Тогда в сумме S слагаемые f(a i + b i) и f(a k + b k) заменятся на f(a i + b k) и f(a k + b i), а все остальные слагаемые останутся без изменений. Из неравенства (2) вытекает, что поэтому сумма S увеличится. f(a i + b k) + f(a k + b i) f(a i + b i) + f(a k + b k) Аналогично можно продолжать сортировку до тех пор, пока не получим одинаково упорядоченные наборы. Полученное значение суммы S будет наибольшим, что и требовалось доказать. Теорема 5 доказывается аналогично. 3. Неравенство Караматы Определение. Невозрастающий набор чисел X = (x, x 2, x) мажорирует невозрастающий набор чисел Y = (y, y 2, y) если выполнены условия x + x x k y + y y k и x + x x = y + y y. Для k =,2 и положительных чисел x, x 2, x и y, y 2, y. Обозначение X Y, если X можарирует Y и X Y, если Y можарирует X. Например. (,0,0,0, 0) (2, 2, 0,0,0, 0) (,) - 2 -

13 If x, x 2, x are positive numbers, i= x i =, then (,) (x, x 2, x) (,0,0,0, 0) “Theorem (Karamata’s Inequality) Let f: (a, b ) R, f is a convex function x, x 2, x, y, y 2, y (a, b) and (x, x 2, x) (y, y 2, y), then f(x) + f(x 2) + f(x) f(y) + f(y 2) + f(y). If f is a concave function, then f(x) + f(x 2) + f(x) f(y) + f(y 2) + f(y)." See the proof in . 4. Solving problems for proving inequalities. In this section, we consider various inequality proof problems that can be solved using Jensen's inequality, permutation inequalities, or Karamata's inequality. Exercise. Prove the inequality where x, x 2, x > 0 Let x + x 2 + x + x x 2 x, f(x) = +x, m i = f(x) = (+ x) f(x) = (+ x ) 2 f(x) = 2(+ x) 3 > 0, x Then Jensen's inequality implies that - 3 -

14 Let us prove that i= + x i + x x 2 x + x + x 2 + +x This is true if and only if + x x 2 x + x + x x + x x 2 x x + x x x x 2 x And the last inequality coincides with the inequality Cauchy. Task 2. Prove that for any a, b > 0 the inequality is true: 2 a + b ab This is equivalent to the inequality 2ab a + b ab 2ab ab(a + b) 2 ab a + b p. Task 3. Prove that for any a, a 2, a > 0 the inequality is true: a a 2 a a a 2 a The inequality can be rewritten as: - 4 -

15 () (a a a 2 a a 2 a) this is the Cauchy inequality. Task 4. Prove that for any a, a 2, a > 0 the inequality is true: Consider the inequality for =3. a + a a + a a 2 a 3 a a a a 2 + a 2 a 3 + a 3 a 3 -true Denote x = a a 2,x 2 = a 2 a 3,x = a a, Then x x 2 x =. Then the inequality takes the form: x + x x This inequality follows from Cauchy's inequality: q.t.d. Task 5. Prove that (x + x x) x x 2 x = si x + si x si x si x + x x, where 0 x i π - 5 -

16 The inequality follows from Jensen's inequality for the function y = si x. The function y = si x is concave on the interval (0, π), since y = si x< 0при x (0, π), Гдеm i =. ч.т.д. Задание 6. si x + si x si x si(x + x x) Доказать,что для любых a, a 2, a >0 the inequality is true: (a + a 2+ +a)(a + a a) 2 The inequality can be rewritten as: this is equivalent to (a + a a) a + a a 2 a +a 2 + +a a + a 2+ +a We consider Jensen's function f(x) = x and obtain this equality. and using the inequality Task 7. Prove that for any x, y, z > 0 the inequality x 5 + y 5 + z 5 x 3 y 2 + y 3 x 2 + z 3 x 2 is true Let's apply the permutation inequality. Let the first set look like Second x 3, y 3, z 3, x 2, y 2, z 2 the value of the expression on the left side of x 5 + y 5 + z 5 is composed of identically ordered sets of numbers. It follows from this that the value obtained - 6 -

17 for all other permutations more value obtained with the "most correct" arrangement of variables. Task 8. Prove that for any x, y, z > 0 the inequality is true: x + x 2 + y + y 2 + z + z 2 x + y 2 + y + z 2 + z + x 2 We can assume that x y z. Let a = x, a 2 = y, a 3 = z, b = + x 2, b 2 = + y 2, b 3 = + z 2 Sets a, a 2, a 3 and b, b 2, b 3 are oppositely ordered, therefore, by the permutation inequality, the sum a b + a 2 b 2 + a 3 b 3 is the smallest among the sums In particular, which is equivalent to a b σ + a 2 b σ2 + a 3 b σ3. a b + a 2 b 2 + a 3 b 3 a b 2 + a 2 b 3 + a 3 b, x + x 2 + y + y 2 + z + z 2 x + y 2 + y + z 2 + z + x 2. Task 9. Prove that for any a, a 2, a > 0 the inequality is true: (+ a 2) (+ a 2 2) (+ a 2) (+ a a 2 a 3 a)(+ a 2) (+ a) Multiply by a a 2 a, we get (a 2 + a 2)(a 3 + a 2 2) (a + a 2) (a + a 2)(a 2 + a 2 2) (a + a 2) - 7 -

18 Let us take the logarithm of the inequality and obtain an equivalent inequality. l(a 2 + a 2) + l(a a 3) + + l(a 2 + a) l(a 2 + a) + l(a a 2) + + l(a 2 + a) (9.) We use the general permutation inequality for the concave function y = l x. Let a i = a i, b i = a i 2. Then the sets b, b 2, b and a, a 2, a are identically ordered, so l(b + a) + l(b 2 + a 2) + + l(b + a ) l(b + a 2) + l(b 2 + a 3) + + l(b + a), which proves inequality (9.). Task 0. Prove that for any positive a, b, c a 2 + b 2 + c 2 ab + bc + ac (0.) Let a b c.. Since the sets (a, b, c) and (a, b , c) are identically ordered, but the sets (a, b, c) and (b, c, a) are not identically ordered, then the inequality (0.) follows from the permutation inequality. Exercise. Prove that if xy + yz + zx =, then Inequality (.) follows from Problem 0. Task 2. Prove that if a, b, c > 0, then x 2 + y 2 + z 2 (.). (a + c)(b + d) ab + cd Since the square root is greater than or equal to zero, we can square the right and left sides. We get: (a + c)(b + d) ab + 2 abcd + cd ab + ad + cb + cd ab + 2 abcd + cd ab + cd 2 abcd - 8 -

19 a 2 d 2 + 2abcd + c 2 d 2 4abcd a 2 d 2 + c 2 d 2 2abcd 0 (ad cd) 2 0 -True Assignment 3, 4. Prove that for any a, a 2, a > 0 the following inequality holds: 3) a 2 + a a 2 (a + a 2 + a) 2 4) a 2 + a a 2 (3.) (4.) where a + a 2 + a = Inequality (4.) follows from ( 3.) for a + a 2 + a =. We will prove inequality (3.). It can be converted to Or a 2 + a a 2 (a + a 2 + a) 2 2 a 2 + a a 2 (a + a a) Let's use Jensen's inequality for a convex function f(x): f(q x + q 2 x 2 + q x) q f(x) + q 2 f(x 2) + q f(x), Where 0 q i, q + q 2 + q =. If we take f(x) = x 2, q i =, i =,2, then we obtain inequality (3.), etc. Task 5. Prove that for any natural number and for any p, q, the inequality () 2 pq + ()(p + q) + ()pq + (5.) - 9 -

20 Let's transform the inequality (5.) to the equivalent form: () 2 pq + ()(p + q) + ()pq + () 2 pq + ()(p + q) + ()pq 0 )[()pq + (p + q) pq] + 0 () () 0 () 0 () 0 always since -natural Let us prove that Note that 0 (5.2) p + q pq = p(q ) (q) = (p)(q) Since p, q, then p 0, q 0, hence inequality (5.2) is valid. Task 6. For any positive numbers x, y, z, the inequality is true: Let x y z xyz (y + z x)(z + x y)(x + y z) (6.)) If y + z x< 0, то неравенство (6.) выполнено 2) Пусть все множители в правой части >0. Then inequality (6.) is equivalent to l x + l y + l z l(y + z x) + l(z + x y) + l(x + y z) Let f(x) = l x. Since f(x)`` = x 2< 0то функция f(x) = l x вогнутая на интервале (0, +) Проверим, что набор (y + z x, x + z y, x + y z) мажорирует набор (x, y, z). Действительно:

21 x + y z x (because y z 0); (x + y z) + (x + z y) = 2x x + y (x + y z) + (x + z y) + (y + z x) = x + y + z Since the function f(x) = l x is concave, then it follows from the Karamata inequality that l(x + y z) + l(x + z y) + l(y + z x) = l x + l y + l z, which proves inequality (6.). Task 7. Prove that for any a, b and c > 0 the inequality is true: a 2 + b 2 + c ab + ac + bc a 2 + 2bc + b 2 + 2ac + c 2 + 2ab This is equivalent to Let a b c. a 2 + b 2 + c 2 + ab + ac + bc + ab + ac + bc ac + b, ab + ac + b) (7.) (a 2 + 2bc, b 2 + 2ac, c 2 + 2ab) (7.2) We need to prove that (7.) majorizes (7.2). Let's use the definition of majorization :) a 2 + b 2 + c 2 a 2 + 2bc (b c) 2 0-correct 2) a 2 + b 2 + c 2 + ab + ac + bc a 2 + 2bc + b 2 + 2ac c 2 bc ac + ab 0 c(c b) a(c b) 0 (c b)(c a) 0-2 -

22 (c b) 0 and (c a) 0, then (c b)(c a) 0 3) 3)a 2 + b 2 + c 2 + ab + ac + bc + ab + ac + bc = a 2 + 2bc + b 2 + 2ac + c 2 + 2ab a 2 + b 2 + c 2 + 2ab + 2ac + 2bc = a 2 + 2bc + b 2 + 2ac + c 2 + 2ab Correct. Thus, the set of numbers (7.) majorizes the set of numbers (7.2). Applying the Karamata inequality for a convex function f(x) = x, we obtain the correct original inequality. Task 8. For a, b, c, d > 0 prove that the inequality a 4 + b 4 + c 4 + d 4 a 2 b 2 + a 2 c 2 + a 2 d 2 + b 2 c 2 + b 2 d 2 + c 2 d 2 Let a b c d y+z+w + e x+y+z+w e 2x+2y + e 2x+2z + e 2x+2w + e 2y+2z + e 2y+2w + e 2z+2w Consider two sets of numbers: (4x, 4y, 4z, 4w, x + y + z + w, x + y + z + w) and (2x + 2y, 2x + 2z, 2x + 2w, 2y + 2z, 2y + 2w, 2z + 2w) sets: (4x, 4y, 4z, x + y + z + w, x + y + z + w, 4w) and (8.) The second remains unchanged: (2x + 2y, 2x + 2z, 2x + 2w, 2y + 2z, 2y + 2w, 2z + 2w) (8.2) Let us prove that (8.) majorizes (8.2)

23 ) 4x 2x + 2y, x y is correct 2) 4x + 4y 4x + 2y + 2z,y z is correct 3) 4x + 4y + 4z 4x + 2y + 2z + 2x + 2w y + z x + w that 2x + 2w 2y + 2z x + w y + z, then case 3) is only possible for x + w = y + z 4) 4x + 4y + 4z + x + y + z + w 4x + 2y + 2z + 2x + 2w + 2y + 2z x + y + z w 0 y + z x + w Similar to the previous case, this inequality is valid for x + w = y + z 5) 4x + 4y + 4z + 2x + 2y + 2z + 2w 2z + 2y + 2w z w is correct 6) 4x + 4y + 4z + 2x + 2y + 2z + 2w + 4w , the set (8.) majorizes the set of numbers (8.2). Using the Karamata inequality for the function f(x) = e x, we obtain the correct inequality. Task 9. For a, b, c > 0, prove that the inequality a 3 + b 3 + c 3 + abc 2 3 (a2 b + b 2 c + c 2 a + ab 2 + bc 2 + ca) is true

24 Let a b c Multiply both sides of the inequality by 3, we get 3a 3 + 3b 3 + 3c 3 + 3abc 2(a 2 b + b 2 c + c 2 a + ab 2 + bc 2 + ca) 2 (9.) We make the change : And write the inequality (9.) in the form: x = l a, y = l b, z = l c e 3x + e 3x + e 3x + e 3y + e 3y + e 3y + e 3z + e 3z + e 3z + e x +y+z + e x+y+z + e x+y+z e 2x+y + e 2x+y + e 2y+z + e 2y+z + e 2z+x + e 2z+x + e x+ 2y + e x+2y + e y+2z + e y+2z + e z+2x + e z+2x + y + z, x + y + z, x + y + z) and (9.2) (2x + y, 2x + y, 2y + z, 2y + z, 2z + x, 2z + x, x + 2y, x + 2y, y + 2z, y + 2z, z + 2x, z + 2x) (9.3) z, x + y + z, 3z, 3z, 3z,) and (9.2) Order the second set: 2x + y z + 2x y z true y + 2z 2z + x y x true Thus, we get the set: (2x + y, 2x + y, z + 2x, z + 2x, 2y + z, x + 2y, x + 2y,2y + z, 2y + z, 2z + x, 2z + x, y + 2z, y + 2z) (9.3) prove that the set of numbers (9.2) majorizes the set of numbers (9.3)) 3 x 2x + y, x y 2) 6x 4x + 2y, x y 3) 9x 6x + 2y + z, 3x 2y + z

25 4) 9x + 3y 4x + 2y + 2z + 4x, x + y 2z, for x = y we get y z 5) 9x + 6y 4x + 2y + 2z + 4x + 2y + x, y z 6) 9x + 9y 4x + 2y + 2z + 4x + 4y + 2x, x + 3y 2z 0 For x = y we get y z 7) 9x + 9y + x + y + z , we get y z 8) 9x + 9y + 2x + 2y + 2z 4x + 2y + 2z + 4x + 4y + 2x + 4y + 2z, x + y + 3z 0 + 2z + 4x + 4y + 2x + 4y + 2z + 2z + x, x + 2y + 3z 0 0) 9x + 9y + 3x + 3y + 3z + 3z 4x + 2y + 2z + 4x + 4y + 2x + 4y + 2z + 4z + 2x, y z) 9x + 9y + 3x + 3y + 3z + 6z 4x + 2y + 2z + 4x + 4y + 2x + 4y + 2z + 4z + 2x + 2z + y, 5y z 2) 9x + 9y + 3x + 3y + 3z + 9z 4x + 2y + 2z + 4x + 4y + 2x + 4y + 2z + 4z + 2x + 4z + 2y 2x + 2y + 2z = 2x + 2y + 2z majorizes the set of numbers (9.3) and by the Karamata inequality for the function f(x) = e x we obtain the correct inequality

26 References) Yu.P. Soloviev. Inequalities. M.: Publishing House of the Moscow Center for Continuous Mathematical Education. 2005. 6s. 2) I.Kh. Sivashinsky. Inequalities in tasks M.: Nauka, p. 3) A. I. Khrabrov. Around the Mongolian inequality, Mat. enlightenment, ser. 3, 7, MTsNMO, M., 2003, p. 4) L. V. Radzivilovskii, Generalization of permutation inequality and the Mongolian inequality, Mat. enlightenment, ser. 3, 0, MCNMO Publishing House, M., 2006, p. 5) V.A.ch Krechmar. Algebra book. Fifth edition M., science, p. 6) D. Nomirovsky Karamata's inequality /D. Nomirovsky // (Quantum)-S

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MOU Grishino - Slobodskaya secondary school

Module program

"Methods for Proving Inequalities"

within the elective course

"Behind the pages of a mathematics textbook"

for students in grades 10-11

Compiled by:

mathematic teacher

Pankova E.Yu.

Explanatory note

“Mathematics is called a tautological science: in other words, mathematicians are said to spend time proving that things are equal to themselves. This statement is highly inaccurate for two reasons. First, mathematics, despite its scientific language, is not a science; rather, it can be called art. Secondly, the main results of mathematics are more often expressed by inequalities than by equalities.”

Inequalities are used in the practical work of a mathematician constantly. They are used to obtain a number of interesting and important extreme properties of "symmetrical" figures: a square, a cube, equilateral triangle, as well as to prove the convergence of iterative processes and calculate some limits. The role of inequalities is also important in various questions of natural science and technology.

The ideas used to prove inequalities are almost as diverse as the inequalities themselves. In specific situations, generic methods often lead to ugly solutions. But the non-obvious combination of several "basic" inequalities is possible only for a few schoolchildren. And, besides, nothing prevents the student in each specific case from looking for a better solution than that obtained by the general method. For this reason, proving inequalities is often relegated to the realm of art. And as in any art, it has its own technical techniques, the set of which is very wide and it is very difficult to master them all, but every teacher should strive to expand the mathematical tool available in his stock.


		Equations and inequalities with a parameter.
		Methods for proving inequalities.
		Equations and inequalities containing the unknown under the module sign.
		Systems of inequalities with two variables.

"Behind the pages of a mathematics textbook"

"Methods for Proving Inequalities"


		Introduction.
		Proof of inequalities based on the definition.
		Method of mathematical induction.
		Application of classical inequalities.
		Graphic method.
		The opposite method.
		A technique for considering inequalities with respect to one of the variables.
		Amplification idea.
		Lesson - control.

Lesson 1. Introduction.

Proving inequalities is a fascinating and challenging topic in elementary mathematics. The lack of a unified approach to the problem of proving inequalities leads to the search for a number of techniques suitable for proving certain types of inequalities. In this elective course, the following methods of proving inequalities will be considered:

Repetition:

Carry out proofs of some properties.

Classical inequalities:

1)
(Cauchy's inequality)

History reference:

Inequality (1) is named after the French mathematician Auguste Cauchy. Number
called arithmetic mean numbers a and b;

number
called geometric mean numbers a and b. Thus, the inequality means that the arithmetic mean of two positive numbers is not less than their geometric mean.

Additionally:

Consider several mathematical sophisms with inequalities.

Mathematical sophism- an amazing statement, in the proof of which imperceptible, and sometimes quite subtle errors are hidden.

Sophisms are false results obtained with the help of reasoning that only seems to be correct, but necessarily contains one or another error.

Example:

four over twelve

Lesson 2. Proof of inequalities based on the definition.

The essence of this method is as follows: in order to establish the validity of the inequalities F(x,y,z)>S(x,y,z) make up the difference F(x,y,z)-S(x,y,z) and prove that it is positive. Using this method, one often singles out the square, cube of the sum or difference, not full square sums or differences. This helps to determine the sign of the difference.

Example. Prove the inequality (x+y)(x+y+2cosx)+2 2sin 2x

Proof:

Consider the difference (x+y)(x+y+2cosx)+2- 2sin 2 x =(x+y)(x+y+2cosx)+2cos 2 x=(x+y)(x+y+2cosx) + cos 2 x +cos 2 x= (x+y) 2 +2(x+y)cosx+ cos 2 x +cos 2 x=((x+y)+cosx) 2 + cos 2 x 0.

Prove the inequality:

1.ab(a+b)+bc(b+c)+ac(a+c) 6abc

4.
>2x-20

6.(a+b)(b+c)(c+a) 8abc

Lesson 3. The method of mathematical induction.

When proving inequalities that include natural numbers, one often resorts to the method of mathematical induction. The method is as follows:

1) check the truth of the theorem for n=1;

2) we assume that the theorem is true for some n=k, and based on this assumption we prove the truth of the theorem for n=k+1;

3) based on the first two steps and the principle of mathematical induction, we conclude that the theorem is true for any n.

Example.

Prove the inequality

Proof:

1) for n=2 the inequality is true:

2) Let the inequality be true for n=k i.e.
(*)

Let us prove that the inequality is true for n=k+1, i.e.
. Let us multiply both parts of the inequality (*) by
we get 3) From item 1. and item 2 we conclude that the inequality is true for any n.

Assignments for classroom and home work

Prove the inequality:

6)
.

Lesson 4 Application of classical inequalities.

The essence of this method is as follows: using a series of transformations, the required inequality is derived using some classical inequalities.

Example.

Prove the inequality:

Proof:

As a reference inequality, we use
.

We bring this inequality to the following form:

, then

But =
, then

Prove the inequality:

1)(p+2)(q+2)(p+q)16pq(for proof we use the inequality
)

2)
(for documentation, the inequality is used)

3) (a+b)(b+c)(c+a) 8abc (the inequality is used for proof)

4)
(for doc-va the inequality is used).

Lesson 5 Graphic method.

The proof of the inequalities by the graphical method is as follows: if we prove the inequality f(x)>g(x)(f(x)

1) build graphs of functions y=f(x) and y=g(x);

2) if the graph of the function y=f(x) is located above (below) the graph of the function y=g(x), then the inequality being proved is true.

Example.

Prove the inequality:

cosx
,x0

Proof:

Let us construct in one coordinate system the graphs of the functions y=cosx and

It can be seen from the graph that at x0 the graph of the function y=cosx lies above the graph of the function y= .

Assignments to work in the classroom and at home.

Prove the inequality:

3)ln(1+x) 0

4)
.

Lesson 6

Example.

Prove the inequality:

Proof:

Assume the opposite, i.e. .

Let's square both parts of the inequality, we get , whence
and beyond

. But this contradicts the Cauchy inequality. So our assumption is wrong, i.e. the inequality is true

Assignments to work in the classroom and at home.

Prove the inequality:

Lesson 7 A technique for considering inequalities with respect to one of the variables.

The essence of the method is to consider the inequality and its solution with respect to one variable.

Example.

Prove the inequality:

Example.

Prove the inequality:

Proof:

Assignments to work in the classroom and at home.

Prove the inequality:

Lesson 9 Lesson - control of students' knowledge.

Work in this lesson can be organized in pairs or if there is a large class size in groups. At the end of the lesson, each student should be assessed. This is the transcript for this course. It is not recommended to carry out control work on this topic. the proof of inequalities, as already mentioned in the explanatory note, belongs to the field of art. At the beginning, students are asked to determine the method of proving the proposed inequalities themselves. If the students have difficulties, then the teacher tells them the rational method, warning the group that this, of course, will affect their assessment.

methods proofinequalities. This is methodproof ofinequalities by introducing helper functions...

Elective Course in Mathematics of Inequality Proof Methods

elective course

unfamiliar, different methodsproof ofinequalities, as well as the application inequalities inequalities via method method for proof ofinequalities, to solve problems...

Elective course in mathematics Inequalities Proof methods Explanatory note

elective course

unfamiliar, different methodsproof ofinequalities, as well as the application inequalities when solving problems of various ... Be able to: evaluate inequalities via method Sturm, apply considered method for proof ofinequalities, to solve problems...

Elective course in mathematics Inequalities Proof methods Explanatory note (1)

elective course

A rare Olympiad does without problems in which it is required to prove some inequality. Algebraic inequalities are proved using various methods, which are based on equivalent transformations and properties of numerical inequalities:

1) if a – b > 0, then a > b; if a - b

2) if a > b, then b a;

3) if a

4) if a

5) if a 0, then ac

6) if a bc; a / c > b / c;

7) if a 1

8) if 0

Let us recall some basic inequalities that are often used to prove other inequalities:

1) a 2 > 0;

2) aх 2 + bx + c > 0, with a > 0, b 2 - 4ac

3) x + 1 / x > 2, for x > 0, and x + 1 / x –2, for x

4) |a + b| |a| + |b|, |a – b| > |a| – |b|;

5) if a > b > 0, then 1 / a

6) if a > b > 0 and x > 0, then a x > b x, in particular, for natural n > 2

a 2 > b 2 and n √ a > n √ b;

7) if a > b > 0 and x

8) if x > 0, then sin x

Many problems of the Olympiad level, and these are not only inequalities, are effectively solved with the help of some special inequalities, with which school students are often not familiar. First of all, they should include:

inequality between the arithmetic mean and the geometric mean of positive numbers (Cauchy's inequality):

Bernoulli's inequality:

(1 + α) n ≥ 1 + nα, where α > -1, n is a natural number;

Cauchy-Bunyakovsky inequality:

(a 1 b 1 + a 2 b 2 + . . . + a n b n) 2 ≤ (a 1 2 + a 2 2 + . . . + a n 2)(b 1 2 + b 2 2 + . . . + b n 2 );

The most "popular" methods of proving inequalities include:

proof of inequalities based on definition;
square selection method;
method of successive assessments;
method of mathematical induction;
use of special and classical inequalities;
use of elements of mathematical analysis;
use of geometric considerations;
the idea of amplification, etc.

Problems with solutions

1. Prove the inequality:

a) a 2 + b 2 + c 2 + 3 > 2 (a + b + c);

b) a 2 + b 2 + 1 > ab + a + b;

c) x 5 + y 5 – x 4 y – x 4 y > 0 for x > 0, y > 0.

a) We have

a 2 + b 2 + c 2 + 1 + 1 + 1 - 2a - 2b - 2c = (a - 1) 2 + (b - 1) 2 + (c - 1) 2 > 0,

which is obvious.

b) The inequality to be proved, after multiplying both parts by 2, takes the form

2a 2 + 2b 2 + 2 > 2ab + 2a + 2b,

(a 2 - 2ab + b 2) + (a 2 - 2a + 1) + (b 2 - 2b + 1) > 0,

(a – b) 2 + (a – 1) 2 + (b – 1) 2 > 0,

which is obvious. Equality takes place only when a = b = 1.

c) We have

x 5 + y 5 - x 4 y - x 4 y = x 5 - x 4 y - (x 4 y - y 5) = x 4 (x - y) - y 4 (x - y) =

\u003d (x - y) (x 4 - y 4) \u003d (x - y) (x - y) (x + y) (x 2 + y 2) \u003d (x - y) 2 (x + y) (x 2 + y 2) > 0.

2. Prove the inequality:

a)	a	+	b		>	2 for a > 0, b > 0;
	b		a

b)	R	+	R	+	R		> 9, where a, b, c are the sides and P is the perimeter of the triangle;
	a		b		c

c) ab(a + b – 2c) + bc(b + c – 2a) + ac(a + c – 2b) > 0, where a > 0, b > 0, c > 0.

a) We have:

a	+	b	– 2 =	a 2 + b 2 - 2ab	=	(a – b) 2		> 0.
b		a		ab		ab

b ) The proof of this inequality follows elementarily from the following estimate:

b+c	+	a+c	+	a+b	=
a		b		c

= (

) + (

) > 6,

Equality is achieved for an equilateral triangle.

c) We have:

ab(a + b - 2c) + bc(b + c - 2a) + ac(a + c - 2b) =

= abc (

– 2 +

– 2 ) =

= abc ((

– 2) + (

– 2) ) > 0,

because the sum of two positive reciprocal numbers is greater than or equal to 2.

3. Prove that if a + b = 1, then the inequality a 8 + b 8 > 1 / 128 holds.

From the condition that a + b = 1, it follows that

a 2 + 2ab + b 2 = 1.

Let us add this equality with the obvious inequality

a 2 - 2ab + b 2 > 0.

We get:

2a 2 + 2b 2 > 1, or 4a 4 + 8a 2 b 2 + 4b 2 > 1.

4a 4 – 8a 2 b 2 + 4b 2 > 0,

we get:

8a 4 + 8b 4 > 1, whence 64a 8 + 128a 4 b 4 + 64b 4 > 1.

Adding this inequality to the obvious inequality

64a 8 – 128a 4 b 4 + 64b 4 > 0,

we get:

128a8 + 128b8 > 1 or a 8 + b 8 > 1/128 .

4. What's more e e π π or e 2 pi?

Consider the function f(x) = x – π log x . Insofar as f'(x) = 1 – π / x , and to the left of the dot X = π f'(x) 0 , and on the right - f'(x) > 0, then f(x) It has smallest value at the point X = π . Thus f(e) > f(π), i.e

e – π ln e = e – π > π – π ln π

e + π log π > 2π .

Hence we get that

e e+ π log π > e 2 pi,

her· e π log π > e 2 π ,

e e π π > e 2 pi.

5. Prove that

log(n + 1) >	lg 1 + lg 2 + . . . + log n	.
	n

Using the properties of logarithms, it is easy to reduce this inequality to an equivalent inequality:

(n + 1) n > n!,

where n! = 1 2 3 . . . · n (n-factorial). In addition, there is a system of obvious inequalities:

n + 1 > 1,

n + 1 > 2,

n + 1 > 3,

. . . . .

n + 1 > n

after term-by-term multiplication of which, we immediately obtain that (n + 1) n > n!.

6. Prove that 2013 2015 2015 2013

We have:

2013 2015 2015 2013 = 2013 2 2013 2013 2015 2013 =

2013 2 (2014 - 1) 2013 (2014 + 1) 2013

Obviously, one can also obtain a general statement: for any natural n, the inequality

(n – 1) n +1 (n + 1) n –1

7. Prove that for any natural number n the following inequality holds:

1	+	1	+	1	+ . . . +	1		2n - 1	.
1!		2!		3!		n!		n

Let us estimate the left side of the inequality:

1	+	1	+	1	+ . . . +	1	=
1!		2!		3!		n!

= 1 +	1	+	1	+	1	+ . . . +	1
	12		1 2 3		1 2 3 4		1 2 3 . . . n

1 +	1	+	1	+	1	+ . . . +	1	=
	12		2 3		3 4		(n – 1) n

= 1 + (1 –	1	) + (	1	–	1	) + (	1	–	1	) + . . . + (	1	–	1	) = 2 –	1	,
	2		2		3		3		4		n - 1		n		n

Q.E.D.

8. Let a 1 2 , a 2 2 , a 3 2 , . . . , and n 2 are the squares of n different natural numbers. Prove that

(1 –	1	) (1	–	1	) (1	–	1	) . . . (1	–	1	) >	1	.
	a 1 2			a 2 2			a 3 2			a n 2		2

Let the largest of these numbers be equal to m. Then

(1 –	1	) (1	–	1	) (1	–	1	) . . . (1	–	1	) >
	a 1 2			a 2 2			a 3 2			a n 2

> ( 1 –	1	) (1	–	1	) (1	–	1	) . . . (1	–	1	) ,
	2 2			3 2			4 2			m2

since factors less than 1 are added to the right side.We calculate the right side by factoring each parenthesis:

2 3 2 4 2 . . . (m - 1) 2 (m + 1)

m + 1

2 2 3 2 4 2 . . . m2

Opening the brackets on the left side, we get the sum

1 + (a 1 + . . . + a n) + (a 1 a 2 + . . . + a n –1 a n) + (a 1 a 2 a 3 + . . . + a n –2 a n –1 a n) + . . . + a 1 a 2 . . . a n .

The sum of the numbers in the second bracket does not exceed (a 1 + . . . + a n) 2 , the sum in the third bracket does not exceed (a 1 + . . . + a n) 3 , and so on. Hence, the whole product does not exceed

1 + 1 / 2 + 1 / 4 + 1 / 8 + . . . + 1/2n = 2 – 1/2n

Method 2.

Let us prove by the method of mathematical induction that for all natural numbers n the following inequality is true:

(1 + a1) . . . (1 + an)

For n = 1 we have: 1 + a 1 1 .

Let for n = k we have:(1 + a 1 ) . . . (1 + a k ) 1 + . . . + a k ).

Consider the case n = k +1:(1 + a 1 ) . . . (1 + a k )(1 + a k +1 )

(1 + 2(a 1 + . . . + a k ) )(1 + a k+1 ) ≤ 1 + 2(a 1 + . . . + a k ) + a k +1 (1 + 2 1 / 2) =

1 + 2(a 1 + . . . + a k + a k +1 ).

By virtue of the principle of mathematical induction, the inequality is proved.

10. Prove the Bernoulli inequality:

(1 + α) n ≥ 1 + nα,

where α > -1, n is a natural number.

Let's use the method of mathematical induction.

For n = 1 we get the true inequality:

1 + α ≥ 1 + α.

Let's assume that the following inequality holds:

(1 + α) n ≥ 1 + nα.

Let us show that then we have

(1 + α) n + 1 ≥ 1 + (n + 1)α.

Indeed, since α > –1 implies α + 1 > 0, then multiplying both sides of the inequality

(1 + α) n ≥ 1 + nα

on (a + 1), we get

(1 + α) n (1 + α) ≥ (1 + nα)(1 + α)

(1 + α) n + 1 ≥ 1 + (n + 1)α + nα 2

Since nα 2 ≥ 0, therefore,

(1 + α) n + 1 ≥ 1 + (n + 1)α + nα 2 ≥ 1 + (n + 1)α.

Thus, according to the principle of mathematical induction, Bernoulli's inequality is true.

Problems without solutions

1. Prove the inequality for positive values variables

a 2 b 2 + b 2 c 2 + a 2 c 2 ≥ abc(a + b + c).

2. Prove that for any a the inequality

3(1 + a 2 + a 4) ≥ (1 + a + a 2) 2 .

3. Prove that the polynomial x 12 – x 9 + x 4 – x+ 1 is positive for all values of x.

4. For 0 e prove the inequality

(e+ x) e– x > ( e– x) e+ x .

5. Let a, b, c be positive numbers. Prove that

a+b

b+c

a+c

≤

Your goal:know the methods of proving inequalities and be able to apply them.

Practical part

The concept of proof of inequality . Some inequalities turn into true numerical inequality for all allowed values variables or on some given set of variable values. For example, the inequalities a 2 ³0, ( a–b) 2 ³ 0 , a 2 +b 2 +c 2 " ³ 0 are true for any real values of the variables, and the inequality ³ 0 – for any real non-negative values a. Sometimes the problem of proving an inequality arises.

To prove an inequality means to show that a given inequality turns into a true numerical inequality for all admissible values of the variables or on a given set of values of these variables.

Methods for proving inequalities. notice, that general method there is no proof of inequality. However, some of them can be specified.

1. A method for estimating the sign of the difference between the left and right parts of an inequality. The difference between the left and right parts inequalities and it is established whether this difference is positive or negative for the considered values of the variables (for non-strict inequalities, it is necessary to establish whether this difference is non-negative or non-positive).

Example 1. For any real numbers a and b there is an inequality

a 2 +b 2³2 ab. (1)

Proof. Compose the difference between the left and right parts of the inequality:

a 2 +b 2 – 2ab = a 2 – 2ab+b 2 = (a-b) 2 .

Since the square of any real number is a non-negative number, then ( a-b) 2 ³ 0, which means that a 2 +b 2³2 ab for any real numbers a and b. Equality in (1) holds if and only if a = b.

Example 2. Prove that if a³ 0 and b³ 0, then ³ , i.e. arithmetic mean of non-negative real numbers a and b less than their geometric mean.

Proof. If a a³ 0 and b³ 0, then

³ 0. Hence, ³ .

2. deductive method proof of inequalities. The essence of this method is as follows: using a series of transformations, the required inequality is derived from some known (reference) inequalities. For example, the following inequalities can be used as reference: a 2 ³ 0 for any aÎ R ; (a-b) 2 ³ 0 for any a and bÎ R ; (a 2 + b 2) ³ 2 ab for any a, bÎ R ; ³ at a ³ 0, b ³ 0.

Example 3. Prove that for any real numbers a and b there is an inequality

a 2 + b 2 + with 2³ ab + bc + ac.

Proof. From the correct inequalities ( a-b) 2 ³ 0, ( b – c) 2 ³ 0 and ( c – a) 2 ³ 0 it follows that a 2 + b 2³2 ab, b 2 + c 2³2 bc, c 2 + a 2³2 ac. Adding all three inequalities term by term and dividing both parts of the new one by 2, we obtain the required inequality.

The original inequality can also be proved by the first method. Indeed, a 2 + b 2 + with 2 –ab-bc-ac= 0,5(2a 2 + 2b 2 + 2with 2 – 2ab- 2bc- 2ac) = = 0,5((a-b) 2 + (a-c) 2 + (b-c) 2)³ 0.

difference between a 2 + b 2 + with 2 and ab + bc + ac greater than or equal to zero, which means that a 2 + b 2 + with 2³ ab + bc + ac(equality is true if and only if a = b = c).

3. Method of estimates in the proof of inequalities.

Example 4. Prove the inequality

+ + + … + >

Proof. It is easy to see that the left side of the inequality contains 100 terms, each of which is not less than. In this case, we say that the left-hand side of the inequality can be estimated from below as follows:

+ + + … + > = 100 = .

4. Full induction method. The essence of the method is to consider all special cases that cover the condition of the problem as a whole.

Example 5. Prove that if x > ï atï , then x > y.

Proof. Two cases are possible:

a) at³ 0 ; then i atï = y, and by condition x >ï atï . Means, x > y;

b) at< 0; then i atï > y and by condition x >ï atï means x > y.

Practical part

Task 0. Take Blank sheet paper and on it write down the answers to all the oral exercises below. Then check your answers against the answers or brief instructions at the end of this learning element under the heading "Your assistant".

oral exercises

1. Compare the sum of the squares of two unequal numbers and with their double product.

2. Prove the inequality:

a) ;

b) ;

in) ;

3. It is known that . Prove that .

4. It is known that . Prove that .

Exercise 1. That more:

a) 2 + 11 or 9; d) + or;

b) or + ; e) - or;

c) + or 2; e) + 2 or + ?

Task 2. Prove that for any real x there is an inequality:

a) 3( x+ 1) + x– 4(2 + x) < 0; г) 4x 2 + 1 ³ 4 x;

b) ( x+ 2)(x+ 4) > (x+ 1)(x+ 5); e) ³ 2 x;

in) ( x– 2) 2 > x(x- 4); f) l + 2 x 4 > x 2 + 2x 3 .

Task 3. Prove that:

a) x 3+1³ x 2 + x, if x³ –1;

b) x 3 + 1 £ x 2 + x, if x£ -1 .

Task 4. Prove that if a ³ 0, b³ 0, with³ 0, d³ 0, then

(a 2 + b 2)(c 2 + d 2) ³ ( ac + bd) 2 .

Task 5. Prove the inequality by selecting a full square:

a) x 2 – 2xy + 9y 2 ³ 0;

b) x 2 +y 2 + 2³2( x+y);

at 10 o'clock x 2 + 10xy + 5y 2 + 1 > 0;

G) x 2 – xy + y 2³0 ;

e) x 2 +y 2 +z 2 + 3³ 2( x + y + z);

e)( x + l)( x- 2y + l) + y 2³0 .

Task 6. Prove that:

a) x 2 + 2y 2 + 2xy + 6y+ l0 > 0 ;

b) x 2 +y 2 – 2xy + 2x – 2at + 1 > 0;

in 3 x 2 +y 2 + 8x + 4y- 2xy + 22 ³ 0;

G) x 2 + 2xy+ 3y 2 + 2x + 6y + 3 > 0.

Task 7. Prove that if n³ k³ 1, then k(n–k+ 1) ³ n.

Task 8. Prove that if 4 a + 2b= 1, then a 2 + b 2³ .

Determine the values a and b, under which equality takes place.

Task 9. Prove the inequality:

a) X 3 + at 3³ X 2 at + hu 2 at x³ 0 and y ³ 0;

b) X 4 + at 4³ X 3 at + hu 3 for any x and at;

in) X 5 + at 5³ X 4 at + hu 4 at x³ 0 and y ³ 0;

G) x n + at n ³ x n-1 y + xy n-1 at x³ 0 and y ³ 0.