Proof of inequalities from geometric considerations. Proof and solution of inequalities

Your aim:know methods for proving inequalities and be able to apply them.

Practical part

Concept of inequality proof . Some inequalities become true numerical inequality in front of everyone acceptable values variables or on some given set of variable values. For example, 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 for a given set of values ​​of these variables.

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

1. A method for estimating the sign of the difference between the left and right sides of an inequality. The difference between the left and right sides of the inequality is compiled 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 inequality

a 2 +b 2³ 2 ab. (1)

Proof. Let's make up the difference between the left and right sides 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 a 2 +b 2³ 2 ab for any real numbers A And b. Equality in (1) occurs 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 no less than their geometric mean.

Proof. If A³ 0 and b³ 0, then

³ 0. So, ³ .

2. Deductive method proofs 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 inequality

A 2 + b 2 + With 2³ ab + bc + ac.

Proof. From the true inequalities ( a–b) 2 ³ 0, ( b – c) 2 ³ 0 and ( c – a) 2 ³ 0 it follows that A 2 + b 2³ 2 ab, b 2 + c 2³ 2 bc, c 2 + a 2³ 2 ac. Adding all three inequalities term by term and dividing both sides of the new one by 2, we obtain the required inequality.

The original inequality can be proved using 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. Estimation method for proving inequalities.

Example 4. Prove inequality

+ + + … + >

Proof. It is easy to see that the left side of the inequality contains 100 terms, each of which is no less. In this case, they say that the left 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 covering the condition of the problem as a whole.

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

Proof. There are two possible cases:

A) at³ 0 ; thenï atï = y, and by condition x >ï atï . Means, x > y;

b) at< 0; thenï 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 given below. Then check your answers against the answers or summary instructions at the end of this educational element in the “Your Assistant” section.

Oral exercises

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

2. Prove the inequality:

A) ;

b) ;

V) ;

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;

V) ( x– 2) 2 > x(x- 4); e) 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 isolating perfect square:

A) x 2 – 2xy + 9y 2 ³ 0;

b) x 2 + y 2 + 2³2( x+y);

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

G) x 2 – xy + y 2 ³ 0 ;

d) x 2 + y 2 + z 2 + 3³ 2( x + y + z);

e) ( x+ l)( x – 2y + l) + y 2 ³ 0 .

Task 6. Prove that:

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

b) x 2 + y 2 – 2xy + 2x – 2at + 1 > 0;

at 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 .

Define Values A And b, at which equality occurs.

Task 9. Prove the inequality:

A) X 3 + at 3³ X 2 at + xy 2 at x³ 0 and y ³ 0;

b) X 4 + at 4³ X 3 at + xy 3 for any x And at;

V) X 5 + at 5³ X 4 at + xy 4 at x³ 0 and y ³ 0;

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

Educational institution: Municipal Educational Institution Lyceum No. 1, Komsomolsk-on-Amur

Head: Budlyanskaya Natalya Leonidovna

If you would like to participate in great life, then fill your head with mathematics while you have the opportunity. She will then provide you with great assistance in all your work. (M.I. Kalinin)

Representation of the left side of the inequality as a sum of non-negative terms (the right side is 0) using identities.

Example 1. Prove that for any xϵR

Proof . 1 way.

Method 2.

for a quadratic function

which means its positivity for any real X.

Example 2. Prove that for any x and y

Proof.

Example 3. Prove that

Proof.

Example 4. Prove that for any a and b

Proof.

2. The opposite method

Here is a good example of using this method.

Prove that for a, b ϵ R.

Proof.

Let's pretend that.

But this clearly proves that our assumption is incorrect.

C.T.D.

Example 5.Prove that for any numbers A, B, C the following inequality is true:

Proof. Obviously, it is enough to establish this inequality for non-negative A, B And WITH, since we will have the following relationship:

, which is the rationale for the original inequality .

Now let there be such non-negative numbers A, B And WITH, for which the inequality holds

, which is impossible under any real A, B And WITH. The assumption made above is refuted, which is proved by the original inequality under study.

Using the properties of the quadratic trinomial

The method is based on the property of non-negativity of a quadratic trinomial if

And.

Example 6. Prove that

Proof.

Let be, a=2, 2>0

=>

Example 7. Prove that for any real x and y the inequality holds

Proof. Consider the left-hand side of the inequality as a quadratic trinomial with respect to X:

, a>0, D

D= => P(x)>0 And

true for any real values X And u.

Example 8. Prove that

for any real values ​​of x and y.

Proof. Let ,

This means that for any real at and inequality

is satisfied for any real X And u.

Method of introducing new variables or substitution method

Example 9. Prove that for any non-negative numbers x, y, z

Proof. Let us use the correct inequality for,

.

We obtain the inequality under study

Using function properties.

Example 10. Let's prove the inequality

for any a and b.

Proof. Let's consider 2 cases:

• If a=b then true

Moreover, equality is achieved only when a=b=0.

2)If

, on R =>

()* ()>0, which proves the inequality

Example 11. Let us prove that for any

Proof.

on R.

If, then the signs of the numbers coincide, which means the difference under study is positive =>

Application of the method of mathematical induction

This method is used to prove inequalities regarding natural numbers.

Example 12. Prove that for any nϵN

• Let's check the truth of the statement when

- (right)

2) Assume the truth of the statement when

(k>1)

3) Let's prove the truth of the statement when n=k+1.

Let's compare and:

We have:

Conclusion: the statement is true for anyone nϵN.

Using remarkable inequalities

• Theorem on averages (Cauchy's inequality)

• Cauchy–Bunyakovsky inequality

• Bernoulli's inequality

Let us consider each of the listed inequalities separately.

Application of the mean value theorem (Cauchy inequality)

The arithmetic mean of several non-negative numbers is greater than or equal to their geometric mean

, Where

The equal sign is achieved if and only if

Let's consider special cases of this theorem:

• Let n=2, then

• Let n=2, a>0, then

• Let n=3, then

Example 13. Prove that for all non-negative a,b,c the inequality holds

Proof.

Cauchy-Bunyakovsky inequality

The Cauchy-Bunyakovsky inequality states that for any; the ratio is valid

The proven inequality has a geometric interpretation. For n=2,3 it expresses the well-known fact that the scalar product of two vectors on the plane and in space does not exceed the product of their lengths. For n=2 the inequality has the form: . For n=3 we get

Example 14.

Proof. Let us write the inequality under study in the following form:

This is a obviously true inequality, since it is a special case of the Cauchy–Bunyakovsky inequality.

Example 15. Prove that for any a,b,c ϵ R the following inequality holds:

Proof. It is enough to write this inequality in the form

and refer to the Cauchy–Bunyakovsky inequality.

Bernoulli's inequality

Bernoulli's inequality states that if x>-1, then for all natural values ​​of n the following inequality holds:

The inequality can be used for expressions of the form

In addition, a very large group of inequalities can be easily proven using Bernoulli's theorem.

Example 16.

Proof. Putting x=0.5 and applying Bernoulli's theorem to express

We obtain the required inequality.

Example 17. Prove that for any n ϵ N

Proof.

by Bernoulli's theorem, as required.

David Gilbert was asked about one of his former students. “Oh, so-and-so?” Hilbert remembered. “He became a poet. He had too little imagination for mathematics.

: Expand your knowledge in the field of proving inequalities. Learn about Cauchy's inequality. Learn to apply the learned methods to prove inequalities.

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State budgetary educational institution

average comprehensive school №655

Primorsky district of St. Petersburg

“Proof of inequalities. Cauchy's inequality"

2014

Li Nina Yurievna

8th grade

Abstract………………………………………………………………………………….3

Introduction………………………………………………………………………………….. 4

Historical background………………………………………………………………………………………..4

Cauchy's inequality………………………………………………………………………………………5

Proof of inequalities………………………………………………………………..7

Conclusions of the study…………………………………………………………………………………..10

References………………………………………………………………………………………11

Li Nina

St. Petersburg, GBOU secondary school No. 655, 8th grade

“Proof of inequalities. Cauchy's inequality".

supervisor: Yulia Vladimirovna Moroz, mathematics teacher

Target scientific work: Expand your knowledge in the field of proving inequalities. Learn about Cauchy's inequality. Learn to apply the learned methods to prove inequalities.

INTRODUCTION

“...the main results of mathematics are more often expressed not by equalities, but by inequalities.”

E. Beckenbach

We deal with solving inequalities throughout the school course. Inequalities can be solved graphically and analytically. To solve any inequality exists specific algorithm action, therefore this task is rather mechanical action, which does not require creativity.

On the contrary, proving inequalities requires an informal, variable approach. Therefore, the proof of inequalities is the most interesting.

However, in school course In mathematics, very little attention is paid to the proof of inequalities. Proving inequalities comes down to one technique - estimating the difference between the parts of the inequality.Meanwhile, at mathematical Olympiads there are often problems on proving inequalities using other methods and techniques (use of support inequalities, estimation method).At olympiads for schoolchildren in mathematics, inequalities are also often proposed, the proof of which better reveals the abilities and capabilities of students, their degree intellectual development. In addition, many tasks increased complexity(from various branches of mathematics) are effectively solved using inequalities.

The relevance of the topic “Proof of inequalities” is undeniable, since inequalities play a fundamental role in most branches of modern mathematics; neither physics, nor astronomy, nor chemistry can do without them. Probability theory, math statistics, financial mathematics, economics - all these interconnected and generalizing sciences, both in the formulation of their basic laws, and in the methods of obtaining them, and in applications, constantly use inequalities.

Proofs of inequalities help develop the skill of understanding and applying techniques for proving inequalities; the ability to apply them when performing various tasks; ability to analyze, generalize and draw conclusions; express thoughts logically; is creative in his work.

The purpose of this work is to expand knowledge in the field of methods and techniques for proving inequalities.

To achieve this research goal, we set ourselves the following tasks:

• collecting information from various sources about techniques and methods for proving inequalities;
• get acquainted with Cauchy's inequality;
• Learn to apply supporting inequalities to prove more complex inequalities.

HISTORICAL REFERENCE

The concepts of “more” and “less”, along with the concept of “equality,” arose in connection with counting objects and the need to compare different quantities. The concepts of inequality were used by the ancient Greeks. Archimedes (3rd century BC), while calculating the circumference, established that “the perimeter of any circle is equal to three times the diameter with an excess that is less than a seventh of the diameter, but more than ten seventy.” In other words, Archimedes indicated the boundaries of the number π.

In 1557, when Robert Record first introduced the equal sign, he motivated his innovation as follows: no two objects can be more equal to each other than two parallel to the segment. Based on Record's equal sign, another English scientist Harriot introduced inequality signs that are still used today, justifying the innovation as follows: if two quantities are not equal, then the segments appearing in the equal sign are no longer parallel, but intersect. The intersection can take place on the right (>) or on the left (

Despite the fact that inequality signs were proposed 74 years after the equal sign proposed by Record, they came into use much earlier than the latter. One of the reasons for this phenomenon is rooted in the fact that printing houses at that time used the Latin letter they already had for inequality signs V, whereas they did not have a typesetting equals sign (=), and it was not easy to make one then.

The signs ≤ and ≥ were introduced by the French mathematician P. Bouguer.

CAUCHY'S INEQUALITY

The ideas used to prove inequalities are almost as varied as the inequalities themselves. In specific situations, general methods often lead to ugly solutions. But only a few succeed in combining several “basic” inequalities in a non-obvious way. And, besides, nothing prevents us in each specific case from looking for a more convenient, better solution than the one received general method. For this reason, proof of inequalities is often relegated to the realm of art. And as in all art there are technique, the set of which is very wide and it is very difficult to master them all.

One of these “basic” inequalities is the Cauchy inequality, which indicates the relationship between two average values ​​– the arithmetic mean and the geometric mean. The arithmetic mean is studied in the fifth grade school course and looks like thisThe geometric mean first appears in the eighth grade geometry course -. IN right triangle Three segments have this property: two legs and a perpendicular dropped from the vertex right angle to the hypotenuse.

There is a surprising relationship between these two quantities that scientists have studied. O. Cauchy, a French mathematician, came to the conclusion that the arithmetic mean of n non-negative numbers is always not less than the geometric mean of these numbers.

Along with Cauchy's inequality, it is useful to know its corollaries:

Equality is achieved when a = b.

The inequalities are true if the conditions a > 0, b > 0 are met.

The algebraic proof of this inequality is quite simple:

(a – c)² ≥ 0;

Let’s apply the “squared difference” formula:

a² - 2av + b² ≥0;

Let us add to both sides of the inequality 4av:

a² + 2av + b² ≥4av;

Let’s apply the “square of the sum” formula:

(a + b)² ≥4av;

Let us divide both sides of the inequality by 4 :

Since a and b are positive by condition, then we extract the square root of both sides of the inequality:

We got the desired expression.

Consider the geometric proof:

Given: ABCD – rectangular, AD = a, AB = b, AK – bisector of angle BAD.

Prove:

Proof:

1. AK is a bisector, therefore, VAL = LAD. LAD and BLA – internal crosswise angles with parallel BC and AD and secant AL, that is BLA = LAD.
2. B = 90°, therefore BAL = LAD = 45°, but BLA = LAD, so ∆ ABL – isosceles, BL = AB = b.
3. ∆AKD – isosceles, since KD┴AD, DAL = 45°, which means AD = KD = a.

It's obvious that , equality is achieved when

a = b , that is, ABCD is a square.

let's replace in the inequality a² by m, b² by n, we get

Or ,

that is, the geometric mean is not greater than the arithmetic mean.

PROOF OF INEQUALITIES

Synthesis method.

This is a method based on obtaining (synthesizing) an inequality (which needs to be justified) from support (basic) inequalities and methods for establishing them.

Let's solve the problem using the synthesis method

Problem 1. Prove that for any non-negative a, b, c inequality is true

Solution. Let us write down three inequalities establishing the relationship between the arithmetic mean and the geometric mean of two non-negative numbers

Let us multiply the resulting inequalities term by term, since their left and right sides are non-negative

Problem 2. Let's apply Cauchy's inequality to prove this inequality:

Method of using identities.

The essence of the method is that this inequality is reduced to an obvious identity through equivalent transformations.

Let's consider solving the problem using this method.

Task. Prove that for any real numbers a and b inequality is true.

Solution. Let us select a complete square on the left side of the inequality

For any valid a and b this expression is non-negative, which means this inequality is also satisfiable, that is.

CONCLUSION

This research work was aimed at solving the following problems:

• collecting information and studying various methods and techniques for proving inequalities;
• acquaintance with the remarkable Cauchy inequality, its proof in an algebraic and geometric way;
• application of acquired knowledge to prove inequalities;
• familiarization with the method of synthesis and use of identities in solving assigned problems.

In the process of solving problems, we achieved our goal research work– finding optimal effective method proofs of inequalities.

BIBLIOGRAPHY

1. Algebra. 8th grade: textbook. for general education students institution/ Yu.N. Makarychev, N.G. Mindyuk, K.I. Neshkov, I.E. Feoktistov. - 13th ed. - M.: Mnemozina, 2013. - 384 p.

1. Algebra. 8th grade. Didactic materials. Guidelines/ I.E.Feoktistov.-3rd ed., ster.-M.: Mnemosyne, 2013.-173 p.

1. Mordkovich A.G. Algebra. 8th grade. At 2 p.m. Part 1. Textbook for students educational institutions/ A.G. Mordkovich. – 10th ed., erased. – M.: Mnemosyne, 2008. – 215 pp., C 185-200.

1. Berkolaiko S.T. The use of Cauchy's inequality in solving problems. - M.: Kvant, 1975. - No. 4.

Rarely does an Olympiad go by without problems that require proving some inequality. Algebraic inequalities are proved using various methods that 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 supporting inequalities that are often used to prove other inequalities:

1) a 2 > 0;

2) aх 2 + bx + c > 0, for 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 at the Olympiad level, and not only inequalities, are effectively solved using some special inequalities that school students are often not familiar with. These, first of all, include:

• inequality between arithmetic mean and geometric mean positive numbers(Cauchy inequality):

• Bernoulli's inequality:

(1 + α) n ≥ 1 + nα, where α > -1, n – 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 for proving inequalities include:

• proof of inequalities based on definition;
• square selection method;
• method of sequential assessments;
• method mathematical induction;
• use of special and classical inequalities;
• use of elements of mathematical analysis;
• use of geometric considerations;
• idea of ​​strengthening, 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 being proved after multiplying both sides by 2 takes the form

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

or

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

or

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

which is obvious. Equality occurs 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) =

= (x – y) (x 4 – y 4) = (x – y) (x – y) (x + y) (x 2 + y 2) = (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 simply from the following estimate:

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

=

b

+

c

+

a

+

c

+

a

+

b

=

a

a

b

b

c

c

= (

b

+

a

) + (

c

+

a

) + (

c

+

b

) > 6,

a

b

a

c

b

c

Equality is achieved for equilateral triangle.

c) We have:

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

= abc (

a

+

b

– 2 +

b

+

c

– 2 +

a

+

c

– 2 ) =

c

c

a

a

b

b

= abc ((

a

+

b

– 2) + (

a

+

c

– 2) + (

b

+

c

– 2) ) > 0,

b

a

c

a

c

b

since 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's add this equality to 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:

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

4. What's more e e · π π or e 2π?

Consider the function f(x) = x – π ln x . Because the f'(x) = 1 – π/x , and to the left of the point X = π f’(x) 0 , and on the right - f’(x) > 0, That f(x) It has smallest value at the point X = π . Thus f(е) > f(π), that is

e – π ln e = e – π > π – π ln π

or

e + π ln π > 2π .

From here we get that

e e + π ln π > e 2π,

her· e π ln π > e 2 π ,

e e · π π > e 2π.

5. Prove that

log(n+1) >	log 1 + log 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 multiplying them term by term, we directly 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, we can also obtain a general statement: for any natural number 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's 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 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			m 2

since in right side added multipliers less than 1.Let's calculate the right-hand side by factoring each bracket:

=	2 · 3 2 · 4 2 · . . . · (m – 1) 2 · (m + 1)	=	m+1	=	1	+	1	>	1	.
	2 2 · 3 2 · 4 2 · . . . m 2 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. This means that the entire product does not exceed 1 + 1/2 + 1/4 + 1/8 + . . . + 1 / 2 n = 2 – 1 / 2 n Method 2. Using the method of mathematical induction, we prove that for all natural n the following inequality is true: (1 + a 1) . . . (1 + a n) For n = 1 we have: 1 + a 1 1 . Let the following hold for n = k:(1 + a 1) . . . (1 + ak) 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 proven. 10. Prove Bernoulli’s 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 us assume that the following inequality holds: (1 + α) n ≥ 1 + nα. Let us show that then it takes place and (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) or (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 holds 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 ≤ 1 + 1 +

MOU Grishino-Slobodskaya secondary school

Module program

"Methods for proving inequalities"

as part of an 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 objects are equal to themselves. This statement is highly inaccurate for two reasons. Firstly, mathematics, despite its inherent scientific language, is not science; rather, it can be called art. Secondly The main results of mathematics are more often expressed by inequalities rather than equalities.”

Inequalities are used in practical work math all the time. They are used to obtain a number of interesting and important extremal properties of “symmetrical” figures: square, 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 issues of natural science and technology.

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

Teaching proofs plays a big role in the development of deductive-mathematical thinking and general thinking abilities of students. How to teach schoolchildren to independently prove 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 varied as the inequalities themselves. In specific situations, general methods often lead to ugly solutions. But only a few schoolchildren succeed in combining several “basic” inequalities in a non-obvious way. And, besides, nothing prevents the student in each specific case from looking for a better solution than the one obtained by the general method. For this reason, proof of inequalities is often relegated to the realm of art. And like any art, there are technical techniques here, the range of which is very wide and it is very difficult to master them all, but every teacher should strive to expand the mathematical tools available to him.

This module is recommended for students in grades 10-11. Not all possible methods for proving inequalities are discussed here (the method of replacing a variable, proving inequalities using a derivative, the method of research and generalization, and the ordering technique are not covered). You can offer to consider other methods at the second stage (for example, in 11th grade), if this module of the course arouses interest among students, and also based 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 an unknown under the modulus sign.
		Systems of inequalities with two variables.

"Behind the Pages of a Mathematics Textbook"

"Methods for proving inequalities"

		Introduction.
		Proof of inequalities based on 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.
		The idea of ​​strengthening.
		Lesson - control.

Lesson 1. Introduction.

Proving inequalities is a fascinating and challenging topic in elementary mathematics. Absence common 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 cover following methods proofs of inequalities:

Repetition:

Prove some properties.

Classic inequalities:

1)
(Cauchy inequality)

2)

3)

4)

Historical reference:

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

number
called geometric mean numbers a and b. Thus, an 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 sophistry- an amazing statement, the proof of which hides imperceptible and sometimes quite subtle errors.

Sophisms are false results obtained through reasoning that only seem correct, but necessarily contain one or another error.

Example:

Four is over twelve

Lesson 2. Proof of inequalities based on 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 isolates a square, a cube of a sum or difference, or an incomplete square of a sum or difference. This helps determine the sign of the difference.

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

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 inequality:

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

3.

4.
>2x-20

5.

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

7.

Lesson 3. Method of mathematical induction.

When proving inequalities that include integers often resort 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 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's multiply both sides of the inequality (*) by
we obtain 3) From item 1. and item 2 we conclude that the inequality is true for any n.

Assignments for work in the classroom and at home

Prove inequality:

1)

2)

3)

4)

5)

6)
.

Lesson4. 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 inequality:

Proof:

As a reference inequality we use
.

Let us reduce this inequality to next view:

, Then

But =
, Then

Prove inequality:

1)(p+2)(q+2)(p+q)16pq (for proof the inequality is used
)

2)
(for docs inequality is used)

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

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

Lesson 5. Graphic method.

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

1) build graphs of the 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 inequality:

cosx
,x0

Proof:

Let us construct graphs of the functions y=cosx and

It is clear from the graph that at x0 the graph of the function y=cosx lies above the graph of the function y=.

Assignments for work in the classroom and at home.

Prove inequality:

1)

3)ln(1+x) 0

4)
.

5)

Lesson 6. The opposite method

The essence of this method is as follows: let you need to prove the truth of the inequality F(x,y,z) S(x,y,z)(1). They assume the opposite, i.e. that for at least one set of variables the inequality F(x,y,z) S(x,y,z) (2) is true. 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 that inequality (2) is true is incorrect, and therefore inequality (1) is true.

Example.

Prove inequality:

Proof:

Let's assume the opposite, i.e.

Let us square both sides of the inequality and obtain , from which
and onwards

. But this contradicts Cauchy's inequality. This means our assumption is incorrect, that is, the inequality is true

Assignments for work in the classroom and at home.

Prove 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 inequality:

Example.

Prove inequality:

Proof:

Assignments for work in the classroom and at home.

Prove inequality:

1)

2)

3)

Lesson9. 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 must be assessed. This is the credit form for this course. It is not recommended to carry out tests on this topic because 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 for proving the proposed inequalities. If students have any difficulties, the teacher tells them the rational method, warning the group that this will, of course, affect their grade.

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