Examples of proof of complex inequalities. Program of the module “Methods of proving inequalities” - Program

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1 FSBEI HE "PETROZAVODSK STATE UNIVERSITY" FACULTY OF MATHEMATICS AND INFORMATION TECHNOLOGY Department of Geometry and Topology Elizaveta Sergeevna Khaltsenen Final qualifying work for a bachelor's degree Methods of proving inequalities Direction: "0.03.0" "Mathematics" Scientific advisor Professor, Doctor f.-m. Sciences, Platonov S.S. (signature of the manager) Petrozavodsk

2 Contents Introduction...3. Jensen's inequality Commutation inequality Karamata's inequality Solving problems to prove inequalities...3 References

3 Introduction A method is a set of sequential actions aimed at solving a specific type of problem. The methods for proving inequalities in this work are aimed at finding a non-standard solution to inequalities that have a certain form. Using such methods, the solution is reduced significantly. The result is the same, but the amount of work is less. The goal of the final work was to study three types of inequalities with the help of which many others can be easily proven. These are Jensen's inequality, commutation inequality, Karamata's inequality. All these inequalities are mathematically beautiful; with the help of these inequalities you can solve school inequalities. This topic is relevant. In my opinion, it could be useful for schoolchildren, including for improving their knowledge in the field of mathematics. Since the methods are not standard, I think that students with a mathematical bent would find them useful and fun. The task is to search and solve thematic inequalities from the proposed literature. The work consists of four paragraphs. The section describes Jensen's inequality, provides its proof and auxiliary definitions. Section 2 contains commutation inequality, its special cases and general commutation inequality. In paragraph 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, commutation inequality and Karamata's inequality

4 . Jensen's inequality Definition. A subset of a plane is called convex if any two points of a 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 a given interval. The set of points (x,y) for which y f(x) is called a subgraph. Definition 3. Consider a function on a certain interval. A function is called convex if on this interval its epigraph is a convex set. A function is called concave if its subgraph is a convex set. Criterion for convexity (concavity) of a function. In order for the 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 for convexity (concavity) of a function. In order for the function y = f(x), 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 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: 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 C 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 C AB also belongs to U, since C AB belongs to the 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 by:) At = 2 the center of mass C A A 2 of the system of points A, A 2 is already determined. We will assume that point C A A 2 has mass m + m 2 2) Let us assume that for the system of points A, A 2 A the center of mass c A, A 2 A has already been determined. 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 assume 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 will assume that the mass of point C A,A 2 A is equal to m B + m = m + m m. From the definition of the center of mass it follows that if all points A, A 2 A belong to the 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 point A i, i =,. If C is the center of mass of the system of points A, A 2 A, then the radius vector r C of point C can be calculated using 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 proven 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 equal to m B = m + m m. By definition, the center of mass C of the system of points A, A 2 A coincides with the center of mass of the 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 a certain interval, x, x 2, x are numbers from this interval; m, m 2, m are positive numbers satisfying the condition m + m m =. Then Jensen’s 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 are 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) . On its graph, consider points A, A 2, A 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 a function is a convex set. Consequently, the center of mass of point A, A 2, A belongs to the epigraph. Let's 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 h.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) Using Jensen’s inequality, we can prove Cauchy’s inequality: For any positive numbers a, a 2, a, the following inequality holds: (a + a a) a a 2 a Let us take the logarithm of inequality (3), and 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 special 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 частный случай неравенства x2-7 -

8 Jensen for a concave function f(x) = l(x). Since inequality (5) is true, then the equivalent inequality (3) is also true. 2. Commutation inequality Definition. A one-to-one correspondence between a set of numbers (,2,3,) and itself is called a permutation of elements. Let us denote the permutation by the letter σ so that σ(), σ(2), σ(3) σ() are the 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 it follows 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 a second largest number from the second set, and so on. Sets a, a 2, a and b, b 2, b are called inversely ordered if, for any numbers i and j, from the fact that a i a j it follows that b i b j. It follows from this that the largest number from the set a, a 2, a corresponds to the 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 set b, b 2 , b and so on. Example.) Let two sets be given such that a a 2 a and b b 2 b, then according to the definitions we have given these sets are equally ordered. 2) Let two sets be given such that a a 2 a and b b 2 b, in which case the sets of numbers a, a 2, a and b, b 2, b will be inversely ordered. Everywhere below, a, a 2, a and b, b 2 , b - positive real numbers “Theorem. (Commutation inequality) Let there be two sets of numbers a, a 2, a and b, b 2, b. Let us consider many of their various 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 reverse ordered. 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 inversely ordered and the value of a a + b b + c c = 3 will be the smallest. Proof of the theorem. Consider two sets of numbers: the first a, a 2, a and the second b, b 2, b. Let's assume that these sets are not ordered in the same way. 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 (this 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 наибольшее значение. Наименьшее значение получается аналогично, только мы делаем сортировку до тех пор, пока наборы не будут обратно упорядоченными. В итоге мы придем к наименьшему значению. «Теорема 2. Рассмотрим два положительных набора a, a 2, a 3 a иb, b 2, b 3 b и все его возможные перестановки. Тогда значение произведения (a i + b σ(i)) будет наибольшим, когда наборы a, a 2, a 3 a иb, b 2, b 3 b одинаково упорядочены, и наименьшим, когда они обратно упорядочены

10 Theorem 3. Consider two sets a, a 2, a 3 a and b, b 2, b 3 b elements of this set are positive. Then the value () a i + b σ(i) will be greatest when the sets a, a 2, a 3 a and b, b 2, b 3 b are equally ordered and least when they are reverse ordered.” Theorems 2 and 3 are special cases of a more general theorem, which is discussed below. General commutation inequality “Theorem 4 (General commutation 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 equally ordered and smallest when the sets are inversely ordered. 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 greatest when the numbers are reverse ordered 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's denote 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, therefore inequality (2) takes the form f(x + k + m) + f(x + k ) f(x + k) + f(x + m) (3) To prove the inequality, we will use the figure. The figure shows a graph of the convex function y = f(x) and the points A(x, f(x)), C(x) are marked on the graph + k, f(x + k)), D(x + m, f(x + m)), B (x + k + m, f(x + k + m)). and on From the convexity of the function f it follows that the chord CD lies below the chord AB. Let K be the midpoint of the chord CD, MZ the midpoint of the chord AB. Note that the abscissas of 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, points K and M lie on the same vertical line, which means 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).” For proof, see . 4. Solving problems to prove inequalities. This section examines various problems for proving inequalities, the solution of which can be solved using Jensen's inequality, commutation 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 it follows from Jensen’s inequality 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 following 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 etc. Task 3. Prove that for any a, a 2, a > 0 the following 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) Let’s make the replacement b i = a i, then the inequality will take the form: (b + b b) (b b 2 b) This inequality is true, because This is Cauchy's inequality. Task 4. Prove that for any a, a 2, a > 0 the following inequality holds: 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 Let us denote a a 2 = x, a 2 a 3 = y, a 3 a = z, xyz = x+y+z 3 3 xyz= -true Let us denote x = a a 2,x 2 = a 2 a 3,x = a a, Then x x 2 x =. Then the inequality will take the form: x + x x This inequality follows from Cauchy’s inequality: h.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 the fact that (a + a a) a + a a 2 a +a 2 + +a a + a 2+ +a Consider the Jensen function f(x) = x, we 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 us apply the commutation inequality. Let the first set have the form Second x 3, y 3, z 3, x 2, y 2, z 2 The expression on the left side will be the largest, because the value of the expression on the left side x 5 + y 5 + z 5 is made up of identically ordered sets of numbers. It follows from this that the value obtained - 6 -

17 for all other permutations is not greater than the value obtained with the “most correct” arrangement of variables. Task 8. Prove that for any x, y, z > 0 the following 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 commutation inequality, the sum a b + a 2 b 2 + a 3 b 3 is the smallest among the sums. In particular, what 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 following inequality holds: (+ 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.) Let’s use general commutation inequality for a 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 equally 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 equally ordered, but the sets (a, b, c) and (b, c, a) are not equally ordered, then inequality (0.) follows from the commutation 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 Task 3, 4. Prove that for any a, a 2, a > 0 the following inequality is true: 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.) with a + a 2 + a =. We will prove inequality (3.). It can be transformed to the form 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 the 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 us transform inequality (5.) to an equivalent form: () 2 pq + ()(p + q) + ()pq + () 2 pq + ()(p + q) + ()pq 0 Bringing similar ones we get: ( )[()pq + (p + q) pq] + 0 () () 0 () 0 () 0 always, since -natural We prove that Note that 0 (5.2) p + q pq = p(q ) (q) = (p)(q) Since p, q, then p 0, q 0, therefore, inequality (5.2) is true. Task 6. For any positive numbers x, y, z the following inequality holds: 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 the inequality 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 (since 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 from Karamata’s inequality it follows 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 following inequality holds: a 2 + b 2 + c ab + ac + bc a 2 + 2bc + b 2 + 2ac + c 2 + 2ab This is equivalent to the fact that Let a b c. a 2 + b 2 + c 2 + ab + ac + bc + ab + ac + bc Consider two sets of numbers a 2 + 2bc + b 2 + 2ac + c 2 + 2ab (a 2 + b 2 + c 2, ab + 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-true 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 Karamata's inequality for the 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 is true 2 d 2 + c 2 d 2 Let a b c d Make the replacement: x = l a, y = l b, z = l c, w = l d and write the original inequality in the form: e 4x + e 4y + e 4z + e 4w + e x+ 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) Let’s order these 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 Since the sets are ordered in this way, that 2x + 2w 2y + 2z I.e. x + w y + z, then case 3) is possible only when 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 correct for x + w = y + z 5) 4x + 4y + 4z + 2x + 2y + 2z + 2w 4x + 2y + 2z + 2x + 2w + 2y + 2z + 2y + 2w z w correct 6) 4x + 4y + 4z + 2x + 2y + 2z + 2w + 4w 4x + 2y + 2z + 2x + 2w + 2y + 2z + 2y + 2w + 2z + 2w 0 = 0 Thus , the set (8.) majorizes the set of numbers (8.2). Using Karamata's 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.) Let’s make the replacement : And we write 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 Consider two sets: (3x, 3x, 3x, 3y, 3y, 3y, 3z, 3z, 3z, x + 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) Let us order these sets: (3x, 3x, 3x, 3y, 3y, 3y, x + y + z, x + y + z, x + y + z, 3z, 3z, 3z,) and (9.2) Let's 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) Need prove that the set of numbers (9.2) majorizes the set of numbers (9.3)) 3x 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 obtain that 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 When x = y we obtain that y z 7) 9x + 9y + x + y + z 4x + 2y + 2z + 4x + 4y + 2x + 2y + z , we get y z 8) 9x + 9y + 2x + 2y + 2z 4x + 2y + 2z + 4x + 4y + 2x + 4y + 2z, x + y + 3z 0 9) 9x + 9y + 3x + 3y + 3z 4x + 2y + 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 Thus, the set of numbers (9.2) majorizes the set of numbers (9.3) and by Karamata’s 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 Continuing Mathematical Education. 2005. 6 p. 2) I.Kh. Sivashinsky. Inequalities in problems M.: Nauka, p. 3) A.I. Khrabrov. Around the Mongolian inequality, Mat. enlightenment, gray 3, 7, MTsNMO, M., 2003, p. 4) L.V. Radzivilovsky, Generalization of the commutation inequality and the Mongolian inequality, Mat. enlightenment, gray 3, 0, Publishing house MTsNMO, M., 2006, p. 5) V.A.ch Kretschmar. Algebra problem book. Fifth edition M., science, p. 6) D. Nomirovsky Karamata inequality /D. Nomirovsky // (Kvant)-S

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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 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 – 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 of 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,

(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 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

= (

) + (

) > 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,

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) has the smallest value at the point X = π . Thus f(е) > f(π), that is

e – π ln e = e – π > π – π ln π

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

2 · 3 2 · 4 2 · . . . · (m – 1) 2 · (m + 1)

m+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)

(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 inequality for positive values of 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

≤

Here we introduce an algebraic interpretation of the relations “greater than” and “less than”, which is illustrated on the coordinate line, and its application to prove inequalities is considered. Here, students are shown how to algebraically prove the properties of inequalities studied in the chapter; previously they were justified geometrically.

Proving inequalities is quite difficult material for students, so depending on the level of preparation of the class, it should be considered with varying degrees of completeness.

A special feature of problems to prove inequalities is the possibility of solving them in various ways. Therefore, it is advisable to show several options for proving inequalities in order to expand students’ ability to solve problems.

There are two main ways to prove inequalities:

1) based on compiling the difference between the right and left sides of the inequality and subsequent comparison of this difference with zero;
2) transition from one inequality to another, equivalent to it, based on the properties of the inequalities.

Both paths are equal. But you need to ensure the accuracy of your records in both cases. It is necessary to explain to students that if the first solution path is chosen, then after compiling the difference, transformations of this expression are performed, which can be written as a chain with the “=” sign. The resulting expression is compared with zero and, based on this, a conclusion is made about the original inequality. If the second path is chosen, then a sequence of equivalent inequalities is written down (as when solving an inequality), and a conclusion is made about the last one - whether it is true or not. Here’s what, for example, the design of a solution to an exercise might look like in both cases.

Prove the inequality a2 + b2 + 2 2 (a + b) .

Let's make up the difference Move 2 (a + b) to the left

left and right sides of the inequality:

inequalities: a2 + b2 + 2 - 2 (a + b) 0;

a2 + b2 + 2 - 2 (a + b) =a2 + b2 + 2 2a - 2b 0;

A2 + b2 + 2 - 2a - 2b == (a2 - 2a + 1) +

= (a2 - 2a + 1) ++ (b2 - 2b + 1) 0;

+ (b2 - 2b + 1) =(a - 1)2 + (b - 1)2 0 -

= (a - 1)2 + (b - 1)2 .correct, therefore,

(a - 1)2 + (b - 1)2 0, the original inequality is also true:

therefore, a2 + b2 + 2 2 (a + b) .

inequality is proven:

a2 + b2 + 2 2 (a + b) .

Note that the “Mandatory Minimum Content” does not include proof of inequalities. Therefore, the corresponding skills can be considered as a result of mastering the topic, but not as the final result of learning. In this regard, tasks on proving inequalities are not included in the exam, and they should not be included in other final tests.

Prove that for positive numbers p and q: p4 + q4 p3q + pq3.

Solution. Let's transform the difference

p4 + q4 - (p3q + pq3).

p4 + q4 - p3q - pq3 = p3 (p - q) + a3 (q - p) =

= (p - q) (p3 - q3) = (p - q)2 (p2 + pq + q2).

Since p 0, q 0, then pq 0, p2 + pq + q2 0, in addition, (p - q)2 0.

The product of a non-negative and a positive number is non-negative, i.e. the difference in question is greater than or equal to zero. Therefore, for p 0, q 0 p4 + q4 p3q + pq3.

Prove in different ways that if a b 0, then a2 + a b2 + b.

Solution. When proving the inequality in the second way,

Let's use the inequality proven in the example: if a and b are positive numbers and a b, then a2 b2. And then, adding the inequality a2 b2 and the inequality a b term by term, we get what we need.

3. In which case will a tourist cover the same distance faster: if he walks along a horizontal road at a constant speed, or if half of the way he walks uphill at a speed 1 km/h less than his speed on a horizontal road , and half the distance - from the mountain at a speed 1 km/h greater than on a horizontal road?

Solution. Let us denote the speed of a tourist on a horizontal road by the letter x, and take the distance as 1. The problem comes down to comparing expressions

1 and ___1___ + ___1___

x 2(x-1) 2(x+1).

Let's make their difference and transform it, we get that

which means that under such conditions, the travel time on a horizontal road is less than on a road with ascent and descent.

Methods for proving inequalities.

Solving inequalities. Equivalent inequalities.

Interval method. Systems of inequalities.

Proof of inequalities. There are several proof methodsinequalities. We will look at them using the example of inequality:

Where a – positive number.

1). Using a known or previously proven inequality.

It is known that ( a– 1 )² 0 .

2). Estimating the sign of the difference between the parts of the inequality .

Consider the difference between the left and right sides:

Moreover, equality occurs only whena = 1 .

3). Proof by contradiction.

Let's assume the opposite:

a, we get: a 2 + 1 < 2 a, i.e.

a 2 + 1 – 2 a < 0 , or ( a– 1 ) 2 < 0, which is not true. (Why?) .

The resulting contradiction proves the validity of

The inequality in question.

4). Method of indefinite inequality.

The inequality is called uncertain if he has a sign\/ or /\ ,

those. when we don't know which waythis sign should be turned

to get a fair inequality.

The same rules apply here asand with ordinary inequalities.

Consider the undefined inequality:

Multiplying both sides of the inequality bya, we get: a 2 + 1 \/ 2 a, i.e.

A 2 + 1 – 2 a \/ 0 , or ( a– 1) 2 \/ 0 , but here we already know how to turn

Sign \/ to get the correct inequality (How?). Turning it

In the right direction along the entire chain of inequalities from bottom to top, we
we obtain the required inequality.

Solving inequalities. Two inequalities containing the same unknowns are called equivalent , if they are valid for the same values of these unknowns. The same definition is used for the equivalence of two systems of inequalities. Solving inequalities is the process of moving from one inequality to another that is equivalent to an inequality. For this purpose they are used basic properties of inequalities(cm. ). In addition, replacing any expression with another that is identical to the given one can be used. Inequalities may be algebraic( containing only polynomials) And transcendental(for example logarithmic ortrigonometric). We will look at one very important method here,often used in solving algebraic inequalities

Interval method. Solve inequality: ( x – 3)( x – 5) < 2( x – 3). Here we cannot divide both sides of the inequality by (x – 3), since we do not know the sign of this binomial (it contains the unknown x ). Therefore we will rescheduleall terms of the inequality to the left side:

(x – 3)( x – 5) – 2( x – 3) < 0 ,

let's factorize it:

(x – 3)( x – 5 – 2) < 0 ,

and we get: ( x – 3)( x – 7) < 0. Теперь определим знак произведения в левой части неравенства в различных числовых интервалах. Заметим, что x= 3 and x = 7 - roots of this expression. Therefore, the entire number line will be divided by theseroots into the following three intervals:

In the interval I(x < 3 ) both factors are negative, therefore, their work positively; V interval II (3 < x< 7 ) first multiplier(x– 3 ) is positive, and the second ( x – 7 ) is negative, so their work negative; in the intervalIII(x> 7) both factors are positive, therefore, their work also positively. Now all that remains is to choose the interval in which our product negative. This is the intervalII, therefore, the solution to the inequality: 3 < x< 7. Last expression- the so-called double inequality. It means thatx must be both greater than 3 and less than 7.

EXAMPLE Solve the following inequality using the interval method:

(x – 1)(x – 2)(x – 3) … (x –100) > 0 .

Solution. The roots of the left side of the inequality are obvious: 1, 2, 3, …, 100.

They split the number line into 101 intervals:

Since the number of parentheses on the left side even(equals 100), then

At x < 1, когда все множители отрицательны, их произведение

Positively. When passing through the root there is a change

Sign of the work. Therefore, at the next interval, inside

Which product is positive, will be (2, 3), then (4, 5),

Then (6, 7), ... , (98, 99) and finally, x >100.

Thus, this inequality has a solution:

x < 1, 2 < x < 3, 4 < x < 5 ,…, x >100.

So, to solve an algebraic inequality, I need to move it allmembers to the left (orright side) and solvethe corresponding equation. After plot the found roots on the number axis; as a result, it is divided into a certain number of intervals. At the last stage of the solution, you need to determine what sign the polynomial has inside each of these intervals, and select the required intervals in accordance with the sign of the inequality being solved.

Note that most transcendental inequalities are reduced to an algebraic inequality by replacing the unknown. It must be solved with respect to the new unknown, and then, by reverse substitution, find a solution to the original inequality.

Systems of inequalities. To solve a system of inequalities, it is necessary to solve each of them and combine their solutions. This combination leads to one of two possible cases: either the system has a solution or it does not.

Example 1. Solve the system of inequalities:

Solution. Solution of the first inequality:x < 4 ; а второго: x > 6.

Thus, this system of inequalities has no solution.

(Why?)

Example 2. Solve the system of inequalities:

Solution: The first inequality, as before, gives:x < 4; но решение

The second inequality in this example:x > 1.

Thus, the solution to the system of inequalities: 1< x < 4.

At the seminar of coordinators of the Kangaroo Olympiad, Vyacheslav Andreevich Yasinsky gave a lecture on how to prove Olympiad symmetric inequalities using his own method of differences of variables.

Indeed, at mathematical Olympiads there are often tasks to prove inequalities, such as, for example, this one from the 2001 International Mathematics Olympiad: $\frac(a)(\sqrt(a^2+8bc))+\frac(b)( \sqrt(b^2+8ac))+\frac(c)(\sqrt(c^2+8ab))\geq 1$ (for positive a,b,c).

Usually, to prove an Olympiad inequality, it must be reduced to one of the basic ones: Cauchy, Cauchy-Bunyakovsky, Jensen, inequality between means, etc. Moreover, you often have to try different versions of the basic inequality before achieving success.

However, often Olympiad inequalities (like the one above) have one feature. When you rearrange the variables (for example, replacing a with b, b with c, and c with a), they will not change.

If a function of several variables does not change with any rearrangement, then it is called symmetric. For a symmetric function f from three variables the equality holds:
f(x,y,z)= f(x,z,y)= f(y ,x ,z )= f(y ,z ,x )= f(z ,x ,y )= f(z,y,x)

If a function does not change only when the variables are cyclically rearranged, it is called cyclic.
f(x,y,z)= f(y,z,x)= f(z,x,y)

For inequalities that are constructed on the basis of symmetric functions, Vyacheslav Andreevich developed a universal proof method.
The method consists of the following steps.
1. Transform the inequality so that there is a symmetric polynomial on the left (let’s denote it D) and 0 on the right.

2. Express the symmetric polynomial D in the variables a, b, c in terms of basic symmetric polynomials.

There are three basic symmetric polynomials in three variables. This:
p = a+b+c - sum;
q = ab+bc+ac - sum of pairwise products;
r = abc - product.

Any symmetric polynomial can be expressed in terms of base polynomials.

3. Since the polynomial D is symmetric, we can, without loss of generality, assume that the variables a, b, c are ordered as follows: $a\geq b\geq c$

4. We introduce two non-negative numbers x and y, so that x = a-b, y = b-c.

5. Transform the polynomial D again, expressing p, q and r in terms of c and x, y. We take into account that
b = y+c
a = (x+y)+c

Then
p = a+b+c = (x+2y)+3c
q = ab+bc+ac = 3c 2 +2(x+2y)c+(x+y)y
r = abc = (x+y)yc + (x+2y)c 2 +c 3

Please note that we do not open parentheses in expressions containing x and y.

6. Now we consider the polynomial D as polynomial in c with coefficients expressed in terms of x and y. Taking into account the non-negativity of the coefficients, it turns out to be easy to show that the inequality sign will be preserved for all admissible values of c.

Let us explain this method with examples.
Example 1. Prove inequality:
$(a+b+c)^2\geq 3(ab+bc+ac)$

Proof
Since the inequality is symmetric (does not change with any permutation of the variables a, b, c), we present it as
$(a+b+c)^2 - 3(ab+bc+ac)\geq 0$

Let's express the polynomial on the left side in terms of basic symmetric ones:
$p^2 - 3q\geq 0$

Since the polynomial is symmetric, we can assume, without loss of generality, that $a\geq b\geq c$ and $x = a-b\geq 0$, $y = b-c\geq 0$.

p 2 -3q = ((x+2y)+3c) 2 -3(3c 2 +2(x+2y)c+(x+y)y) = (x+2y) 2 +6(x+2y)c +9c 2 -9c 2 -6(x+2y)c-3(x+y)y

After bringing similar ones, we obtain an inequality that does not contain a variable at all with
$(x+2y)^2-3(x+y)y\geq 0$

Now you can open the brackets
$x^2+4xy+4y^2-3xy-3y^2\geq 0$
$x^2+xy+y^2\geq 0$ - which is true both for negative x, y and for any.

Thus, the inequality is proven.

Example 2(from the 1999 British Mathematical Olympiad)
Prove that $7(ab+bc+ac)\leq 2+9abc$ (for positive numbers, if a+b+c = 1)

Proof
Before we begin to reduce everything to the left side, let us note that the degrees of the parts of the inequality are not balanced. If in Example 1 both sides of the inequality were polynomials of the second degree, then here the polynomial of the second degree is compared with the sum of the zero and third polynomials. We use the fact that the sum a+b+c by condition is equal to 1 and multiply the left side by one, and the two from the right side by one cubed.

$7(ab+bc+ac)(a+b+c)\leq 2(a+b+c)^3+9abc$

Now let’s move everything to the left and imagine the left side as a symmetric polynomial of a, b, c:
$7(ab+bc+ac)(a+b+c)- 2(a+b+c)^3-9abc\leq 0$

Let's express the left-hand side in terms of basic symmetric polynomials:
$7qp- 2p^3-9r\leq 0$

Let's express the left-hand side in terms of x, y and c, presenting it as a polynomial with respect to c.
7qp- 2p 3 -9r = 7(3c 2 +2(x+2y)c+(x+y)y)((x+2y)+3c)-2((x+2y)+3c) 3 -9( (x+y)yc + (x+2y)c 2 +c 3) = 7 (3(x+2y)c 2 +2(x+2y) 2 c+(x+2y)(x+y)y+ 9c 3 +6(x+2y)c 2 +3(x+y)yс) - 2 ((x+2y) 3 +9(x+2y) 2 c+27(x+2y)c 2 +27c 3 ) - 9((x+y)yc + (x+2y)c 2 +c 3) = 21(x+2y)c 2 +14(x+2y) 2 c +7(x+2y)(x+ y)y+63c 3 +42(x+2y)c 2 +21(x+y)yс -2(x+2y) 3 -18(x+2y) 2 c -54(x+2y)c 2 - 54c 3 -9(x+y)yc -9(x+2y)c 2 -9c 3

The main thing is to carry out the transformations carefully and carefully. As Vyacheslav Andreevich said, if he performs transformations and someone distracts him, he throws away the sheet of paper with the formulas and starts again.

For the convenience of reducing similar ones in the final polynomial, they are highlighted in different colors.

All terms with c 3 will be destroyed: 63c 3 -54c 3 -9c 3 = 0
The same will happen with the second degree with: 21(x+2y)c 2 +42(x+2y)c 2 -54(x+2y)c 2 -9(x+2y)c 2 = 0

Let's transform the terms with the first degree with: 14(x+2y) 2 c+21(x+y)yс-18(x+2y) 2 c-9(x+y)yc= -4(x+2y) 2 c+12(x+y)yс = (12 (x+y)y - 4 (x+2y) 2 )c = (12xy+12y 2 - 4x 2 -16xy-16 y 2 )c = (- 4x 2 -4xy-4 y 2 )c = -4 (x 2 +xy+ y 2 )c - this expression will never be positive.

And free terms: 7(x+2y)(x+y)y-2(x+2y) 3 = 7(x+2y)(xy+y 2) - 2(x+2y)(x 2 +4xy+ 4y 2) = (x+2y) (7xy+7y 2 -2x 2 -8xy-8y 2) = - (x+2y)(2x 2 +xy+y 2) - and this expression too.

Thus, the original inequality will always be satisfied, and it will turn into equality only if a=b=c.

At his lecture, Vyacheslav Andreevich discussed many more interesting examples. Try to use this method to prove Olympiad inequalities. Perhaps it will help you get several valuable points.