INEQUALITIES. Theorem 1. Let x 1,...,x n > 0 be positive real numbers. Then their geometric mean is no greater then their arithmetic mean, i.e.

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1 INEQUALITIES Books: 1. Hardy, Littlewood, Polya Iequalities.. M. Steele The Cauchy-Schwarz Master Class AM-GM (Arithmetic mea geometric mea iequality). Theorem 1. Let x 1,...,x > 0 be positive real umbers. The their geometric mea is o greater the their arithmetic mea, i.e. (x 1...x ) 1/ x x. Moreover, the equality holds iff all the umbers are equal to each other, x 1 = = x. About the proof. Method I: Iductio (o powers of ). First, cosider the case =. The iequality becomes x 1 x x 1 +x. Algebraic proof: Rewrite the iequality i the form 4x 1 x (x 1 + x ), which is equivalet to (x 1 x ) 0. Geometric proof: Costruct a circle of diameter d = x 1 +x. Let AB be a diameter of this circle, ad C be the poit o this diameter so that AC = x 1 ad CB = x. Let D be the poit o the circle so that CD is a lie segmet perpedicular to AB. The usig elemetary geometry oe ca easily see that CD = x 1 x. O the other had, this legth is clearly o greater the the radius, i.e., x 1 x x 1+x. The equality holds oly i the case that x 1 = x (ad is equal to the radius of the circle). Now cosider the case whe = k, where k 0 is a iteger ad proceed by iductio. The case k = 1 is already doe. Assume that the iequality holds for = k 1 ad to prove it for = k. This is doe by rewritig the arithmetic mea as follows: x 1 + +x x 1 + x + + x k 1 k + x k x k = k 1 k 1 k ad applyig the iequality first to each of the arithmetic meas i the umerator, ad the to the arithmetic mea of the two resultig geomteric meas. (Carry out the details as a exercise). 1

2 eglishinequalities Fially, we eed to deal with the case whe is ot a power of. I this case, there is a k such that < k = N. Cosider the set of N umbers so that the first of them are x 1,...,x ad the rest are all equal to the arithmetic mea of these umbers, x +1 = = x N = x 1 + +x. The A. = x x = x x N N N x 1 x x +1 x N = = N x 1 x A N. This implies that A x 1 x, which establishes the AM-GM iequality for umbers. Method : (Aother iductive proof). First, ote that if the AM- GM iequality is true for x 1,..., x, the it is also true for αx 1,...,αx. This observatio allows us to rescale the give umbers so that we ca assume that x 1 x = 1. Now, assume that at least oe of the umbers is strictly bigger the 1, ad at least oe is strictly smaller the 1. For example, let x 1 > 1 ad x < 1. By iductio assumptio, x 1 x + x x 1 (x 1 x )x 3...x = 1, which implies that x 1 x + x x 1. Fiish this proof as a exercise. Method 3: Take the atural logarithm of both sides of the iequaltiy. The cosider the cocavity of the fuctio l(x). (See Jese s iequality below). Method 4: (Lagrage multipliers) Cosider the fuctio of variables which is just the product of these variables: P(x 1,...,x ) = x 1 x x. Look for the maximum of this fuctio uder the costrait that g(x 1,, x ) = x 1 + +x S = 0, where S is a costat (the fixed arithmetic mea). By applyig the method of Lagage multipliers, you will see that P is maximal iff x 1 = = x = S/. This implies the AM-GM iequality. Some geeralizatios of this iequality iclude the Power Mea iequality ad the Jese s iequality (see below). Here are several problems from the Putam exam, which ca be solved usig the AM-GM iequality. (Note that some of the problems ca be solved by differet methods too). Problem 1. Prove or disprove: if x ad y are real umbers with y 0 ad y(y + 1) (x + 1), the y(y 1) x.

3 eglishinequalities 3 Problem. Let A, B, C deote three distict poits with iteger coordiates i R. Prove that if ( AB + BC ) < 8 [ABC] + 1, (where AB deotes the legth of AB, ad [ABC] deoes the area of the triagle ABC), the A, B, C are three vertices of a square. Problem 3. Fid the miimal value of the expressio for x > 0. (x + 1/x) 6 (x 6 + 1/x 6 ) (x + 1/x) 3 + (x 3 + 1/x 3 ) Problem 4. (1968, A6) Determie all polyomials of the form k=0 a kx k with all a k = ±1 (0 k, 1 < ) such that each has oly real zeros. Problem 5. (1975, B6) Show that if s = , the (a) ( + 1) 1/ < + s for > 1; (b) ( 1) 1/ 1 < s for >. 0.. Power Mea Iequality. Let x 1,...,x be positive real umbers. For r 0, let P r = 1 + +a r 1/r ( ) a r be the r th power mea for ad let P 0 = lim r 0 P r = (a 1...a ) 1/. Let also P = mi{x 1,..., x } ad P = max{x 1,...,x }. The the followig Power Mea Iequality holds: P r P s, for r < s. The followig are the special cases of this iequality: P 1 P 0 is the AM-GM iequality; P 0 P 1 is the GM-HM iequality, where P 1 = the so-called harmoic mea of the umbers a 1,...,a a 1 a is

4 eglishinequalities Iequalities for covex fuctios. Recall that a fuctio f(x) is called covex if for ay real umbers a < b, the segmet joiig the poits (a, f(a)) ad (b, f(b)) lies etirely above the graph {(x, f(x)) : x [a, b]} of the fuctio. This coditio ca be writte as follows: f((1 t) a + t b) (1 t) f(a) + t f(b) for all t [0, 1] ad all a < b. If this iequality holds strictly, the fuctio is called strictly covex. A fuctio whose egative is covex is called cocave. (I.e., f(x) is cocave if f(x) is covex). Jese s iequality: If f(x) is a covex fuctio o a iterval I, the ( ) a1 + + a f f(a 1) + + f(a ), where a 1,...,a are poits o the iterval I. Of course, if f(x) is a cocave futio, the iequality is reversed. Oe possible proof is by iductio (try to carry it out!) Exercise: what do you get whe you apply the Jese s iequality to fuctios l(x), e x, x, cos(x) (for = )? A geeralizatio of Jese s iequality: if f(x) is covex, ad µ 1,...,µ are positive weights (so that µ i = 1), the ( ) f µ i x i µ i f(x i ). A simple but useful property of covex fuctios: A fuctio which is covex o a iterval reaches its maximum o the iterval is reached at oe (or both) of the eds Cauchy-Schwarz Iequality. For a i > 0, b i > 0 for i = 1,...,, the Cauchy-Schwarz Iequality states ( ) ( ) ( ) a i b i. To prove this iequality, cosider the followig polyomial i variable x: P(x) = (a i x b i ). a i b i

5 eglishinequalities 5 Sice P(x) 0 (with the equality takig place oly if a i /b i has the same value for all i), it follows that the discrimiat of P(x) = 0 ca ot be positive. This is equivalet to the Cauchy-Schwarz iequality. As a exercise, cosider the case = ad fid a relatio betwee the Cauchy-Schwarz ad the AM-GM iequality Various Putam Exam problems ivolvig iequalities: Problem 6. (1986, A1) Fid the maximum value of f(x) = x 3 3x o the set of all real umbers satisfyig x x. Problem 7. (1991, B6) Let a ad b be positive umbers. Fid the largest umber c, i terms of a ad b, such that a x b 1 x sih ux a sih u + bsih u(1 x) sih u for all u with 0 u c ad for all x, 0 < x < 1. (Note that sih u = (e u e u )/). Problem 8. (1993, B1) Fid the smallest positive iteger such that for every iteger m with 0 < m < 1993, there is a iteger k such that m 1993 < k < m (Oe way to solve this problem is to use the followig so-called Mediat property: For positive umbers a, b, c, d such that a/b < c/d we have a/b < (a + c)/(b + c) < c/d. ) Problem 9. (1996, B) Show that for every positive iteger, we have ( ) 1 ( ) < ( 1) <. e e Problem 10. (1999, B4) Let f be a real fuctio with a cotuuous third derivative such that f(x), f (x), f (x) ad f (x) are positive for all x. Suppose that f (x) f(x) for all x. Show that f (x) < f(x). Problem 11. (1991, A5) Fid the maximum value of y 0 x4 + (y y ) dx for y [0, 1].

6 eglishinequalities 6 Problem 1. (1966, B3) Show that if the series 1 =1 p, where p are positive real umbers, is coverget, the the series =1 (p p ) p is also coverget. (Hit: Cauchy-Schwarz iequality).

Convexity, Inequalities, and Norms

Convexity, Inequalities, and Norms Covexity, Iequalities, ad Norms Covex Fuctios You are probably familiar with the otio of cocavity of fuctios. Give a twicedifferetiable fuctio ϕ: R R, We say that ϕ is covex (or cocave up) if ϕ (x) 0 for