Inequalities. Putnam Notes, Fall 2006 University of Utah

Size: px

Start display at page:

Download "Inequalities. Putnam Notes, Fall 2006 University of Utah"

Bruce Berry
5 years ago
Views:

1 Iequalities Putam Notes, Fall 2006 Uiversity of Utah There are several stadard methods for provig iequalities, ad there are also some classical iequalities you should kow about. Method 1: Good old calculus If y = f(x) is defied ad differetiable o some iterval I ad has f (x) > 0 [f (x) < 0] except possibly at the edpoits of I, the f is icreasig [decreasig] ad so a < b implies f(a) < f(b) [f(a) > f(b)]. To see how this works we prove: Theorem (Beroulli s iequality). Let x > 1. If r > 1 or 1 < r < 0 the (1 + x) r 1 + rx If 0 < r < 1 the (1 + x) r 1 + rx As a warm-up, first prove this (by iductio) i the case that r = is a positive iteger. Proof. Cosider the fuctio f(x) = (1 + x) r rx 1 defied for x 1. Note that f(0) = 0, so our task is to show that f has its global miimum at 0 whe r > 1 or 1 < r < 0 ad that it has its global maximum whe 0 < r < 1. This will be accomplished if we argue that f is decreasig o ( 1,0] ad icreasig o [0, ) i the first case, ad the opposite i the secod case. We have ( ) f (x) = r(1 + x) r 1 r = r (1 + x) r 1 1 ad there are four cases to look at: r > 1 or r < 0 ad x < 0. The f (x) < 0. 1

2 r > 1 or r < 0 ad x > 0. The f (x) > 0. 0 < r < 1 ad x < 0. The f (x) > 0. 0 < r < 1 ad x > 0. The f (x) < 0. The details of these cases are left as a exercise. Method 2: Jese s iequality Let y = f(x) be a covex 1 fuctio defied o some iterval i R. By this we mea that f is at least twice differetiable ad that f (x) > 0 for all x. For example, y = x p for p > 1 ad y = log x are covex fuctios. The mai thig to otice about covex fuctios is that if you take two poits o the graph ad draw the chord joiig these two poits, the the etire chord will be above the graph of the fuctio (except at the edpoits). For example, the statemet that the midpoit of the chord is above the graph is: ( x1 + x ) 2 f f(x 1) + f(x 2 ) (1) 2 2 (with equality oly if x 1 = x 2 ). More geerally, take values x 1,,x ad cosider the covex polygo with vertices at ((x 1,y 1 ),,(x,y )) where y i = f(x i ). A poit P iside this covex polygo is a average of the vertices, i.e. there are weights w 1,,w [0,1] with w w = 1 so that P = w 1 (x 1,y 1 ) + + w (x,y ) Now P is above the graph of f, i other words f(w 1 x w x ) w 1 f(x 1 ) + + w f(x ) (2) with equality oly if w i w j (x i x j ) = 0 for all i,j. Iequality (2) geeralizes (1) ad both are called Jese s iequality. A importat special case is whe all weights w i are equal to 1/ ad the ( x1 + + x ) f 1 Calculus books call this property covex up. f(x 1) + + f(x ) 2

3 Similarly, a fuctio is cocave 2 if f (x) < 0. If f is covex the f is cocave. For cocave fuctios the reverse of the above iequalities hold. To illustrate this, let s show Theorem (AM-GM). Let x 1,...,x be positive real umbers. The x x with equality if ad oly if x 1 = = x. x 1 x, Proof. The fuctio f(x) = log x is cocave o (0, ) ad so ( x1 + + x ) f i.e. log x x After expoetiatig we have f(x 1) + + f(x ) log x log x x x (x 1 x ) 1/ The art of applyig this method is figurig out what fuctio to use. Method 3: Covex fuctio o a polytope If f is a covex fuctio defied o a closed iterval [a,b] the f achieves its maximum at either a or b. More geerally, if f is a (multi-variable cotiuous) fuctio defied o a polytope, ad the restrictio of f to ay straight lie is covex, the f achieves its maximum at oe of the vertices. Oe usually does ot have to check covexity o every lie. For example, if f is defied o the square [0,1] [0,1] the it suffices to check covexity o horizotal ad vertical lies, i.e. covexity i each variable separately. Problem. Whe 0 a,b,c 1 the a b + c b c + a Covex dow i calculus. c + (a 1)(b 1)(c 1) 1 a + b + 1 3

4 Solutio. Cosider the fuctio F(a,b,c) = a b + c b c + a c + (a 1)(b 1)(c 1) a + b + 1 defied o the cube 0 a,b,c 1. If we fix two of the variables, say b,c, ad let a ru over [0,1] we get a covex fuctio. So if there is some (a,b,c) where F(a,b,c) > 1 the also either F(0,b,c) > 1 or F(1,b,c) > 1. We ca cotiue this util all 3 variables are either 0 or 1. I other words, we oly eed to check the iequality o the 8 vertices of the cube where each variable is 0 or 1. This is easy to do. Method 4: Smoothig This method cosists of repeatedly replacig pairs of variables ultimately reducig the iequality to checkig oe special case. To illustrate it, I give a proof of the AM-GM iequality (so by ow you have 3 proofs of this really importat iequality). This proof is due to Kira Kedlaya. Theorem (AM-GM). Let x 1,...,x be positive real umbers. The x x with equality if ad oly if x 1 = = x. x 1 x, Proof (Kedlaya). We will make a series of substitutios that preserve the left-had side while strictly icreasig the right-had side. At the ed, the x i will all be equal ad the left-had side will equal the right-had side; the desired iequality will follow at oce. (Make sure that you uderstad this reasoig before proceedig!) If the x i are ot already all equal to their arithmetic mea, which we call a for coveiece, the there must exist two idices, say i ad j, such that x i < a < x j. (If the x i were all bigger tha a, the so would be their arithmetic mea, which is impossible; similarly if they were all smaller tha a.) We will replace the pair x i ad x j by x i = a,x j = x i + x j a; by desig, x i ad x j have the same sum as x i ad x j, but sice they are closer together, their product is larger. To be precise, a(x i + x j a) = x i x j + (x j a)(a x i ) > x i x j 4

5 because x j a ad a x i are positive umbers. 3 By this replacemet, we icrease the umber of the x i which are equal to a, preservig the left-had side of the desired iequality ad icreasig the right-had side. As oted iitially, evetually this process eds whe all of the x i are equal to a, ad the iequality becomes equality i that case. It follows that i all other cases, the iequality holds strictly. Note that we made sure that the replacemet procedure termiates i a fiite umber of steps. If we had proceeded more aively, replacig a pair of x i by their arithmetic mea, we would get a ifiite procedure, ad the would have to show that the x i were covergig i a suitable sese. (They do coverge, but makig this precise requires some additioal effort which our alterate procedure avoids.) Here is aother example i the same vei. This oe is more subtle because Jese does t apply. Problem. Let a 0,...,a be umbers i the iterval (0,π/2) such that ta(a 0 π/4) + ta(a 1 π/4) + + ta(a π/4) 1. Prove that ta a 0 ta a 1 ta a +1. Solutio (Kedlaya). Let x i = ta(a i π/4) ad y i = ta a i = (1 + x i )/(1 x i ), so that x i ( 1,1). The claim would follow immediately 4 from Jese s iequality if oly the fuctio f(x) = log(1 + x)/(1 x) were covex o the iterval ( 1,1), but alas, it is t. It s cocave o ( 1,0] ad covex o [0,1). So we have to fall back o the smoothig priciple. What happes if we try to replace x i ad x j by two umbers that have the same sum but are closer together? The cotributio of x i ad x j to the left side of the desired iequality is 1 + x i 1 x i 1 + x j 1 x j = x i x j +1 x i +x j 1. The replacemet i questio will icrease x i x j, ad so will decrease the above quatity provided that x i + x j > 0. So all we eed to show is that we 3 There is a calculus way of doig this. Cosider the fuctio f(x) = (x i+x)(x j x). The derivative is f (x) = x j x i 2x so f is icreasig o [0, x j x i ]. Say a x i+x j, i.e. 2 2 a x i x j x i. But the f(a x 2 i) > f(0) meas exactly that a(x i + x j a) > x ix j. If a x i+x j, i.e. x 2 j a x j x i use f(x 2 j a) > f(0). 4 Well, also usig that f is mootoically icreasig. 5

6 ca carry out the smoothig process so that every pair we smooth satisfies this restrictio. Obviously there is o problem if all of the x i are positive, so we cocetrate o the possibility of havig x i < 0. Fortuately, we ca t have more tha oe egative x i, sice x x 1 ad each x i is less tha 1. (So if two were egative, the sum would be at most the sum of the remaiig 1 terms, which is less tha 1.) Let s say x 0 < 0. The x 0 + x i 0 for all i 1 for a similar reaso, ad we ca t have x 0 +x i = 0 for all i (otherwise addig all these up we would coclude ( 1)x 0 +(x 0 + +x ) = 0 which is impossible). So choose i so that x 0 + x i > 0 ad we ca replace these two by their arithmetic mea. Now all of the x j are oegative ad smoothig (or Jese) may cotiue without further restrictios, yieldig the desired iequality. Rearragemet iequality Let x 1 x 2... x ad y 1 y 2... y be two sequeces of positive umbers. If z 1,z 2,...,z is ay permutatio of y 1,y 2,...,y the x 1 z 1 + x 2 z x z x 1 y 1 + x 2 y x y. You ca thik about this as follows (for = 2). You are sellig apples at $1 a piece ad hot dogs at $2 a piece. Do you receive more moey if you sell 10 apples ad 15 hot dogs, or if you sell 15 apples ad 10 hot dogs? 5 You ca prove the rearragemet iequality i the spirit of smoothig. If z i < z j for some i < j exchage z i ad z j ad watch the sum grow. Likewise, x i z i is miimized over permutatios of the z i s whe z 1 z 2 z. Problems Let x 1,...,x be positive real umbers. Prove the AM-HM (arithmetic mea - harmoic mea) iequality 1 x x x. x 5 Sorry for beig crass. 6 The hard part is figurig out to what sequece to apply AM-GM or to what fuctio to apply Jese. You are ot expected to solve these i few miutes, but thik about oe or two over a period of time ad experimet with differet choices. You ca also peek at the Hits. 6

7 2. (Putam 2005; B2) Fid all positive itegers, k 1,...,k such that k k = 5 4 ad = 1. k 1 k 3. Let H = Show that ( + 1) 1/ < + H for all. 4. Let s = (1 + 1 ) ad S = 1/(1 1 ). Show: (a) s < s +1. (b) S > S +1. (c) s < S ad sequeces s ad S are coverget, say to e 1 e 2. (d) S /s 1 so that e 1 = e Let a 1,,a (0,π) with arithmetic mea µ. Prove that i si a i a i ( si µ µ ) 6. Let p,q > 1 with 1 p + 1 q = 1. If a,b 0 show that ab ap p + bq q with equality oly if a p = b q. 7. Suppose si 2+2 A si 2 B + cos2+2 A cos 2 B = 1 for some 0, 1. Show that the equality holds for all. 8. The quadratic mea of a 1,,a is QM(a 1,,a ) = a a 2 Show that AM(a 1,,a ) QM(a 1,,a ) with equality oly if a 1 = = a. 7

8 9. More geerally, let M α (a 1,,a ) = ( a α a α ) 1/α for α > 0. So M 1 = AM ad M 2 = QM. Show that α 1 < α 2 M α1 (a 1,,a ) M α2 (a 1,,a ) 10. Give real umbers x 1,...,x, what is the miimum value of x x x x? 11. For a,b,c > 0, prove that a a b b c c (abc) (a+b+c)/ Let x 1,...,x be positive umbers whose sum is 1. Prove that x 1 x x1 1 x Let A,B,C be the agles of a triagle. Prove that (1) si A + si B + sic 3 3/2; (2) cos A + cos B + cos C 3/2; (3) si A/2si B/2si C/2 1/8; (Beware: ot all of the requisite fuctios are covex everywhere!) 14. If a 1,,a,b 1,,b > 0 the (a 1 a ) 1/ + (b 1 b ) 1/ [ ] 1/ (a 1 + b 1 ) (a + b ) 15. (Cauchy s iequality) If u i,v i > 0 for i = 1,2,, the u 1 v 1 + u 2 v u v u u u2 v1 2 + v v2 The sigificace of this iequality is that u v u v for ay two vectors u,v R. 8

9 16. (Hölder s iequality) If u i,v i > 0 for i = 1,2,, ad p,q > 1 with 1 p + 1 q = 1 the u 1 v 1 + u 2 v u v (u p 1 + up up )1/p (v q 1 + vq vq )1/q This geeralizes Cauchy s iequality for p = q = 2. I terms of orms Hölder s iequality says u v u p v q, where p has the obvious meaig. That this is really a orm follows from 17. (Mikowski s triagle iequality) If u i,v i > 0 for i = 1,2,, ad p > 1 the ( (u 1 +v 1 ) p +(u 2 +v 2 ) p + +(u +v ) p) 1/p (u p 1 +up 2 + +up ) 1/p +(v p 1 +vp 2 + +vp ) 1/p I other words u + v p u p + v p. 18. Let x 1,...,x ( 2) be positive umbers satisfyig 1 x x x = Prove that x1 x 2 x 1 (Agai, beware of ocovexity.) (XVII IMO) Let x 1 x 2... x ad y 1 y 2... y be two sequeces of positive umbers. If z 1,z 2,...,z is ay permutatio of y 1,y 2,...,y the (x i y i ) 2 i=1 k=1 (x i z i ) (XX IMO) Suppose that a 1,...,a be a sequece of distict positive itegers. Prove that a k k 2 1 k. i=1 k=1 21. Show that ( x + 1 x x 2 ) x+1. 9

10 22. For a positive iteger x ad a iteger prove that k=1 23. Let be a positive iteger such that Show that si 2 (A) = cos 2 (A) = 1/ Let a,b,c > 0. Show that [kx] k [x]. si 2+2 (A) + cos 2+2 (A) = 1 2. a b + c + b a + c + c a + b (XXV IMO) Let x,y ad z be o-egative real umbers such that x + y + z = 1. Show that xy + xz + yz 2xyz

11 Hits. 1. Apply the AM-GM iequality to the right had side ad to the iverse of the left had side. 2. AM-HM iequality. 3. AM-GM iequality for 2, 3 2, 4 3,, (a) AM-GM for + 1 umbers 1 + 1/,1 + 1/,,1 + 1/,1. (b) AM-GM for + 1 umbers 1 1/,1 1/,,1 1/,1. (d) Use Beroulli. Of course, the commo limit is e = ( ) 5. f(x) = log si x x is cocave o (0,π). 6. Jese for f(x) = log x with weights 1/p ad 1/q applied to a p ad b q. 7. Iequality from the previous problem. Figure out what p, q should be from si 2 B + cos 2 B = Jese for x x Whe α 1 = 1 cosider x x α 2. I geeral, by substitutios show that M α1 M α2 follows from M 1 M α 2 α The fuctio is piecewise liear. What are the slopes o each piece? 11. For x > 0 x xlog x is covex ad x log x is cocave. Put the two iequalities together. 12. x x/ 1 x is covex for 0 < x < (1) x si x is cocave for 0 x π. (2) x cos x is cocave o [0,π/2] but that s ot quite eough. If say C > π/2 you should decrease C ad simultaeously icrease say A by the same amout. Thus show that x cos(a + x) + cos(c x) is icreasig o [0,C π/2]. (3) x log si x is cocave for 0 < x π/ Apply AM-GM to up. 15. Special case of Hölder. a 1 a 1 +b 1,, a a +b ad also to b 1 a 1 +b 1,, b a +b ad add 11

12 16. This geeralizes Cauchy s iequality for p = q = 2. To derive it from Jese s iequality with weights, let f(t) = t q, the weights are w i = u p i / j up j ad the argumets are x i = u i v i /w i. 17. This ca be proved usig Jese s iequality with weights with f(x) = (1 + x 1/p ) p, w i = v p i / j vp j ad x i = (u i /v i ) p. 18. It is coveiet to make the substitutio y = 1 x+1998 so that x = 1 y It would be cool if the fuctio y log( 1 y 1998) were covex o (0, ) but ufortuately it is covex oly o the first half of this iterval. Apply the smoothig techique. 19. Rearragemet iequality. 20. Rearragemet iequality. 21. Use the covexity of x log x. 22. Rearragemet iequality. 23. Similar to # Covexity of x x/(1 x). Alteratively, use the rearagemet iequality twice. This is due to Masaki Iio. Assume that a b c the a + b a + c b + c Smoothig priciple. Assume that x y z ad the replace x ad z by their arithmetic mea. Alteratively, use Lagrage multipliers. Show that whe umber 2 is replaced by 2.25 the maximum is still attaied at the ceter of the triagle. 12

INEQUALITIES BJORN POONEN

INEQUALITIES BJORN POONEN 1 The AM-GM iequality The most basic arithmetic mea-geometric mea (AM-GM) iequality states simply that if x ad y are oegative real umbers, the (x + y)/2 xy, with equality if ad