INEQUALITIES BJORN POONEN

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1 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 oly if x = y The last phrase with equality meas two thigs: first, if x = y 0, the (x + y)/2 = xy (obvious); ad secod, if (x + y)/2 = xy for some x, y 0, the x = y It follows that if x, y 0 ad x y, the iequality is strict: (x + y)/2 > xy Here s a oe-lie proof of the AM-GM iequality for two variables: x + y xy = 1 ( ) 2 x y The AM-GM iequality geeralizes to oegative umbers AM-GM iequality: If x 1,, x 0, the x 1 + x x with equality if ad oly if x 1 = x 2 = = x x 1 x 2 x 2 The power mea iequality Fix x 1,, x 0 For r 0 (assume r > 0 if some x i are zero), the r-th power mea P r of x 1,, x is defied to be the r-th root of the average of the r-th powers of x 1,, x : ( ) x r P r := x r 1/r This formula yields osese if r = 0, but there is a atural way to defie P 0 too: it is simply defied to be the geometric mea 1 : P 0 := x 1 x 2 x Oe also defies P = max{x 1,, x } sice whe r is very large, P r is a good approximatio to the largest of x 1,, x For a similar reaso oe uses the otatio Here are some examples: P = mi{x 1,, x } P 1 = x x is the arithmetic mea, x 2 P 2 = x 2 is sometimes called the root mea square For x 1,, x > 0, 1 P 1 = 1 x x is called the harmoic mea Power mea iequality: Let x 1,, x 0 Suppose r > s (ad s 0 if ay of the x i are zero) The P r P s, with equality if ad oly if x 1 = x 2 = = x

2 2 BJORN POONEN 3 Covex fuctios A fuctio f(x) is covex if for ay real umbers a < b, each poit (c, d) o the lie segmet joiig (a, f(a)) ad (b, f(b)) lies above or at the poit (c, f(c)) o the graph of f with the same x-coordiate Algebraically, this coditio says that (1) f((1 t)a + tb) (1 t)f(a) + tf(b) wheever a < b ad for all t [0, 1] (The left had side represets the height of the graph of the fuctio above the x-value x = (1 t)a + tb which is a fractio t of the way from a to b, ad the right had side represets the height of the lie segmet above the same x-value) Those who kow what a covex set i geometry is ca iterpret the coditio as sayig that the set S = {(x, y) : y f(x)} of poits above the graph of f is a covex set Loosely speakig, this will hold if the graph of f curves i the shape of a smile istead of a frow For example, the fuctio f(x) = x 2 is covex, as is f(x) = x for ay positive eve iteger Oe ca also speak of a fuctio f(x) beig covex o a iterval I This meas that the coditio (1) above holds at least whe a, b I (ad a < b ad t [0, 1]) For example, oe ca show that f(x) = x 3 is covex o [0, ), ad that f(x) = si x is covex o ( π, 0) Fially oe says that a fuctio f(x) o a iterval I is strictly covex, if f((1 t)a + tb) < (1 t)f(a) + tf(b) wheever a, b I ad a < b ad t (0, 1) I other words, the lie segmet coectig two poits o the graph of f should lie etirely above the graph of f, except where it touches at its edpoits For coveiece, here is a brief list of some covex fuctios I these, k represets a positive iteger, r, s represet real costats, ad x is the variable I fact, all of these are strictly covex o the iterval give, except for x r ad x r whe r is 0 or 1 x 2k, o all of R x r, o [0, ), if r 1 x r, o [0, ), if r [0, 1] x r, o (0, ), if r 0 log x, o (0, ) si x, o [0, π] cos x, o [ π/2, π/2] ta x, o [0, π/2) e x, o all of R r/(s + x) o ( s, ), if r > 0 A sum of covex fuctios is covex Addig a costat or liear fuctio to a fuctio does ot affect covexity Remarks (for those who kow about cotiuity ad derivatives): If oe wats to prove rigorously that a fuctio is covex, istead of just guessig it from the graph, it is ofte easier to use oe of the criteria below istead of the defiitio of covexity

3 INEQUALITIES 3 1 Let f(x) be a cotiuous fuctio o a iterval I The f(x) is covex if ad oly if (f(a) + f(b))/2 f((a + b)/2) holds for all a, b I Also, f(x) is strictly covex if ad oly if (f(a) + f(b))/2 > f((a + b)/2) wheever a, b I ad a < b 2 Let f(x) be a differetiable fuctio o a iterval I The f(x) is covex if ad oly if f (x) is icreasig o I Also, f(x) is strictly covex if ad oly if f (x) is strictly icreasig o the iterior of I 3 Let f(x) be a twice differetiable fuctio o a iterval I The f(x) is covex if ad oly if f (x) 0 for all x I Also, f(x) is strictly covex if ad oly if f (x) > 0 for all x i the iterior of I 4 Iequalities for covex fuctios A covex fuctio f(x) o a iterval [a, b] is maximized at x = a or x = b (or maybe both) Example (USAMO 1980/5): Prove that for a, b, c [0, 1], a b + c b c + a c + (1 a)(1 b)(1 c) 1 a + b + 1 Solutio: Let F (a, b, c) deote the left had side If we fix b ad c i [0, 1], the resultig fuctio of a is covex o [0, 1], because it is a sum of fuctios of the type f(a) = r/(s + a) ad liear fuctios Therefore it is maximized whe a = 0 or a = 1; ie, we ca icrease F (a, b, c) by replacig a by 0 or 1 Similarly oe ca icrease F (a, b, c) by replacig each of b ad c by 0 or 1 Hece the maximum value of F (a, b, c) will occur at oe of the eight vertices of the cube 0 a, b, c 1 But F (a, b, c) = 1 at these eight poits, so F (a, b, c) 1 wheever 0 a, b, c 1 Jese s Iequality: Let f be a covex fuctio o a iterval I If x 1,, x I, the f(x 1 ) + + f(x ) ( x1 x 2 x ) f If moreover f is strictly covex, the equality holds if ad oly if x 1 = x 2 = = x Hardy-Littlewood-Polyà majorizatio iequality: Let f be a covex fuctio o a iterval I, ad suppose a 1,, a, b 1,, b I Suppose that the sequece a 1,, a majorizes b 1,, b : this meas that the followig hold: a 1 a b 1 b a 1 b 1 a 1 + a 2 b 1 + b 2 a 1 + a a 1 b 1 + b b 1 a 1 + a a 1 + a = b 1 + b b 1 + b

4 4 BJORN POONEN (Note the equality i the fial equatio) The f(a 1 ) + + f(a ) f(b 1 ) + + f(b ) If f is strictly covex o I, the equality holds if ad oly if a i = b i for all i 5 Iequalities with weights May of the iequalities we have looked at so far have versios i which the terms i a mea ca be weighted uequally Weighted AM-GM iequality: If x 1,, x > 0 ad w 1,, w 0 ad w w = 1, the w 1 x 1 + w 2 x w x x w 1 1 x w 2 2 x w, with equality if ad oly if all the x i with w i 0 are equal Oe recovers the usual AM-GM iequality by takig equal weights w 1 = w 2 = = w = 1/ Weighted power mea iequality: Fix x 1,, x > 0 ad weights w 1,, w 0 with w w = 1 For ay ozero real umber r, defie the r-th (weighted) power mea by the formula ( ) w1 x r w x r 1/r P r := Also let P 0 be the weighted geometric mea (usig the same weights): P 0 := x w 1 1 x w The P r is a icreasig fuctio of r R Moreover, if the x i with w i 0 are ot all equal, the P r is a strictly icreasig fuctio of r Weighted Jese s Iequality: Let f be a covex fuctio o a iterval I If x 1,, x I, w 1,, w 0 ad w w = 1, the w 1 f(x 1 ) + + w f(x ) f (w 1 x w x ) If moreover f is strictly covex, the equality holds if ad oly if all the x i with w i 0 are equal 6 Symmetric fuctio iequalities Give umbers a 1,, a ad 0 i, the i-th elemetary symmetric fuctio σ i is defied to be the coefficiet of x i i (x + a 1 ) (x + a ) For example, for = 3, σ 0 = 1 σ 1 = a 1 + a 2 + a 3 σ 2 = a 1 a 2 + a 2 a 3 + a 3 a 1 σ 3 = a 1 a 2 a 3

5 INEQUALITIES 5 The i-th elemetary symmetric mea S i is the arithmetic mea of the moomials appearig i the expasio of σ i ; i other words, S i := σ i / ( i) I the example above, S 0 = 1 S 1 = a 1 + a 2 + a 3 3 S 2 = a 1a 2 + a 2 a 3 + a 3 a 1 3 S 3 = a 1 a 2 a 3 Newto s iequality: For ay real umbers a 1,, a, we have S i 1 S i+1 S 2 i Maclauri s iequality: For a 1,, a 0, we have S 1 S 2 3 S 3 S Moreover, if the a i are positive ad ot all equal, the the iequalities are all strict 7 More iequalities Cauchy(-Schwartz-Buiakowski) iequality: If x 1,, x, y 1,, y are real umbers, the (x x 2 )(y y 2 ) (x 1 y x y ) 2 Chebychev s iequality: If x 1 x 0 ad y 1 y 0, the ( ) ( ) x 1 y x y x1 + + x y1 + + y with equality if ad oly if oe of the sequeces is costat Chebychev s iequality with three sequeces: If x 1 x 0, y 1 y 0, ad z 1 z 0, the ( ) ( ) ( ) x 1 y 1 z x y z x1 + + x y1 + + y z1 + + z with equality if ad oly if at least two of the three sequeces are costat or oe of the sequeces is all zero You ca probably guess what the four-sequece Chebychev iequality looks like Hölder s iequality: Let a 1,, a, b 1,, b, α, β > 0 ad suppose that α + β = 1 The with equality if ad oly if (a a ) α (b b ) β (a α 1 b β a α b β ), a 1 b 1 = a 2 b 2 = = a b

6 6 BJORN POONEN Jese s extesio of Hölder s Iequality: Suppose a 1,, a, b 1,, b,, l 1,, l, α, β,, λ > 0, ad α + β + + λ 1 The ( ) α ( ) β ( ) λ ( ) a i b i l i a α i b β i lλ i Rearragemet iequality: Suppose a 1 a ad b 1 b are real umbers If π is a permutatio of 1, 2,,, the a 1 b + a 2 b a b 1 a 1 b π(1) + + a b π() a 1 b a b Mikowski s iequality: Suppose a 1,, a, b 1,, b 0, ad r is a real umber If r > 1, the r a r a r + r b r b r r (a 1 + b 1 ) r + + (a + b ) r If 0 < r < 1, the the iequality is reversed Beroulli s iequality: If x > 1 ad 0 < a < 1, the (1 + x) a 1 + ax, with equality if ad oly if x = 0 The iequality reverses for a < 0 or a > 1 8 Problems There are a lot of problems here Just do the oes that iterest you 1 Prove that for ay a, b, c > 0, (a + b)(b + c)(c + a) 8abc, ad determie whe equality holds 2 Prove! < ( ) +1 2 for all itegers > 1 3 Prove that the sum of the legs of a right triagle ever exceeds 2 times the hypoteuse of the triagle 4 Prove 2 x 3 1/x for x > 0 5 Prove that if a > b > 0 ad 1 is a iteger, the 6 Let E be the ellipsoid a b > (a b)(ab) ( 1)/2 x 2 a + y2 2 b + z2 2 c = 1, 2 for some a, b, c > 0 Fid, i terms of a, b, ad c, the volume of the largest rectagular box that ca fit iside E, with faces parallel to the coordiate plaes 7 Amog all plaes passig through a fixed poit (a, b, c) with a, b, c > 0 ad meetig the positive parts of the three coordiate axes, fid the oe such that the tetrahedro bouded by it ad the coordiate plaes has miimal area 8 Amog all rectagular boxes with volume V, fid the oe with smallest surface area 9 Now cosider ope boxes, with oly five faces Agai fid the oe with smallest surface area with a give volume V

7 INEQUALITIES 7 10 Let T be the tetrahedro with vertices (0, 0, 0), (a, 0, 0), (0, b, 0), ad (0, 0, c) for some a, b, c > 0 Let V be the volume of T, ad let l be the sum of the legths of the six edges of T Prove that l 3 V 6( ) 3 11 Let g = a 1 a be the geometric mea of the umbers a 1,, a > 0 Prove that (1 + a 1 )(1 + a 2 ) (1 + a ) (1 + g) 12 Suppose x, y, z > 0 ad x + y + z = 1 Prove that ( ) ( ) ( ) 64 x y z 13 Show that oe ca derive the AM-GM iequality for positive umbers from Jese s iequality with f(x) = log x 14 Prove x x ( ) x+1 x+1 2 for x > 0 (Hit: the fuctio x log x is covex o (0, )) 15 Use Jese s iequality to show that amog all covex -gos iscribed i a fixed circle, the regular -gos have the largest perimeter 16 Give a, b, c, p, q, r > 0 with p + q + r = 1, prove a + b + c a p b q c r + a r b p c q + a q b r c p 17 Show that by takig some of the a i to be equal i the AM-GM iequality, oe ca deduce the weighted AM-GM iequality at least i the case where the weights are oegative ratioal umbers (To deduce from this the geeral weighted AM-GM iequality, oe ca the use a limit argumet) Ca oe similarly deduce the weighted power mea iequality ad weighted Jese s iequality from the uweighted versios? 18 Prove that if a, b, c are sides of a triagle, the (a + b c) a (b + c a) b (c + a b) c a a b b c c 19 Give a, b, c, d > 0 such that (a 2 + b 2 ) 3 = c 2 + d 2, prove a 3 c + b3 d 1 20 What well-kow iequality does oe obtai by takig oly the ed terms i Maclauri s iequality? 21 Prove (bc + ca + ab)(a + b + c) 4 27(a 3 + b 3 + c 3 ) 2 for a, b, c 0 22 Prove that if x, y, z, a, b, c > 0, the x 4 a + y4 3 b + z4 (x + y + z)4 3 c3 (a + b + c) 3 23 Prove that if a, b, c are sides of a triagle, the a 2 b(a b) + b 2 c(b c) + c 2 a(c a) 0 24 Derive Chebychev s iequality from the rearragemet iequality 25 Derive the 3-sequece Chebychev iequality from the 2-sequece Chebychev iequality

8 8 BJORN POONEN 26 Suppose that 0 θ 1,, θ π/2 ad θ θ = 2π Prove that 4 si(θ 1 ) + + si(θ ) si(2π/) 27 Show that Jese s iequality is a special case of the Hardy-Littlewood-Polyà majorizatio iequality 28 What is the geometric meaig of Mikowski s iequality whe r = 2 ad = 3? 29 (IMO 1975/1) Let x i, y i (i = 1, 2,, ) be real umbers such that x 1 x 2 x ad y 1 y 2 y Prove that if z 1, z 2,, z is ay permutatio of y 1, y 2,, y, the (x i y i ) 2 (x i z i ) 2 30 (USAMO 1977/5) Suppose 0 < p < q, ad a, b, c, d, e [p, q] Prove that ( 1 (a + b + c + d + e) a + 1 b + 1 c + 1 d + 1 ) ( p q e q + p ad determie whe there is equality 31 (IMO 1964/2) Prove that if a, b, c are sides of a triagle, the a 2 (b + c a) + b 2 (c + a b) + c 2 (a + b c) 3abc 32 (USAMO 1981/5) If x is a positive real umber, ad is a positive iteger, prove that x x 1 + 2x + + x 2, where t deotes the greatest iteger less tha or equal to t 33 Make your ow iequality problems ad give them to your frieds (or eemies, depedig o the difficulty!) May of the problems above were draw from otes from the US traiig sessio for the Iteratioal Mathematics Olympiad Others are from the USSR Olympiad Problem Book May of the iequalities themselves are treated i the book Iequalities by Hardy, Littlewood, ad Polyà, which is a good book for further readig ) 2 c Berkeley Math Circle Departmet of Mathematics, Uiversity of Califoria, Berkeley, CA , USA address: pooe@mathberkeleyedu

PUTNAM TRAINING INEQUALITIES

PUTNAM TRAINING INEQUALITIES (Last updated: December, 207) Remark This is a list of exercises o iequalities Miguel A Lerma Exercises If a, b, c > 0, prove that (a 2 b + b 2 c + c 2 a)(ab 2 + bc 2 + ca