PUTNAM TRAINING INEQUALITIES

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1 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 2 ) 9a 2 b 2 c 2 ( ) + 2 Prove that! <, for = 2,, 4,, 2 If 0 < p, 0 < q, ad p + q <, prove that (px + qy) 2 px 2 + qy 2 4 If a, b, c 0, prove that (a + b + c) a + b + c 5 Let x, y, z > 0 with xyz = Prove that x + y + z x 2 + y 2 + z 2 6 Show that a 2 + b 2 + a b a 2 + b 2 (a + a2 + + a) 2 + (b + b2 + + b) 2 7 Fid the miimum value of the fuctio f(x, x 2,, x ) = x + x x, where x, x 2,, x are positive real umbers such that x x 2 x = 8 Let x, y, z 0 with xyz = Fid the miimum of S = x2 y + z + y2 z + x + z2 x + y 9 If x, y, z > 0, ad x + y + z =, fid the miimum value of x + y + z 0 Prove that i a triagle with sides a, b, c ad opposite agles A, B, C (i radias) the followig relatio holds: a A + b B + c C π a + b + c

2 PUTNAM TRAINING INEQUALITIES 2 (Putam, 200) Let a, a 2,, a ad b, b 2,, b oegative real umbers Show that (a a 2 a ) / + (b b 2 b ) / ((a + b )(a 2 + b 2 ) (a + b )) / 2 The otatio! (k) meas take factorial of k times For example,! () meas ((!)!)! What is bigger, 999! (2000) or 2000! (999)? Which is larger, or ? 4 Which is larger, log 2 or log 5? 5 Prove that there are o positive itegers a, b such that b 2 + b + = a 2 6 (Ispired i Putam 968, B6) Prove that a polyomial with oly real roots ad all coefficiets equal to ± has degree at most 7 (Putam 984) Fid the miimum value of ( 2 (u v) 2 + u2 9 v for 0 < u < 2 ad v > 0 ( ) ( ) ( ) 2 8 Show that ) 2 < 2 9 (Putam, 2004) Let m ad be positive itegers Show that (m + )! m!! < (m + ) m+ m m 20 Let a, a 2,, a be a sequece of positive umbers, ad let b, b 2,, b be ay permutatio of the first sequece Show that a + a a b b 2 b 2 (Rearragemet Iequality) Let a, a 2,, a ad b, b 2,, b icreasig sequeces of real umbers, ad let x, x 2,, x be ay permutatio of b, b 2,, b Show that a i b i a i x i 22 Prove that the p-mea teds to the geometric mea as p approaches zero I other other words, if a,, a are positive real umbers, the ( /p ( ) / lim a p p 0 k) = a k k= k=

3 PUTNAM TRAINING INEQUALITIES 2 If a, b, ad c are the sides of a triagle, prove that a b + c a + b c + a b + c a + b c 24 Here we use Kuth s up-arrow otatio: a b = a b, a b = a (a ( a)), so }{{} b copies of a eg 2 = 2 (2 2)) = 2 22 What is larger, 2 20 or 200? 25 Prove that e /e + e /π 2e / 26 Prove that the fuctio f(x) = a + + a (x a i ) 2 attais its miimum value at x = a = 27 Fid the positive solutios of the system of equatios x + x 2 = 4, x 2 + x =,, x 99 + x 00 = 4, x 00 + x = 28 Prove that if the umbers a, b, ad c satisfy the iequalities a b c, b c a, c a b, the oe of those umbers is the sum of the other two 29 Fid the miimum of si x/ cos x + cos x/ si x, 0 < x < π/2 0 Let a i > 0, i =,,, ad s = a + + a Prove a + a a s a s a 2 s a Fid the maximum value of f(x) = si 4 (x) + cos 4 x for x R ( a+b 2 Let a, b, c be positive real umbers Prove that 2 ) ab 2 Whe is equality attaied? ( a+b+c ) abc (Geeralizatio of previous problem) Let a, =, 2,, be a sequece of positive real umbers Prove that b = a + + a a a is a icreasig sequece of o-egative real umbers 4 Let f : [0, ) R a fuctio with the followig properties: (a) f is subadditive, ie, f(x + y) f(x) + f(y), for every x, y [0, ) (b) f is (right) differetiable at 0 (c) f(0) = 0 Prove f(x) xf (0) for every x [0, )

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5 PUTNAM TRAINING INEQUALITIES 5 Hits Oe way to solve this problem is by usig the Arithmetic Mea-Geometric Mea iequality o each factor of the left had side 2 Apply the Arithmetic Mea-Geometric Mea iequality to the set of umbers, 2,, Power meas iequality with weights 4 Power meas iequality 5 p p+q ad q p+q 6 This problem ca be solved by usig Mikowski s iequality, but aother way to look at it is by a appropriate geometrical iterpretatio of the terms (as distaces betwee poits of the plae) 7 May miimizatio or maximizatio problems are iequalities i disguise The solutio usually cosists of guessig the maximum or miimum value of the fuctio, ad the provig that it is i fact maximum or miimum I this case, give the symmetry of the fuctio a good guess is f(,,, ) =, so try to prove f(x, x 2,, x ) Use the Arithmetic Mea-Geometric Mea iequality o x,, x x 8 Apply the Cauchy-Schwarz iequality to the vectors ( y+z, ad choose appropriate values for u, v, w 9 Arithmetic-Harmoic Mea iequality y z+x, 0 Assume a b c, A B C, ad use Chebyshev s Iequality z x+y ) ad (u, v, w), Divide by the right had side ad use the Arithmetic Mea-Geometric Mea iequality o both terms of the left 2 Note that! is icreasig ( < m =! < m!) Look at the fuctio f(x) = (999 x) l (999 + x) 4 Use the defiitio of logarithm 5 The umbers b 2 ad (b + ) 2 are cosecutive squares 6 Use the Arithmetic Mea-Geometric Mea iequality o the squares of the roots of the polyomial 7 Thik geometrically Iterpret the give expressio as the square of the distace betwee two poits i the plae The problem becomes that of fidig the miimum distace betwee two curves

6 PUTNAM TRAINING INEQUALITIES 6 8 Cosider the expressios P = ( ( ) ( 2) 4 2 ) ( 2 ad Q = 2 ) ( 4 ) ( ) Note that k < k, for k =, 2, k k+ 9 Look at the biomial expasio of (m + ) m+ 20 Arithmetic Mea-Geometric Mea iequality 2 Try first the cases = ad = 2 The use iductio 22 Take logarithms ad use L Hôpital 2 Set x = b + c a, y = c + a b, z = a + b c 24 We have 2 22 = 6 < 27 = 25 Show that f(x) = e /x for x > 0 is decreasig ad covex 26 Prove that f(x) f(a) 0 27 By the AM-GM iequality we have x + x 2 2 x x 2, Try to prove that those iequalities are actually equalities 28 Square both sides of those iequalities 29 Rearragemet iequality 0 Rearragemet iequality Note that si x, so what which is smaller, si 2 x or si 4 x? (Same with cos x) 2 Arithmetic Mea-Geometric Mea iequality Arithmetic Mea-Geometric Mea iequality 4 Look at the limit of f(x/) x/ as

7 PUTNAM TRAINING INEQUALITIES 7 Solutios Usig the Arithmetic Mea-Geometric Mea Iequality o each factor of the LHS we get (a ) ( ) 2 b + b 2 c + c 2 a ab 2 + bc 2 + ca 2 ( ) ( ) a b c a b c = a 2 b 2 c 2 Multiplyig by 9 we get the desired iequality Aother solutio cosists of usig the Cauchy-Schwarz iequality: (a 2 b + b 2 c + c 2 a)(ab 2 + bc 2 + ca 2 ) = ((a b) 2 + (b c) 2 + (c a) 2 ) (( b c) 2 + ( c a) 2 + ( a b) 2 ) (abc + abc + abc) 2 = 9a 2 b 2 c 2 2 This result is the Arithmetic Mea-Geometric Mea applied to the set of umbers, 2,, : (+) < = = + 2 Raisig both sides to the th power we get the desired result The simplest solutio cosists of usig the weighted power meas iequality to the p (weighted) arithmetic ad quadratic meas of x ad y with weights ad q : p+q p+q p p + q x + q p p + q y p + q x2 + q p + q y2, hece (px + qy) 2 (p + q)(px 2 + qy 2 ) Or we ca use the Cauchy-Schwarz iequality as follows: (px + qy) 2 = ( p p x + q q y) 2 ( { p} 2 + { q} 2) ( { p x} 2 + { q y} 2) = (p + q)(px 2 + qy 2 ) Fially we use p + q to obtai the desired result (Cauchy-Schwarz) 4 By the power meas iequality: a + b + c } {{ } M (a,b,c) From here the desired result follows ( a + b + c ) 2 } {{ } M /2 (a,b,c)

8 PUTNAM TRAINING INEQUALITIES 8 5 We have: x + y + z = (x + y + z) xyz (xyz = ) (x + y + z)2 (AM-GM iequality) x 2 + y 2 + z 2 (power meas iequality) 6 The result ca be obtaied by usig Mikowski s iequality repeatedly: a 2 + b 2 + a b a 2 + b 2 (a + a 2 ) 2 + (b + b 2 ) a 2 + b 2 (a + a 2 + a ) 2 + (b + b 2 + b ) a 2 + b 2 (a + a a ) 2 + (b + b b ) 2 Aother way to thik about it is geometrically Cosider a sequece of poits i the plae P k = (x k, y k ), k = 0,,, such that (x k, y k ) = (x k + a k, y k + b k ) for k =,, The the left had side of the iequality is the sum of the distaces betwee two cosecutive poits, while the right had side is the distace betwee the first oe ad the last oe: d(p 0, P ) + d(p, P 2 ) + + d(p, P ) d(p 0, P ) 7 By the Arithmetic Mea-Geometric Mea Iequality = x x 2 x x + x x Hece f(x, x 2,, x ) O the other had f(,,, ) =, so the miimum value is 8 For x = y = z = we see that S = /2 We will prove that i fact /2 is the miimum value of S by showig that S /2 Note that ( ) 2 ( ) 2 ( ) 2 x y z S = + + y + z z + x x + y Hece by the Cauchy-Schwarz iequality: S (u 2 + v 2 + w 2 ) ( xu y + z + yv z + x + Writig u = y + z, v = z + x, w = x + y we get S 2(x + y + z) (x + y + z) 2,, zw x + y ) 2

9 PUTNAM TRAINING INEQUALITIES 9 hece, dividig by 2(x + y + z) ad usig the Arithmetic Mea-Geometric Mea iequality: S 2 (x + y + z) 2 xyz = 2 9 By the Arithmetic Mea-Harmoic Mea iequality: + + x + y + z =, x y z hece 9 x + y + z O the other had for x = y = z = / the sum is 9, so the miimum value is 9 0 Assume a b c, A B C The 0 (a b)(a B) + (a c)(a C) + (b c)(b C) = (aa + bb + cc) (a + b + c)(a + B + C) Usig A + B + C = π ad dividig by (a + b + c) we get the desired result - Remark: We could have used also Chebyshev s Iequality: ( ) ( ) a A + b B + c C a + b + c A + B + C Assume a i + b i > 0 for each i (otherwise both sides are zero) The by the Arithmetic Mea-Geometric Mea iequality ( ) / a a ( a + + a ), (a + b ) (a + b ) a + b a + b ad similarly with the roles of a ad b reversed Addig both iequalities ad clearig deomiators we get the desired result (Remark: The result is kow as superadditivity of the geometric mea) 2 We have that! is icreasig for, ie, < m =! < m! So 999! > 2000 = (999!)! > 2000! = ((999!)!)! > (2000!)! = = 999! (2000) > 2000! (999) Cosider the fuctio f(x) = (999 x) l (999 + x) Its derivative is f (x) = l (999 + x) x, which is egative for 0 x, because i that iterval x 999 x = l e < l (999 + x) x Hece f is decreasig i [0, ] ad f(0) > f(), ie, 999 l 999 > 998 l 2000 Cosequetly > Let x = log 2 ad y = log 5, so 2 x =, y = 5 The, 27 = = (2 x ) = 8 x, ad 25 = 5 2 = ( y ) 2 = 9 y, hece 8 x > 9 y, but 8 < 9, hece x > y, ie, log 2 > log 5

10 PUTNAM TRAINING INEQUALITIES 0 5 We have b 2 < b } 2 + {{ b + } < b 2 + 2b + = (b + ) 2 But b 2 ad (b + ) 2 are cosecutive a 2 squares, so there caot be a square strictly betwee them 6 We may assume that the leadig coefficiet is + The sum of the squares of the roots of x + a x + + a is a 2 2a 2 The product of the squares of the roots is a 2 Usig the Arithmetic Mea-Geometric Mea iequality we have a 2 2a 2 a 2 Sice the coefficiets are ± that iequality is ( ± 2)/, hece Remark: x x 2 x + = (x + )(x ) 2 is a example of th degree polyomial with all coefficiets equal to ± ad oly real roots 7 The give fuctio is the square of the distace betwee a poit of the quarter of circle x 2 + y 2 = 2 i the ope first quadrat ad a poit of the half hyperbola xy = 9 i that quadrat The tagets to the curves at (, ) ad (, ) separate the curves, ad both are perpedicular to x = y, so those poits are at the miimum distace, ad the aswer is ( ) 2 + ( ) 2 = 8 8 Let ( ) ( ) ( ) 2 P =, Q = ( 2 ) ( ) 4 5 ( ) We have P Q = 2 Also < 2 < < 4 < < 2, hece 2P Q, so P 2 P Q =, ad from here we get P 2 4 O the other had we have P < Q < Q, hece P 2 < P Q =, ad from here 2 P < 2 9 The give iequality is equivalet to ( ) (m + )! m + m m = m m < (m + ) m+, m!! which is obviously true because the biomial expasio of (m + ) m+ icludes the term o the left plus other terms 20 Usig the Arithmetic Mea-Geometric Mea iequality we get: { a + a a } a b b 2 b a2 a = b b 2 b From here the desired result follows

11 PUTNAM TRAINING INEQUALITIES 2 We prove it by iductio For = the result is trivial, ad for = 2 it is a simple cosequece of the followig: 0 (a 2 a )(b 2 b ) = (a b + a 2 b 2 ) (a b 2 + a 2 b ) Next assume that the result is true for some 2 We will prove that is is true for + There are two possibilities: If x + = b +, the we ca apply the iductio hypothesis to the first terms of the sum ad we are doe 2 If x + b +, the x j = b + for some j +, ad x + = b k for some k + Hece: 22 We have + a i x i = = a i x i + a j x j + a + x + i j a i x i + a j b + + a + b k i j (usig the iequality for the two-term icreasig sequeces a j, a + ad b k, b + ) a i x i + a j b k + a + b + i j This reduces the problem to case l ( k= a p k ) /p = l ( ) k= ap k p Also, a k as p 0, hece umerator ad deomiator ted to zero as p approaches zero Usig L Hôpital we get l ( ) k= lim ap k k= = lim ap k l a ( k k= p 0 p = l a ) / k = l a k p 0 k= ap k From here the desired result follows 2 Set x = b + c a, y = c + a b, z = a + b c The triagle iequality implies that x, y, ad z are positive Furthermore, a = (y + z)/2, b = (z + x)/2, ad c = (x + y)/2 The LHS of the iequality becomes: y + z 2x + z + x 2y + x + y 2z = 2 k= ( x y + y x + y z + z y + x z + z ) x

12 PUTNAM TRAINING INEQUALITIES 2 24 We have that 2 = 2 22 = 6 < 27 = = 2 The usig a ( + ) = a a we get 2 ( + ) < for 2, ad from here it follows that 2 20 < Cosider the fuctio f(x) = e /x for x > 0 We have f (x) = e /x < 0, f (x) = x 2 e /x ( 2 + ) > 0, hece f is decreasig ad covex x x 4 By covexity, we have 2 (f(e) + f(π)) f ( ) e+π 2 O the other had we have (e + π)/2 <, ad sice f is decreasig, f( e+π 2 ) > f(), ad from here the result follows 26 We have: f(x) f(a) = hece f(x) f(a) for every x = = (x a i ) 2 (a a i ) 2 { (x ai ) 2 (a a i ) 2} (x 2 2a i x a 2 + 2a i a) = x 2 2ax + a 2 = (x a) 2 0, 27 By the Geometric Mea-Arithmetic Mea iequality x + x 2 2 x x 2,, x 00 + x 2 x00 x Multiplyig we get ) ) ) (x + (x x2 2 + x (x 00 + x 2 00 From the system of equatios we get ) ) ) (x + (x x2 2 + x (x 00 + x = 2 00, so all those iequalities are equalities, ie, x + ( x x = 2 = ) 2 = 0 = x =, x 2 x 2 x2 x 2 ad aalogously: x 2 = /x,, x 00 = /x Hece x = /x 2, x 2 = /x,, x 00 = /x, ad from here we get x = 2, x 2 = /2,, x 99 = 2, x 00 = /2

13 PUTNAM TRAINING INEQUALITIES 28 Squarig the iequalities ad movig their left had sides to the right we get Multiplyig them together we get: 0 c 2 (a b) 2 = (c + a b)(c a + b) 0 a 2 (b c) 2 = (a + b c)(a b + c) 0 b 2 (c a) 2 = (b + c a)(b c + a) 0 (a + b c) 2 (a b + c) 2 ( a + b + c) 2, hece, oe of the factors must be zero 29 The aswer is I fact, the sequeces (si x, cos x) ad (/ si x, / cos x) are oppositely sorted, hece by the rearragemet iequality: si x/ cos x + cos x/ si x si x/ si x + cos x/ cos x Equality is attaied at x = π/4 = si 2 x + cos 2 x = 0 By the rearragemet iequality we have for k = 2,,, (assume a +i = a i ): a + a a a a 2 a s a s a 2 s a s a k s a k+ s a k Addig those iequalities we get ( a ( ) + a a ) s a s a 2 s a ad the result follows s a s a + s a 2 s a s a s a =, The aswer is Sice si x we have si 4 (x) si 2 (x), ad aalogously cos 4 (x) cos 2 (x) Hece f(x) = si 4 (x) + cos 4 x si 2 (x) + cos 2 x = O the other had the value s attaied eg at x = 0 2 Let u = v = ab, w = c By the AGM iequality we have u + v + w uvw abc 2 ab + c abc (a + b + c) 2 ab (a + b) The last iequality is equivalet to the desired result Equality happes precisely for u = v = w, ie, c = ab The AGM iequality applied to a,, a shows that b 0 Also, lettig u k = a a, k =,,, u + = a + ad usig agai the AGM iequality we get + u u u + u + + u + u a a a + a a + a + + Multiplyig both sides by + ad subtractig a + +a +a + we get b + b, which is equivalet to the desired result

14 PUTNAM TRAINING INEQUALITIES 4 4 By subadditivity we have that for every positive iteger ad ay y > 0, f(y) f(y) Writig x = y, we get f(x) f(x/) = x f(x/) xf (0), ad the result x/ follows

15 PUTNAM TRAINING INEQUALITIES 5 Mai Iequalities Arithmetic Mea, Geometric Mea, Harmoic Mea Iequalities Let a,, a be positive umbers The the followig iequalities hold: + + a a 2 a a a }{{}}{{} Harmoic Mea Geometric Mea I all cases equality holds if ad oly if a = = a a + + a }{{} Arithmetic Mea 2 Power Meas Iequality The AM-GM, GM-HM ad AM-HM iequalities are particular cases of a more geeral kid of iequality called Power Meas Iequality Let r be a o-zero real umber We defie the r-mea or rth power mea of positive umbers a,, a as follows: ( ) /r M r (a,, a ) = a r i The ordiary arithmetic mea is M, M 2 is the quadratic mea, M is the harmoic mea Furthermore we defie the 0-mea to be equal to the geometric mea: ( ) / M 0 (a,, a ) = a i The, for ay real umbers r, s such that r < s, the followig iequality holds: Equality holds if ad oly if a = = a M r (a,, a ) M s (a,, a ) 2 Power Meas Sub/Superadditivity Let a,, a, b,, b be positive real umbers (a) If r >, the the r-mea is subadditive, ie: M r (a + b,, a + b ) M r (a,, a ) + M r (b,, b ) (b) If r <, the the r-mea is superadditive, ie: M r (a + b,, a + b ) M r (a,, a ) + M r (b,, b ) Equality holds if ad oly if (a,, a ) ad (b,, b ) are proportioal Cauchy-Schwarz ( ) 2 ( a i b i a 2 i )( b 2 i )

16 PUTNAM TRAINING INEQUALITIES 6 4 Hölder If p > ad /p + /q = the ( ) /p ( ) /q a i b i a i p b i p (For p = q = 2 we get Cauchy-Schwarz) 5 Mikowski If p > the ( ) /p ( ) /p ( ) /p a i + b i p a i p + b i p Equality holds iff (a,, a ) ad (b,, b ) are proportioal 6 Norm Mootoicity If a i > 0 (i =,, ), s > t > 0, the ( a s i ) /s ( /t ai) t 7 Chebyshev Let a, a 2,, a ad b, b 2,, b be sequeces of real umbers which are mootoic i the same directio (we have a a 2 a ad b b 2 b, or we could reverse all iequalities) The ( a i b i )( a i b i ) 8 Rearragemet Iequality For every choice of real umbers x x ad y y, ad ay permutatio x σ(),, x σ() of x,, x, we have x y + + x y x σ() y + + x σ() y x y + + x y If the umbers are differet, eg, x < < x ad y < < y, the the lower boud is attaied oly for the permutatio which reverses the order, ie σ(i) = i +, ad the upper boud is attaied oly for the idetity, ie σ(i) = i, for i =,, 9 Schur If x, y, x are positive real umbers ad k is a real umber such that k, the x k (x y)(x z) + y k (y x)(y z) + z k (z x)(z y) 0 For k = the iequality becomes x + y + z + xyz xy(x + y) + yz(y + z) + zx(z + x)

17 PUTNAM TRAINING INEQUALITIES 7 0 Weighted Power Meas Iequality Let w,, w be positive real umbers such that w + + w = Let r be a o-zero real umber We defie the rth weighted power mea of o-egative umbers a,, a as follows: ( ) /r Mw(a r,, a ) = w i a r i As r 0 the rth weighted power mea teds to: ( ) Mw(a 0,, a ) = a w i i which we call 0th weighted power mea If w i = / we get the ordiary rth power meas The for ay real umbers r, s such that r < s, the followig iequality holds: M r w(a,, a ) M s w(a,, a ) Covexity A fuctio f : (a, b) R is said to be covex if f(λx + ( λ)y) λf(x) + ( λ)f(y) for every x, y (a, b), 0 λ Graphically, the coditio is that for x < t < y the poit (t, f(t)) should lie below or o the lie coectig the poits (x, f(x)) ad (y, f(y)) f(y) f(x) f(t) x t y Figure Covex fuctio

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