Inequalities of Olympiad Caliber. RSME Olympiad Committee BARCELONA TECH

Transcription

1 Ineulities of Olymid Clier José Luis Díz-Brrero RSME Olymid Committee BARCELONA TECH

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3 José Luis Díz-Brrero RSME Olymi Committee UPC BARCELONA TECH Bsi fts to rove ineulities Herefter, some useful fts for roving ineulities re resented:. If x y nd y z then holds x z for ny x, y, z R.. If x y nd then x y for ny x, y,, R.. If x y then x z y z for ny x, y, z R. 4. If nd > 0 then for ny, R. 5. If nd < 0 then for ny, R. 6. If x y nd then x y for ny x, y R or, R. 7. Let,, R suh tht then. 8. Let x, y, z, t R suh tht x y z t then xy z xy t xz t y z t. 9. If x R then x 0 with eulity if nd only if x = If A i R nd x i R, i n then holds A x A x A x... A nx n 0, with eulity if nd only if x = x = x =... = x n = 0.

4 José Luis Díz-Brrero RSME Olymi Committee UPC BARCELONA TECH Ineulities Wrm-u. Prove the following sttements:. If > 0, then holds. If < 0, then holds.. Let >, <. Then, > holds.. Let, e ositive numers, then holds. 4. Let, e ositive numers, then holds If >, then > holds. Pietro Mengoli Let, e ositive numers suh tht =, then holds 5. Prove the following sttements:. Let,, e nonnegtive rel numers, then holds Let,, e ositive numers suh tht, then holds / / /. Let,, e the length of the sides of tringle ABC. Then, holds. 4. Let,, e ositive numers suh tht =, then holds Let,, e ositive numers, then holds Nesitt Let,, e ositive numers lying in the intervl 0, ]. Then holds I hoe you enjoy solving the reeding roosls nd e sure tht I will e very lesed heking nd disussing your nie solutions.

5 José Luis Díz-Brrero RSME Olymi Committee UPC BARCELONA TECH SET. Let x, y, z e stritly ositive rel numers. Prove tht x y z y xyz z x z xyz x y xyz. Let x, y, nd z e three distint ositive rel numers suh tht x y z = z y x Prove tht 40xz <.. Let,, nd e ositive rel numers. Prove tht Let,, e ositive rel numers suh tht =. Prove tht Let,, e three ositive numers suh tht =. Prove tht 4 6. Let,, e three ositive numers suh tht =. Prove tht 5 /5 5 5 I hoe you enjoy solving the reeding roosls nd e sure tht I will e very lesed heking nd disussing your nie solutions.

6 José Luis Díz-Brrero RSME Olymi Committee UPC BARCELONA TECH Solutions Wrm-u. Prove the following sttements:. If > 0, then holds. If < 0, then holds.. Let >, <. Then, > holds.. Let, e ositive numers, then holds. 4. Let, e ositive numers, then holds If >, then > holds. Pietro Mengoli Let, e ositive numers suh tht =, then holds 5 Solution. From the ineulity x 0 or x x immeditely follows x /x fter division y x > 0. Eulity holds when x =. Likewise, from x 0 or x x we get x /x fter division y x < 0 ft 5. Eulity holds when x =. An lterntive roof n e otined finding the mximum nd minimum of the funtion fx = x /x. Finlly, utting x = / in the reeding the ineulities limed re roven. Solution. From > nd < or > 0 nd > 0, we hve tht = = > 0 nd we re done. Solution. We hve nd = on ount of men ineulities. Multilying u the reeding we get the ineulity limed. Eulity holds when =. Solution.4 We oserve tht 4 or euivlently 4 = 0 4

7 whih trivilly holds with eulity when = /. Solution.5 Note tht the ineulity given is euivlent to >. Now lying HM-AM ineulity to the ositive numers nd yields = with eulity if nd only if = whih is imossile. So, the ineulity given is strit. Solution.6 On ount tht x y xy, we hve [ ] = = 5 euse from = immeditely follows = nd 4. Eulity holds when = = / nd we re done. Notie tht this ineulity is generliztion of the well-known ineulity sin x sin os x 5 x os x. Prove the following sttements:. Let,, e nonnegtive rel numers, then holds Let,, e ositive numers suh tht, then holds / / /. Let,, e the length of the sides of tringle ABC. Then, holds. 4. Let,, e ositive numers suh tht =, then holds Let,, e ositive numers, then holds Nesitt Let,, e ositive numers lying in the intervl 0, ]. Then holds 5

8 Solution. Alying AM-GM ineulity, we hve = Eulity holds when = = nd we re done. Solution. We egin roving tht. Indeed, this ineulity is eui- vlent to whih trivilly holds on ount of the ft tht 0 <. Next we rove tht To do it, we need the ineulity whih follows immeditely dding u the trivil ineulities,,, nd the identity = tht n e esily heked. Comining the reeding results, we get 0. Now utting = x, = y, = z, we hve x y z xyz x y z xyz Alying this ineulity to the ositive numers /, / nd / yields Inverting terms immeditely follows ineulity terms.. To rove tht we sure oth sides nd we get or euivlently,. It holds s ws roven efore. Finlly, the trivilly holds fter suring its oth sides nd rerrnging Solution. Sine,, re the lengths of the sides of tringle, then >, > nd >. Then, we n write =, =, =. Alying CBS ineulity to the vetors u =,, nd v =,, yields = from whih the sttement follows. Eulity holds when = = nd we re done. 6

9 Solution.4 We hve = = 64 on ount tht from =, nd lying men ineulities, immeditely follows 7, 9 nd = = 7. Eulity holds when = = = /. Solution.5 We hve or 6 from whih immeditely follows Eulity holds when = =. 6 Solution.6 Sine =, then Likewise, we hve Therefore, = =, =. = Eulity holds when t lest two of the,, re eul to one, nd we re done. 7

10 José Luis Díz-Brrero RSME Olymi Committee UPC BARCELONA TECH Solutions. Let x, y, z e stritly ositive rel numers. Prove tht x y z y xyz z x z xyz x y xyz Solution. Alying AM-QM ineulity, yields [ x y z y xyz z x z xyz x from whih follows x y 9 x y y z z x x y z xyz z y xyz z x y y z z x x y z xyz = x z xyz x x y z xyz y ] xyz y xyz x y y z z x x y z xyz euse x y z on ount of AM-GM ineulity. Eulity holds when x = xyz y = z nd we re done. Solution. Setting = x xyz, = y xyz nd = To rove the reeding ineulity we set u = into the CBS ineulity nd we otin z xyz into the sttement yields,,, v =,, 8

11 = [ ] [ ] Now, tking into ount tht = nd lying the AM-GM ineulity twie, we get Therefore, on ount of the reeding, we hve [ ] = 6 nd we re done. Note tht eulity holds when x = y = z.. Let x, y, nd z e three distint ositive rel numers suh tht x y z = z y x Prove tht 40xz <. Solution. From x y z = z y x we otin x z = y x y z nd x z y x y z = y x y z y x y z = x z Sine x z, dividing y oth sides of the reeding exression y x z yields x z y x y z = On the other hnd, sine x, y, nd z re ositive, then 4 x < y x nd 4 z < y z. Therefore, x z 4 x 4 z < x z y x y z = Alying AM-GM ineulity, we otin 4 xz x z, nd 8 xz 4 x 4 z. Multilying u the lst two ineulities, we get 4 8 xz x z 4 x 4 z < 9

12 Thus, xz /8 < 4 nd 8/ xz < = 4 from whih follows 40xz < < = 40. Let,, nd e ositive rel numers. Prove tht Solution. WLOG we n ssume tht from whih immeditely follows tht nd. Sine the first nd the lst seuenes re sorted in the sme wy, y lying rerrngement ineulity, we get Adding u the reeding ineulities, yields 5 6 from whih we otin Tking into ount AM-QM ineulity, we hve [ 5 5 ] from whih the sttement follows. Eulity holds when = = nd we re done. 4. Let,, e ositive rel numers suh tht =. Prove tht Solution. Suring oth terms, we get

13 , Alying CBS to the vetors u =, nd v = 4, 4, 4 we otin From = nd the well-known ineulity immeditely follows 4/. Eulity holds when = = = /. On the other hnd, 4 = = 4/ The lst ineulity holds in ount of HM-GM ineulity. Likewise, 4 nd 4 Adding the reeding ineulities, we otin 4 Therefore, 4 4 = = Eulity holds when = = = /. From the reeding, we hve nd the sttement follows. Eulity holds when = = = /, nd we re done. 5. Let,, e three ositive numers suh tht =. Prove tht 4

14 Solution. First we oserve tht from nd the onstrin, immeditely follows tht. Multilying nd dividing the LHS of the ineulity limed y we hve 7 7 = 8 = 8 In the reeding we hve used the ineulity f/ f, where w α w α w α fα = w w w is the men owered ineulity with weights w = nd w = re done. /α, w =, resetively. Eulity holds when = = = /, nd we Solution. First, using the onstrin, we write the ineulity limed in the most onvenient form 4/ = 4/ To rove the reeding ineulity, we onsider the funtion f : [0, R defined y ft = t 7/ whih is onvex, s n e esily heked. Alying Jensen s ineulity, with = A, = A, = A, where A = ; nd x =, x =, x =, we hve f f f f, or euivlently, A 7/ A 7/ A 7/ A 7/

15 Rerrnging terms, nd fter simlifition, we otin [ ] 7/ [ ] 4/ [ ] 7/ [ ] 4/ = 4/ 4 on ount of the well-known ineulity. Eulity holds when = = = /, nd we re done. 6. Let,, e three ositive numers suh tht =. Prove tht 5 /5 5 5 Solution. To rove the ineulity limed we onsider the funtion f : R R defined y ft = t 5. This funtion is onvex in [0, s n e esily heked. Alying Jensen s ineulity, nmely k fx k f k x k k= k= with nd yields =, = x =, x = [, =, x = 5, 5 5 ] or euivlently, where we hve used the onstrin =. Now, it is esy to see tht 5

16 Indeed, WLOG we n ssume tht from whih follows nd. Now, lying rerrngement ineulity, we get Adding the reeding exressions, we otin Eulity holds when = =. Nesitt s Ineulity Finlly, sustituting this result in the reeding the sttement follows. holds when = = = /, nd we re done. Eulity Solution. We rell tht Hölder s ineulity n e stted s follows: Let,,..., n nd,,..., n, n, e seuenes of ositive rel numers nd let, R suh tht =, then the following ineulity holds n i= i n i= i n i i i= Prtiulrizing the reeding result for n = with = 5/4, = 5 for whih / / =, we get i i i= 5/4 i Putting in the ove result = 8/5, = 8/5, = 8/5, = /5, = /5 yields i= 5 i i= /5, = [ 8/5 5/4 8/5 5/4 8/5 ] [ 5/4 4/5 5 5 ] 5 /5 /5 /5 /5 or euivlently, 4/5 5 /5 5 5 from whih the sttement follows on ount of the onstrin nd Nesitt s ineulity. Eulity holds when when = = = /, nd we re done. Solution y Emilio Fernández Morl. On ount of men ineulities we hve 5 /5 /

17 = 5 /5/ /5 euse = nd /0 /5 = /0 /0 = = nd /5//0 =. Eulity holds when = = = /. /0 5

18 José Luis Díz-Brrero RSME Olymi Committee UPC BARCELONA TECH Some Results Herefter, some lssil disrete ineulities re stted nd roven. We egin with Theorem Generl Men Ineulities Let,,..., n e ositive rel numers. Then, the funtion f : R R defined y α fα = α... α n n /α is nonderesing nd the following limits hold } lim fα = min { i, lim fα = n... n, α i n α 0 } lim fα = mx { i α i n Proof. Let 0 < <. Consider the funtion g : [0, R defined y gα = α /. Sine g α = α / 0, α 0, then g is onvex in [0,. Alying Jensen s ineulity, we get g n α j n gα j n n or n j= n j= α j / from whih follows f f. For negtive vlues of α onsider the funtion hα = α / whih is onvex in, 0. Let L = j= n α lim = lim α... /α α n. Tht is, α α n ln L = [ α lim α α ln α... ] α n n = α lim α ln... α n ln n α α α... α n = ln min i n { i } n j= α j 6

19 from whih follows L = } { i mx i n } min { i. Likewise, we otin tht L = lim fα = i n α. Finlly, denoting y L = lim α 0 fα, we hve [ α ln L = lim α 0 α ln α... ] α n n = lim α 0 α ln α ln... α n ln n α α... α n = ln ln... ln n n = ln n... n from whih immeditely follows lim fα = n... n nd the roof is omlete. α 0 In 8 Cuhy ulished his fmous ineulity s the seond of the two notes on the theory of ineulities tht formed the finl rt of his ook Cours d Anlyse Algériue. Nmely, Theorem Let,,..., n,,,..., n e ny rel numers. Then, the following ineulity holds: n n n k k k= k= Proof. Consider the udrti olynomil Ax = Ax = x n k= k x n k= k k= k n k x k. Then, k= k k n k 0 Sine the reeding eution is nonnegtive its disriminnt must e less or eul thn zero. Tht is, n n n k k 0 k= k= from whih the sttement follows. Eulity holds when the n-tules,,..., n nd,,..., n re roortionl. This omletes the roof. A generliztion of CBS ineulity is the well known ineulity of Hölder. It is stted nd roved in the following theorem. Theorem Let,,..., n nd,,..., n, n e seuenes of ositive rel numers nd let, R suh tht =, then the following ineulity holds n i= i n i= i k k= k= k n i i. Proof. We will rgue y indution. Assume tht the ineulity holds for n nd we hve to rove it for n. In ft, n i= i i = n n i i n n i= i= i i= n i= i n n 7

20 n i n i= n i n i= = n i= i n To omlete the indutive roess, we must verify tht the se when n = lso holds. Indeed, the ineulity, immedite follows from the ft tht the funtion fx = x x, is stritly ositive for ll x > 0. This omletes the roof. Proof. First, we write the ineulity in the most onvenient form n i= i i n i= n i i= i or euivlently, n i= i n i= i Now, using the ower men ineulity, i n i= i x α y β α x β y,. i= i. we hve n i= i n i= i i n i= i n [ i= i n i= i ] i n i= i = n i= i n i= i n i= i n i= i = =, s desired. Eulity holds if nd only if the n-tules,,..., n nd,,..., n re roortionl. Notie tht for = =, we get the ineulity... n n... n... n This is the Cuhy-Bunykowski-Shwrz ineulity. Using Hölder s ineulity n e esily roven the following ineulity of Minkowski. Theorem 4 Let,,..., n nd,,..., n, n e seuenes of ositive rel numers nd let >, e rel numer, then n i i i= n i= i n i= i When < ineulity reverses. 8

21 Proof. Let e = n i= i, then n i i = i= =, nd lying Hölder s ineulity, we otin n i i i i= n n i i i= n = i= i n i= i i= i n i i i i= n i i i= n i i Multilying the reeding ineulity y n i= i i, we get i= n i i i= n i= i n i= i nd the roof is omlete. Eulity holds if nd only if,,..., n nd,,..., n re roortionl. A most useful result is resented in the following Theorem 5 Rerrngement Let,,, n nd,,, n e seuenes of ositive rel numers nd let,,, n e ermuttion of,,, n. The sum S = n n is mximl if the two seuenes,,, n nd,,, n re sorted in the sme wy nd miniml if the two seuenes re sorted oositely, one inresing nd the other deresing. Proof. Let i > j. Consider the sums S = i i j j n n S = i j j i n n We hve otined S from S y swithing the ositions of i nd j. Then S S = i i j j i j j i = i j i j Therefore, This omletes the roof. i > j S > S nd i < j S < S. In the seuel, fter defining the onets of onvex nd onve funtion, the ineulity of Jensen is resented. Definition. Let f : I R R e rel funtion. We sy tht f is onvex onve on I if for ll, I nd for ll t [0, ], we hve ft t f tf When f is onve the ineulity reverses. 9

22 Theorem 6 Let f : I R R e onvex funtion. Let,,..., n e nonnegtive rel numers suh tht... n =. Then for ll k I k n holds n n f k k k f k k= Eulity holds when = =... = n. If f is onve the reeding ineulity reverses. Proof. We will rgue y mthemtil indution. For n =, we hve k= f f f The reeding ineulity holds euse f is onvex nd =. Assume tht n n f k k k f k nd we hve to see tht f k= n k= k k k= n k= k f k when... n n =. Let = n n k= k k. Then, we hve =, k= n k k I nd n =. Now, tking into ount the se when n = k= nd the indutive hyothesis, yields f n k= k k = f = f [ n n n k= n k= k k= k k= k k f k k n n ] k n f n f k n f n nd y the rinile of mthemtil indution PMI the sttement is roven. Eulity holds when = =... = n. 0

23 Think out Pulhr diuntur ue vis lent euty is tht whih, eing seen, leses. This definition, lies well to Mthemtil euty in whih lk of understnding is so often resonsile of lk of lesure. Thoms Auin s definition of Beuty