Lecture 23 Rearrangement Inequality

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1 Lecture 23 Rearragemet Iequality Holde Lee 6/4/ The Iequalities We start with a example Suppose there are four boxes cotaiig $0, $20, $50 ad $00 bills, respectively You may take 2 bills from oe box, 3 bills from aother, 4 bills from aother, ad 5 bills from the remaiig box What is the maximum amout of moey you ca get? Clearly, you d wat to take as may bills as possible from the box with largest-value bills! So you would take 5 $00 bills, 4 $50 bills, 3 $20 bills, ad 2 $0 bills, for a grad total of 5 $ $ $ $0 = $780 () Suppose istead that your arch-emesis (who is t very good at math) is pickig the bills istead, ad he asks you how may bills he should take from each box I this case, to miimize the amout of moey he gets, you d wat him to take as may bills as possible from the box with lowest-value bills So you tell him to take 5 $0 bills, 4 $20 bills, 3 $50 bills, ad 2 $00 bills, for a grad total of 5 $0 + 4 $ $ $00 = $480 (2) The maximum is attaied whe the umber of bills take ad the deomiatios are similarly sorted as i () ad the miimum is attaied whe they are oppositely sorted as i (2) The Rearragemet Iequality formalizes this observatio Theorem (Rearragemet): Let x, x 2,, x ad y, y 2,, y be real umbers (ot ecessarily positive) with x x 2 x, ad y y y, ad let σ be a permutatio of {, 2,, } (That is, σ seds each of, 2,, to a differet value i {, 2,, }) The the followig iequality holds: x y + x 2 y x y x y σ + x 2 y σ2 x y σ x y + x 2 y 2 x y Proof We prove the iequality o the right by iductio o The statemet is obvious for = Suppose it true for Let m be a iteger such that σm = Sice x x m ad y y σ, 0 (x x m )(y y σ ) (3) = x m y + x y σ x m y σ + x y

2 OMC 20 Rearragemet Iequality Lecture 23 Hece x y σ x m y }{{} σm x y σ x y σ x m y σ x y (4) y By the iductio hypothesis, x y σ x m y σ x y σ( ) x y x m y m x y Thus the RHS of (4) is at most x y x y + x y, as eeded To prove the LHS, apply the above with y i istead of y i (otig that egatig a iequality reverses the sig) Remark 2: Equality is attaied o the RHS if ad oly if, for every r, the followig are equal as multisets: {y m x m = r} = {y σm x m = r} To see this, ote that otherwise, usig the procedure above, at some time we will have to switch two uequal umbers y k, y m with uequal correspodig x s, x k x m, ad we get iequality (see (3)) Similarly, equality is attaied o the LHS if ad oly if for every r, {y + m x m = r} = {y σm x m = r} I particular, if the a,, a are all distict ad b,, b are all distict, the equality o the right-had side occurs oly whe σ(m) = m for all m, ad equality o the left-had side occurs oly whe σ(m) = + m for all m The rearragemet iequality ca be used to prove the followig Theorem 3 (Chebyshev): Let a a 2 a ad b b 2 b be two similarly sorted sequeces The a b + a 2 b a b a + a 2 a b + b 2 b a b a b Proof Add up the followig iequalities (which hold by the Rearragemet Iequality): a b + a 2 b 2 a b a b + a 2 b 2 a b a b 2 + a 2 b 3 a b a b + a 2 b 2 a b a b + a 2 b a b a b + a 2 b 2 a b After factorig the left-had side ad dividig by 2, we get the right-had iequality By replacig b i with b i ad usig the above result we get the left-had iequality 2 Problems Give that a, b, c 0, prove a 3 + b 3 + c 3 a 2 b + b 2 c + c 2 a 2 Powers: For a, b, c > 0 prove that 2

3 OMC 20 Rearragemet Iequality Lecture 23 (a) a a b b c c a b b c c a (b) a a b b c c (abc) a+b+c 3 3 Suppose a, a 2,, a > 0 ad let s = a a Prove that a s a I particular, coclude Nesbitt s Iequality for a, b, c > 0 a 4 Prove the followig for x, y, z > 0: (a) x2 y + y2 x x + y (b) x2 + y2 + z2 x + y + z y 2 z 2 x 2 z x y (c) xy z 2 + yz x 2 + zx y 2 x y + y z + z x b a + c + a s a c a + b (IMO 978/2) Let a,, a be pairwise distict positive itegers Show that a 2 + a a (modified ISL 2006/A4) Prove that for all positive a, b, c, ab a + b + bc ac a + c 3(ab + bc + ca) 2(a + b + c) 7 Prove that for ay positive real umbers a, b, c the followig iequality holds: a 2 + bc b + c + b2 + ac c + a + c2 + ab a + b a + b + c 8 (MOSP 2007) Let k be a positive iteger, ad let x, x 2,, x be positive real umbers Prove that ( ) ( ) ( ) ( x k+ ) i x i + x i= i + x i= i= i x k i= i 9 The umbers to 00 are writte o a 0 0 board ( 0 i the first row, etc) We are allowed to pick ay umber ad two of its eighbors (horizotally, vertically, or diagoally but our choice must be cosistet), icrease the umber with 2 ad decrease the eighbors by, or decrease the umber by 2 ad icrease the eighbors by At some later time the umbers i the table are agai, 2,, 00 Prove that they are i the origial order 3

4 OMC 20 Rearragemet Iequality Lecture 23 3 Solutios The sequeces a, b, c ad a 2, b 2, c 2 are similarly sorted Therefore, by the rearragemet iequality, a 2 a + b 2 b + c 2 c a 2 b + b 2 c + c 2 a 2 Sice l is a icreasig fuctio, we take the l of both sides to fid that the iequalities are equivalet to a l a + b l b + c l c b l a + c l b + a l c a l a + b l b + c l c a + b + c (l a + l b + l c) 3 Note the sequeces (a, b, c) ad (l a, l b, l c) are similarly sorted, sice l is a icreasig fuctio The the first iequality follows from Rearragemet ad the secod from Chebyshev 3 Sice both sides are symmetric, we may assume without loss of geerality that a a The s a s a ad s a s a By Chebyshev s ( iequality with (a,, a ) ad s a,, s a ), we get a s a a ( s a (a a ) ) s a s a = ( s s ) s a s a = ( a s ) + s a s a This gives ( a a ) = s a s a a a s a s a 4 (a) Without loss of geerality, x y The x 2 y 2 ad, ie y x (x2, y 2 ) ad (, ) are similarly sorted Thus y x x 2 y + y2 x x2 x + y2 y (b) Lettig a = x, b = y, c = z, the iequality is equivalet to y z x a 2 + b 2 + c 2 ab + bc + ca This is true by the rearragemet iequality applied to the similarly sorted sequeces (a, b, c) ad (a, b, c) 4

5 OMC 20 Rearragemet Iequality Lecture 23 (c) Let a = x 3 y 3 z 2 3, b = x 3 z 3 y 2 3 which was proved i problem, ad c = y 3 z 3 The the iequality to prove becomes x 2 3 a 3 + b 3 + c 3 a 2 b + b 2 c + c 2 a 5 Let b,, b be the umbers a,, a i icreasig order Sice b b ad, by the rearragemet iequality, 2 2 a 2 + a a 2 b 2 + b b 2 However, sice the positive itegers b m are distict ad i icreasig order, we must have b m m This gives the RHS is at least Sice the iequality is symmetric we may assume a b c The We claim that a + b a + c b + c (5) ab a + b ac a + c Ideed, the two iequalities are equivalet to both of which hold a 2 b + abc a 2 c + abc abc + ac 2 abc + bc 2 Thus by Chebyshev applied to (5) ad (6), we get ( ab a + b + ac a + c + bc b + c (6) bc ) ((a + b) + (a + c) + (b + c)) 3(ab + bc + ca) (7) b + c Dividig by 2(a + b + c) gives the desired iequality (Note: The origial problem asked to prove i<j a i a j a i + a j 2(a a ) a i a j whe a,, a This ca be proved by summig (7) over all 3-elemet subsets {a, b, c} of (the multiset) {a,, a }, the dividig This is a rare istace of a geeral iequality followig directly from the 3-variable case!) 7 Sice the iequality is symmetric, we may assume without loss of geerality that a b c The a 2 b 2 c 2 b + c a + c a + b 5 i<j

6 OMC 20 Rearragemet Iequality Lecture 23 Hece by the rearragemet iequality, a 2 Addig bc b+c + ac c+a + ab a+b a 2 + bc b + c b2 c + a + c2 a + b to both sides gives + b2 + ac c + a + c2 + ab a + b b2 c2 c + a + a2 a + b b2 + bc c2 + ac c + a + a2 + ab a + b b(b + c) c(c + a) a(a + b) = + + b + c c + a a + b = a + b + c 8 We apply Chebyshev s iequality twice: ( ) ( ) [ ( ) ( )] ( ) x k i x i x + x i= i x k i i= i= i + x i= i i= ( ) [ ( ) ( )] x k i = x x k i i= i + x i= i i= ( ) ( ) x k+ i + x i i= x k i Ideed, without loss of geerality x x I the first applicatio of Chebyshev we use that the followig are oppositely sorted: i= x k x k x k + x xk + x The secod iequality comes from the fact that f(x) = xk is a icreasig fuctio +x for k, x 0: if x y the x k y k ad x k y k together give x k + x k y y k + xy k x k + x yk + y (A simple derivative calculatio also does the trick) I the secod applicatio of Chebyshev we use that the followig are similarly sorted: x k + x xk + x x x 9 Look for a ivariat! Let a ij = 0(i ) + j, the origial umber i the (i, j) positio i the array Let b ij be the umbers after some trasformatios The P = a ij b ij i,j 0 6

7 OMC 20 Rearragemet Iequality Lecture 23 is ivariat (Ideed, two opposite eighbors of a ij are a ij ± d for some d; the sum chages by ±(2a ij (a ij d) (a ij + d)) = 0 at each step) Iitially, P = i,j 0 a2 ij Suppose that b ij are a permutatio of the a ij s The i,j 0 a 2 ij = i,j 0 a ij b ij By the equality case of the rearragemet iequality, sice the a ij are all distict ad the b ij are all distict, the a ij ad b ij must be sorted similarly, ie a ij = b ij for all i, j Refereces [] A Egel Problem-Solvig Strategies Spriger, 998, New York 7