Journal of Inequalities in Pure and Applied Mathematics

Journal of Inequalities in Pure and Applied Mathematis A NEW ARRANGEMENT INEQUALITY MOHAMMAD JAVAHERI University of Oregon Department of Mathematis Fenton Hall, Eugene, OR 97403. EMail: javaheri@uoregon.edu URL: http://www.uoregon.edu/ javaheri volume 7, issue 5, artile 162, 2006. Reeived 29 April, 2006; aepted 17 November, 2006. Communiated by: G. Bennett Abstrat Home Page 2000 Vitoria University ISSN eletroni): 1443-5756 126-06

Abstrat In this paper, we disuss the validity of the inequality x i x a i x b i+1 ) 2 x 1+a+b)/2 i, where 1, a, b are the sides of a triangle and the indies are understood modulo n. We show that, although this inequality does not hold in general, it is true when n 4. For general n, we show that any given set of nonnegative real numbers an be arranged as x 1, x 2,..., x n suh that the inequality above is valid. 2000 Mathematis Subjet Classifiation: 26Dxx. Key words: Inequality, Arrangement. I would like to thank Omar Hosseini for bringing to my attention the ase a = b = 1 of the inequality 1.2). I would also like to thank Professor G. Bennett for reviewing the artile and making useful omments. 1 Main Statements..................................... 3 2 Proofs.............................................. 5 Referenes Page 2 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

1. Main Statements Let a, b, x 1, x 2,..., x n be nonnegative real numbers. If a + b = 1 then, by the Rearrangement inequality [1], we have 1.1) x a i x b i+1 x i, where throughout this paper, the indies are understood to be modulo n. In an attempt to generalize this inequality, we onsider the following 2 1.2) x a i x b i+1 xi), x i where = a + b + 1)/2. It turns out that if a + b 1 then the inequality 1.2) is false for n large enough f. Prop. 2.2). However, we show that if 1.3) b a + 1, a b + 1, 1 a + b, then the inequality 1.2) is true in the ase of n = 4 f. Prop. 2.1). Moreover, under the same onditions on a, b as in 1.3), we show that one an always find a permutation µ of {1, 2,..., n} suh that f. Prop. 2.4) 2 1.4) x a µi)x b µi+1) xi). x i The onditions in 1.3) annot be ompromised in the sense that if for all nonnegative x 1, x 2,..., x n there exists a permutation µ suh that the onlusion Page 3 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

1.4) holds, then a, b must satisfy 1.3). To see this, let x 1 = x > 0 be arbitrary and x i = 1, i = 2,..., n. Then, for any permutation µ, the inequality 1.4) reads the same as: 1.5) x + n 1)x a + x b + n 2) x + n 1) 2. If the above inequality is true for all x and n, by omparing the oeffiients of n on both sides of the inequality 1.5), we should have x a + x b + x 3 2x 2. Sine x > 0 is arbitrary, 1, a, b and onditions 1.3) follow. The ase of a = b = 1 of 1.2) is partiularly interesting: 1.6) x i x i x i+1 x 3/2 i ) 2. There is a ounterexample to 1.6) when n = 9, e.g. take 1.7) x 1 = x 9 = 8.5, x 2 = x 8 = 9, x 3 = x 7 = 10, x 4 = x 6 = 11.5, x 5 = 12, and subsequently the inequality 1.6) is false for all n 9 f. prop. 2.2)). Proposition 2.1 shows that the inequality 1.6) is true for n 4, and there seems to be a omputer-based proof [2] for the ases n = 5, 6, 7 whih, if true, leaves us with the only remaining ase n = 8. Page 4 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

2. Proofs Applying Jensen s inequality [1, 3.14] to the onave funtion log x gives 2.1) u r v s w t ru + sv + tw, where u, v, w, r, s, t are nonnegative real numbers and r + s + t = 1. If, in addition, we have r, s, t > 0 then the equality ours iff u = v = w. However, if t = 0 and r, s, w > 0 then the equality ours iff u = v. We use this inequality in the proof of the proposition below. Proposition 2.1. Let a, b 0 suh that a + 1 b, b + 1 a and a + b 1. Then for all nonnegative real numbers x, y, z, t, 2.2) x + y + z + t)x a y b + y a z b + z a t b + t a x b ) x + y + z + t ) 2, where = a + b + 1)/2. The equality ours if and only if {a, b} = {0, 1} or x = y = z = t. Proof. We apply the inequality 2.1) to 2.3) and obtain: 2.4) x a y b z u = yz), v = xz), w = xy), r = 1 a, s = 1 b, t = 1 1, 1 a ) yz) + 1 b ) xz) + 1 1 ) xy). Page 5 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

Notie that the assumptions on a, b in the lemma are made exatly so that r, s, t are nonnegative. Similarly, by replaing z with t in 2.4), we have: 2.5) x a y b t 1 a ) yt) + 1 b ) tx) + 1 1 ) xy). Next, apply 2.1) to 2.6) u = x 2, v = xy), w = 1, r = 1 b, s = b, t = 0, and get 2.7) x a+1 y b 1 b ) x 2 + b xy). Similarly, by interhanging a and b, one has 2.8) x a y b+1 1 a ) x 2 + a xy). Adding the inequalities 2.4), 2.5), 2.7) and 2.8) gives: 2.9) Sx a y b 1 x2 + 4 3 ) xy) + 1 a ) yz) + + 1 a ) yt) + 1 b ) tx) 1 b ) xz), Page 6 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

where S = x+y+z+t. There are three more inequalities of the form above that are obtained by replaing the pair x, y) by y, z), z, t) and t, x). By adding all four inequalities or by taking the yli sum of 2.9)) we have 2.10) ST 1 x 2 + 4 2 ) x + z )y + t ) + 2 { xz) + yt) }, where ST stands for the left hand side of the inequality 2.2). The right hand side of the above inequality is equal to 2.11) ) ) 2 x 1 {x + 1 + z ) 2 + y + t ) 2 2x + z )y + t ) }, whih is less than or equal to x ) 2, sine 1. This onludes the proof of the inequality 2.2). Next, suppose the equality ours in 2.2) and so the inequalities 2.4) 2.8) are all equalities. If a = 0 then we have x x b = x ) 2 and so, by the equality ase of Cauhy-Shwarz, the two vetors x, y, z, t) and x b, y b, z b, t b ) have to be proportional. Then either b = = 1 or x = y = z = t. Thus suppose a, b 0. Sine = a = b is impossible, without loss of generality suppose that b. Sine the inequality 2.7) must be an equality, x 2 = x y f. the disussion on the equality ase of 2.1)). Similarly y 2 = y z, z 2 = z t and t 2 = t x. It is then not diffiult to see that x = y = z = t. Let Na, b) denote the largest integer n for whih the inequality 1.2) holds for all nonnegative x 1, x 2,..., x n. By the above proposition, we have Na, b) 4. Page 7 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

Proposition 2.2. Let a, b 0 suh that a + b 1. Then Na, b) <. Moreover, if n Na, b) then the inequality 1.2) is valid for all nonnegative x 1,..., x n. Proof. The proof is divided into two parts. First we show that the inequality 1.2) annot be true for all n. Proof is by ontradition. If a = b = 0 then 1.2) is false for n = 2 e.g. take x 1 = 1, x 2 = 2). Thus, suppose a + b > 0 and that the inequality 1.2) is true for all n. Let f be a non-onstant positive ontinuous funtion on the interval I = [0, 1] suh that f0) = f1). Let ) i 1 2.12) x i = f, y i = x a i x b n i+1) 1/a+b), i = 1,..., n. Sine y i is a number between x i and x i+1 possibly equal to one of them), by the Intermediate-value theorem [3, Th 3.3], there exists t i I i suh that ft i ) = y i. By the definition of integral we have: 2.13) I fx)dx f a+b 1 x)dx = lim I n n 2 1 = lim n n 2 1 lim n n 2 x i x i y a+b i x a i x b i+1 2 xi) = I f x)dx) 2, where we have applied the inequality 1.2) to the x i s. On the other hand, by Page 8 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

the Cauhy-Shwarz inequality for integrals, we have ) 2 fx)dx f a+b x)dx f 1 a+b 2.14) 2 x)f 2 x)dx I I = I I f x)dx) 2, with equality iff f and f a+b are proportional. The statements 2.13) and 2.14) imply that the equality indeed ours. Sine a + b 1 and f is not a onstant funtion, the two funtions f and f a+b annot be proportional. This ontradition implies that 1.2) ould not be true for all n i.e. Na, b) <. Next, we show that 1.2) is valid for all n N. It is suffiient to show that if the inequality 1.2) is true for all ordered sets of k+1 nonnegative real numbers, then it is true for all ordered sets of k nonnegative real numbers. Let y 1,..., y k be nonnegative real numbers and set 2.15) S = k y i, A = k yi a yi+1, b P = k yi. Without loss of generality we an assume P = 1. For eah 1 i k, define an ordered set of k + 1 nonnegative real numbers by setting: y j 1 j i + 1 x j = y j 1 i + 2 j k + 1 Applying the inequality 1.2) to x 1,..., x k+1 gives 2.16) S + y i )A + y a+b i ) P + yi ) 2 = 1 + yi 2 + 2yi. Page 9 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

Adding these inequalities for i = 1,..., k, yields: 2.17) ksa + S i y a+b i + AS k + 2. On the other hand, by the Rearrangement inequality [1] we have 2.18) k yi a yi+1 b k y a+b i, and the lemma follows by putting together the inequalities 2.17) and 2.18). The inequality 1.1) translates to Na, b) = when a + b = 1. We expet that Na, b) as a + b 1. The following proposition supports this onjeture. We define 2.19) A n a, b) = sup x i 2 x a i x b i+1 xi) max x i = 1 1 i n. This number roughly measures the validity of the inequality 1.2). Also let 2.20) σ t = 1 x t n i. By the Hölder inequality [1], if α, β > 0 and α + β = 1 then for any s, t > 0 we have: 2.21) σ α s σ β t σ αs+βt. Page 10 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

Proposition 2.3. Nu, u) is a non-inreasing funtion of u 1/2. Moreover, for all n and a, b 0 2.22) lim a+b 1 A na, b) = 0. Proof. Suppose u > v > 1/2. We show that Nu, u) Nv, v). Without loss of generality we an assume: 2.23) u v < 1 4. By the definition of N = Nv, v), there must exist N + 1 nonnegative integers x 1,..., x N+1 suh that the inequality 1.2) is false and so N+1 N+1 N+1 ) 2 2.24) x i x v i x v i+1 >. x v+1/2 i We show that the nonnegative numbers y i = x u/v i, i = 1,..., N + 1 give a ounterexample to 1.2) when a = b = u. In light of 2.24), one just needs to show N+1 ) 2 / N+1 N+1 ) 2 / N+1 2.25) x i. To prove this, first let 2.26) α = x u+1/2v i x u/v i x u+1/2 i u + 1/2v) u/v u + 1/2v) 1, β = u/v 1 u + 1/2v) 1, s = 1, t = u + 1 2v. Page 11 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

The numbers above are simply hosen suh that α + β = 1 and αs + βt = u/v. We briefly hek that α, β > 0. The denominator of frations above is positive, sine u + 1/2v) v + 1/v)/2 1. This implies β > 0. Now the positivity of α > 0 is equivalent to u1 v) < 1/2. If v 1 then u1 v) 0 < 1/2. So suppose v 1. By using 2.23), we have: 2.27) u1 v) v + 1 ) 1 v) = v 2 + 3 4 4 v + 1 4 < 1 2, for all v 0. Now we an safely plug α, β, s, t in 2.21) and get 2.28) σ α 1 σ β u+1/2v σ u/v. Next, let α = 1 α)/2 and β = 1 β/2. Sine α + β = 1 and α, β > 0, we an use Hölder s inequality 2.21) with α, β instead of α and β and the same s, t as before) and get this time α s + β t = u + 1/2): 2.29) σ 1 α)/2 1 σ 1 β/2 1+1/2v σ u+1/2. Now we square the above inequality and multiply it with 2.28) to obtain: 2.30) σ 1 σ 2 1+1/2v σ u/v σ 2 u+1/2, whih is equivalent to the inequality 2.25). So far we have shown the existene of a ounterexample to 1.2) for a = b = u when n = N + 1. Then Prop. 2.2 gives Nu, u) N = Nv, v) and this onludes the proof of the monotoniity of N. Page 12 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

It remains to prove that A n a, b) onverges to 0 as a+b 1. To the ontrary, assume there exists ɛ > 0 and a sequene a j, b j ) suh that A n a j, b j ) > ɛ and a j + b j 1. Then by definition, for eah j, there exists an n-tuple X j = x 1j,..., x nj ) suh that max x ij = 1 and 2.31) x ij x a j ij xb j i+1j x j ij ) 2 ɛ 2, where j = a j + b j + 1)/2. Sine X j is a bounded sequene, it follows that, along a subsequene j k, the X jk s onverge to some X = x 1,..., x n ). On the other hand, along a subsequene of j k denoted again by j k ), a jk a and b jk b for some a, b 0. Sine a j + b j 1, we have a + b = 1. By taking the limits of the inequality 2.31) along this subsequene, we should have ) 2 2.32) x a i x b i+1 x i ɛ 2 > 0, x i whih ontradits the inequality 1.1). This ontradition establishes the equation 2.22). The next proposition shows that the inequality 1.2) holds if one mixes up the order of the x i s. The proof is simple and makes use of the monotoniity of σ t ) 1/t where σ t is defined by the equation 2.20). It is well-known that σ t ) 1/t is a non-dereasing funtion of t [1, Th. 16]. Proposition 2.4. Let a, b, be as in Proposition 2.1. Then for any given set of n nonnegative real numbers there exists an arrangement of them as x 1,..., x n suh that the inequality 1.2) holds. Page 13 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

Proof. Equivalently, we show that if x 1, x 2,..., x n are nonnegative then there exists a permutation µ of the set {1, 2,..., n} suh that the inequality 1.4) holds. Let 2.33) S = x i, T = x a i x b j. j i Then ST = nσ 1 n 2 σ a σ b nσ a+b ) = n 3 σ 1 σ a σ b n 2 σ 1 σ a+b. Now by the Cauhy- Shwarz inequality [4], σ 2 σ 1 σ a+b. On the other hand by the monotoniity of σ 1/t t, we have σ 1 σ 1/, σ a σ a/, σ b σ b/, and so σ 1 σ a σ b σ 2. It follows from these inequalities that 2.34) ST n 2 n 1)σ 2. Now for a permutation µ of 1, 2,..., n, let: 2.35) A µ = x a µi)x b µi+1). We would like to show that SA µ nσ ) 2 for some permutation µ. It is suffiient to show that the average of SA µ over all permutations µ is less than or equal to nσ ) 2. To show this, observe that the average of SA µ is equal to ST/n 1) and so the laim follows from the inequality 2.34). The symmetri group S n ats on R n in the usual way, namely for µ S n and x 1,..., x n ) R n let 2.36) µ x 1,..., x n ) = x µ1),..., x µn) ). Page 14 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

Let R be a region in R n that is invariant under the ation of permutations i.e. µ R R for all µ). define: 2.37) λr) = x ) 2 1,..., x n ) R x i x a i x b i+1 x i. By Proposition 2.4: 2.38) R µ S n µ λr). In partiular, by taking the Lebesgue measure of the sides of the inlusion above, we get 2.39) vol λr) vol R. n! We prove a better lower bound for vol λr) when n is a prime number similar but weaker results an be proved in general). Proposition 2.5. Let a, b be as in Proposition 2.1 and n be a prime number. Let R R n + be a Lebesgue-measurable bounded set that is invariant under the ation of permutations. Let λr) denote the set of all x 1,..., x n ) R for whih the inequality 1.2) holds. Then 2.40) vol λr) vol R n 1. Page 15 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

Proof. For m {1, 2,..., n 1} let µ m S n and denote the permutation 2.41) µ m i) = mi, where all the numbers are understood to be modulo n in partiular µ m n) = n for all m). Now reall the definition of A µ from the equation 2.35) and observe that: 2.42) n 1 m=1 A µm = = n 1 m=1 x a mix b mi+m = n 1 x a j j=1 m=1 x b j+m = n 1 m=1 j=1 j=1 x a j x a j x b j+m x b i. i j Then, the same argument in the proof of Prop. 2.4 implies that, for some m {1,..., n 1}, we have A µm nσ ) 2. We onlude that 2.43) R n 1 m=1 whih in turn implies the inequality 2.40). µ m λr), Page 16 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006

Referenes [1] G.H. HARDY, J.E. LITTLEWOOD AND G. PÓLYA, Inequalities, Cambridge University Press, 2nd ed. 1988). [2] O. HOSSEINI, Private Communiation, Fall 2006. [3] M.H. PROTTER AND C.B. Jr MORREY, A First Course in Real Analysis, Springer 1997). [4] M.J. STEELE, The Cauhy-Shwarz Master Class: An Introdution to the Art of Mathematial Inequalities, Cambridge University Press 2004). Page 17 of 17 J. Ineq. Pure and Appl. Math. 75) Art. 162, 2006