Symmetric polynomials and symmetric mean inequalities

Symmetric polynomials and symmetric mean inequalities Karl Mahlburg Department of Mathematics Louisiana State University Baton Rouge, LA 70803, U.S.A. mahlburg@math.lsu.edu Clifford Smyth Department of Mathematics and Statistics University of North Carolina Greensboro Greensboro, NC 27402, U.S.A. cdsmyth@uncg.edu Submitted: Nov 22, 2012; Accepted: Aug 27, 2013; Published: Sep 6, 2013 Mathematics Subect Classifications: 05E05, 26E60, 60C05 Abstract We prove generalized arithmetic-geometric mean inequalities for quasi-means arising from symmetric polynomials. The inequalities are satisfied by all positive, homogeneous symmetric polynomials, as well as a certain family of nonhomogeneous polynomials; this family allows us to prove the following combinatorial result for marked square grids. Suppose that the cells of a n n checkerboard are each independently filled or empty, where the probability that a cell is filled depends only on its column. We prove that for any 0 l n, the probability that each column has at most l filled sites is less than or equal to the probability that each row has at most l filled sites. Keywords: symmetric means; symmetric polynomials; arithmetic-geometric mean inequality 1 Introduction Let n be a positive integer. We define an n-variable orthant function to be a continuous function F : R n 0 R 0 such that F x = F x 1,..., x n is monotonically increasing in each x i and that also has a strictly increasing diagonal restriction, f F y = F y,..., y. Given an n-variable orthant function F, we define the following n-variable orthant functions associated with F : the quasi-arithmetic mean, the quasi-geometric mean, and the Partially supported by an NSF Postdoctoral Fellowship administered by the Mathematical Sciences Research Institute through its core grant DMS-0441170 and by NSF Grant DMS-1201435.. the electronic ournal of combinatorics 203 2013, #P34 1

quasi-mean; respectively, they are A F x := f 1 F G F x := f 1 F ff x 1 + + f F x n n n f F x 1 n, =1 M F x := f 1 F F x., 1.1 Note that these means have been studied classically see [2], Chapter III. Some care is needed to verify that these definitions are well-defined. One must note that since f F y is strictly increasing and continuous, its range R := f F R 0 is of the form R = [f F 0, M or [f F 0, +, according to the value M = lim y + f F y. Furthermore, f F is a biection and f 1 F is strictly increasing and continuous. Since R is closed under taking arithmetic and geometric means of its elements, A F x and G F x are well-defined. Since F is monotonically increasing in each variable, it also satisfies f F x F x f F x where x := min{x i : 1 i n} and x := max{x i : 1 i n}. This implies that F R n 0 = R and so M F x is well-defined also. If M = A F, G F, or M F, we therefore have that M is a quasi-mean. In particular, M satisfies the following usual properties of a mean: it is continuous and monotonically increasing in each variable, x Mx x for all x R 0, and My,..., y = y for all y 0. Note that the f 1 F in the definition of M ensures the final identity property, My,..., y = y. However, M is not necessarily linearly homogeneous, i.e. we do not necessarily have Mλx = λmx for all λ 0 and x R n 0. The function F x = 1 + x 1 1 + x 2 provides a simple counterexample for each type of M. Since f F and hence f 1 F are strictly increasing, A F and G F are strictly increasing in each variable. An arithmetic-geometric mean inequality between G F and A F also easily follows: for all x R n 0, G F x A F x, with equality if and only if all x i are equal. This is also Theorem 85 of [2], with ψ = log f F, χ = f F, and q 1/n. Note that the classical arithmetic-geometric mean inequality can be recovered from this by setting F x = x 1 x n 1/n. In this paper we study functions whose quasi-means provide a refinement of the preceding arithmetic-geometric mean inequality. Namely, we are interested in S, which we define to be the set of all orthant functions F for which G F x M F x A F x, for all x R n 0. 1.2 Proposition 1.1. The set S satisfies the following properties. 1. If F x 1,..., x n is a homogeneous symmetric polynomial with positive coefficients, then F S. the electronic ournal of combinatorics 203 2013, #P34 2

2. If F i x 1,..., x n S for i = 1, 2, then F 1 F 2 S. Remark 1.2. The set S is not closed under addition. For example, F x = 1 + x 1 x 2 is the sum of two homogeneous functions but G F M F fails to hold when x 1 x 2. For integers l and n with 0 l n we define η l x := l x I, 1.3 I [n] I = where we have adopted the common notation x I := i I x i, along with the convention x = 1. Note that η l is an orthant function if and only if l 1. We denote the diagonal restriction of η l by l n µ l y := η l y,..., y = y. 1.4 In Section 3 we will prove the following result, which states that η l S for 1 l n. Theorem 1.3. If l and n are integers with 0 l n and n 1, then n 1/n µ l x i η l x 1 n µ l x i, for all x R n 0. Equality holds throughout if and only if l = 0, l = n, or x 1 = = x n. These quasi-mean inequalities have an appealing application to combinatorial probability. Let {X i, 1 i, n} be a collection of independent Bernoulli random variables with probability p ; in other words, PX i = 1 = p and PX i = 0 = 1 p. Using these, we further define the random variables C := R i := X i for 1 n, X i for 1 i n. =1 Using Theorem 1.3, we prove bounds relating the distributions of the C s and R i s. Theorem 1.4. Suppose that p 1,..., p n [0, 1]. If l is an integer with 0 l n, then P max {C } l P max {R i} l 1 n 1 i n the electronic ournal of combinatorics 203 2013, #P34 3

Theorem 1.4 has two immediate combinatorial reformulations: one to marked square grids and another to random bipartite graphs. First, suppose that markers are independently placed in the squares of an n by n grid such that the probability that a marker is placed in a square depends only on that square s column. Then for all integers l with 0 l n, the probability that every column has at most l markers is less than or equal to the probability that each row has at most l markers. Alternatively, suppose that G is a finite complete bipartite graph with bipartition A, B, where A = B = n. Let H be a random subgraph of G where each edge e of G is independently selected to belong to H with a probability that depends only on the left vertex e A. If l 0, then P max a A {d Ha} l P max {d Hb} l. b B Remark 1.5. In the grid formulation, both events in the inequality require that at most nl squares be occupied. Similarly, both events in the bipartite graph formulation require that there be at most nl edges. The remainder of the paper is as follows. In Section 2 we study the basic properties of homogeneous symmetric polynomials and the functions in S, and also prove Proposition 1.1. In Section 3 we use Lagrange multipliers and polynomial inequalities to prove Theorem 1.3. We conclude in Section 4, where we describe the relationship between the quasi-mean inequalities and combinatorial probability, and prove Theorem 1.4. 2 Properties of symmetric polynomials and S As found in Chapter 7 of [4], one can define various homogeneous graded bases for the ring of symmetric polynomials on n variables; we will make explicit use of the elementary symmetric polynomials {e x 0 n}, where the degree polynomial is defined as e x := {x I : I [n], I = }. We will also use the monomial symmetric polynomials, which are defined as follows. For a positive integer n, let n := {λ R n 0 : λ 1 λ n }, and define the weight of such a vector λ as λ := λ1 + + λn. If λ n, the monomial symmetric polynomial associated to λ is defined as M λ x := 1 n! σ S n x λ1 σ1 xλn σn. 2.1 Note that this is homogeneous of degree λ. We now recall various inequalities between symmetric polynomials. Suppose that λ i = λ i 1,..., λ i n n for i {1, 2}. We say that λ 1 maorizes λ 2 if and only if λ 1 = λ 2, and λ 1 1 + + λ 1 λ 2 1 + + λ 2 for all 1 < n. In this case, we write λ 1 λ 2. Muirhead s inequalities can be concisely stated as follows. the electronic ournal of combinatorics 203 2013, #P34 4

Theorem Muirhead [3]. Suppose that λ 1, λ 2 n. The inequality is satisfied for all x R n 0 if and only if λ 1 λ 2. M λ1 x M λ2 x 2.2 Note that [3] only contains the case where both λ i have integral parts, while [2] Theorem 45, p. 45 contains the general result above. We also recall the following elementary result from the general theory of series inequalities; see [2] Theorem 368, p. 261. Theorem Rearrangement Inequality. Suppose that a 1... a n 0 and b 1... b n 0. If σ S n is a permutation, then a i b i a i b σi. 2.3 Our final preliminary observations address the inequalities in 1.2 individually. Let L denote the set of orthant functions F that satisfy the left inequality: G F x M F x, for all x R n 0, and let R denote the set of orthant functions that satisfy the right inequality: M F x A F x, for all x R n 0. Clearly we have S = L R. definitions of the quasi-means. The following properties follow immediately from the Proposition 2.1. The classes L and R have the following closure properties: 1. If F i L for i = 1, 2, then F a 1 F b 2 L for all a, b 0. 2. If F i R for i = 1, 2, then af 1 + bf 2 R for all a, b 0. The preceding facts now allow us to prove our first result about S. Proof of Proposition 1.1. We first prove statement 1. Suppose F is a homogeneous symmetric polynomial with positive coefficients and total degree w. It can be written as F x = k a i M λi x, 2.4 where a i > 0, λ i n and λ i = w for all i. Writing A := k a i > 0, Muirhead s theorem 2.2 then implies that A M w/n,...,w/n x F x A M w,0,...,0 x, for all x R n 0. 2.5 the electronic ournal of combinatorics 203 2013, #P34 5

We finish the proof by showing that this is equivalent to the statement F S. Since f F y = F y,..., y = Ay w, we find that the leftmost expression in 2.5 is the same as A 1 n 1/n n! x 1 x n w/n n! = f F x i. Similarly, the rightmost expression in 2.5 equals A 1 n! xw 1 + + x w n n 1! = 1 n f F x i. This completes the first part of the proof. We now turn to statement 2. Throughout this part of the proof we write f i as shorthand for f Fi. By part 1 of Proposition 2.1 we need only show that if F i S for i = 1, 2, then F 1 F 2 R. This is equivalent to showing that F 1 F 2 x 1 n f F1 F 2 x, 2.6 and we can immediately rewrite f F1 F 2 = f 1 f 2. Using the assumption that F i R, we find that the left side of 2.6 satisfies 1 1 F 1 F 2 x f 1 x i f 2 x i = 1 f n n n 2 1 x i1 f 2 x i2. 2.7 i 1,i 2 =1 Define the one-step shift cyclical permutation by σi := i + 1 for 1 i n 1, and σn := 1. Reordering the x i if necessary so that x 1... x n 0, we then further rewrite the sum from 2.7 as 1 n 2 i 1,i 2 =1 f 1 x i1 f 2 x i2 = 1 n 1 n 2 f 1 x i f 2 xσ i. The Rearrangement Inequality, 2.3, now implies that the largest term in the outer summation occurs when = 0, so F 1 F 2 x 1 n f 1 f 2 x i, 2.8 which verifies 2.6. the electronic ournal of combinatorics 203 2013, #P34 6

3 Proof of Theorem 1.3 3.1 Overview In this section we prove Theorem 1.3, devoting most of our effort to the left inequality. In particular, we consider the level sets of η l x and then apply the technique of Lagrange multipliers in order to determine the extremal behavior of Ux := n µ l x 1 n. =1 For each R in the range of η l on R n 0, define the surface ΩR = {x R n 0 : η l x = R}. Given d > 0 and an integer k with 1 k n, we define c k d = de 1 + + e k R n, where e i is the i-th standard basis vector. We refer to a point in R n 0 as a k-diagonal point if it is any coordinate permutation of a point of the form c k d for some d > 0. Lemma 3.1. If 1 l n 1, 1 < R <, and z is a maxima of Ux on ΩR then z is a k-diagonal point for some k with 1 k n. We will prove Lemma 3.1 in Section 3.2 using the method of Lagrange multipliers see Theorem 3.3. By the symmetry of f and η l, if we restrict our attention to a single point x R n 0, we may assume that there is some 0 k n such that x 1,..., x k > 0 and x k+1 = = x n = 0. With this in mind, we generalize the functions defined in 1.3 and 1.4 by setting η l,k x := η l x 1,..., x k, 0,..., 0, 3.1 µ l,k y := η l y,..., y, 0,..., 0. }{{} k times These are related to our earlier definitions by η l = η l,n and µ l = µ l,n, and we also have the further relations η l,k x = I [k] I l x I, µ l,k y = η l,k y,..., y = and finally, η l,k x = η k,k x = k 1 + x i if l k. Lemma 3.2. If 0 l n, 1 k n, and y 0, then µ k l,ny µ n l,ky. The inequality is tight if and only if l = 0, l = n, y = 0 or k = n. l k x, 3.2 We will prove Lemmas 3.1 and 3.2 in Sections 3.2 and 3.3, respectively. We now show how they imply Theorem 1.3. the electronic ournal of combinatorics 203 2013, #P34 7

Proof of Theorem 1.3. If l = 0 or x = 0, then all terms in the two inequalities are equal to 1. If x 1 = = x n = y then all terms are identically equal to µ l y. If l = n, the left inequality is an identity and the right inequality is an application of the arithmeticgeometric mean inequality. In general, the right inequality follows from Proposition 1.1 part 1 and Proposition 2.1 part 2. So it suffices to prove the left inequality in the case where x R n 0, x 0 and 1 l n 1. Since η l is strictly increasing in each variable, a non-zero x is contained in the surface ΩR with R = Rx = η l x > η l 0 = 1. Since η l x 1 + x i for all 1 i n, ΩR is bounded. Since η l x is continuous, ΩR is also closed, and hence compact. This means that there exists at least one point z ΩR at which fx takes its maximum value on ΩR. Since R > 1, z 0 and, by Lemma 3.1, z = c k d, for some 1 k n and d > 0. By Lemma 3.2, if k < n, then fz = µ l,n d k n < µl,k d = η l z = R. However, if k = n, we have fz = µ l d = η l z = R. 3.2 Lagrange multipliers and maxima In this section we prove Lemma 3.1 using Proposition 3.3.1 of [1], p.284. We restate this result as Theorem 3.3, a form more suitable to our purposes. Theorem 3.3 The method of Lagrange multipliers with inequality constraints.. Let f, h 1,..., h m, g 1,..., g r : R n R be continuously differentiable functions. Suppose z is a point at which fx has a local maximum over Ω = {x : h 1 x = = h m x = 0, g 1 x,..., g r x 0}. Suppose also that z is regular, i.e. { h 1 z,..., h m z} { g i z : i Az} is a linearly independent set where Ax := {1 r : g x = 0}. Then there exist unique Lagrange multiplier vectors λ R m and ρ R r 0, such that L x i z, λ, ρ = 0, 1 i n, ρ = 0, Az where Lx, λ, ρ := fx + m λ ih i x + r =1 ρ g x. Proof of Lemma 3.1. Theorem 3.3 applies to the present setting with ΩR as the set Ω, constraint function hx := η l x R, and inequality constraints g i x := x i for 1 i n. The Lagrangian function is then Lx, λ, ρ := Ux + λη l x R + ρ i x i. Let z ΩR be a point at which Ux attains its maximum value over ΩR. Since R > 1, z has at least one non-zero coordinate. By symmetry we may assume z 1, z 2,..., z k > 0 and z k+1 =... = z n = 0 for some k 1. Since hz trivially has the electronic ournal of combinatorics 203 2013, #P34 8

positive coordinates, hz together with g i z = e i, for k + 1 i n, form a linearly independent set. Thus z is regular, and the conditions of Theorem 3.3 are met. The theorem statement now implies that there is a constant λ such that 1 µ l,n z i n µ l,n z i Uz + η λ l z = 0, 1 i k. x i It is trivial to check that Uz 1 and η l x i z 1 for all i with 1 i n. Thus, if we define µ l,n z i γ i := 1 n µ l,n z i η l x i z then γ 1 = = γ k = λ /fz. Note that, 1 i k, γ i = l 1 µ l,n z i l 1 n 1 z i I [k] i I = z I, 1 i k. If, k are non-negative integers, define Z,k := I [k]\{1,2} I = Suppose that z 3,..., z k are fixed. We will show that the equality γ 1 = γ 2 holds if and only if z 1 = z 2. By symmetry, this will then prove that z 1 = = z k, completing the proof that z is a k-diagonal point. Observe that we can write z I. γ 1 = l 1 n 1 z 1 µ l,n z 1 1 + z 2, 3.3 l 2 Z,k + Z l 1,k and γ 2 = l 1 n 1 z 2 µ l,n z 2 1 + z 1. 3.4 l 2 Z,k + Z l 1,k Comparing 3.3 and 3.4, it is now clear that γ 1 = γ 2 if and only if Γ l,k z 1 = Γ l,k z 2, where l 1 n 1 y 1 + y l 2 Z,k + Z l 1,k Γ l,k y :=. 3.5 µ l,n y the electronic ournal of combinatorics 203 2013, #P34 9

To conclude, we show that Γ l,k y is strictly decreasing on [0, and, hence, one-toone. Note that Γ l,k y = Γ l,k y A + B, 1 + y where Γ l,k y := l 1 n 1 y 1 + y µ l,n y and where A, B > 0 are constants. We show that Γ l,k y < 0. This implies Γ l,k y is strictly decreasing and hence Γ l,k y is strictly decreasing as well. Simple algebra shows that n 1 l y l Γ l,k y = 1 n, y and thus n 1 Γ l,ky = l The numerator simplifies to and the proof is complete. z l l lyl 1 l n y + y l l n y 1. µ l,n y 2 l n l y < 0, 3.3 Maorization and polynomial inequalities In this section we use a partial order on polynomials in order to prove Lemma 3.2. Let fy = n 0 cnyn and gy = n 0 dnyn be polynomials in y with real coefficients. We say that f is dominated by g in the coefficient partial order, denoted f g, if and only if cn dn for all n 0. If f g, and cn < dn for some n 0, then we denote this by f g. We write 0 f if and only if f has all coefficients non-negative and 0 f if and only f has all coefficients non-negative and at least one coefficient positive. Proposition 3.4. If a > b 0 are integers, then µ l,a yµ l,b x µ l,a 1 xµ l,b+1 x. We have µ l,a xµ l,b x µ l,a 1 xµ l,b+1 x if and only if, additionally, a b+2, a 1 l, and l 1. The proof of Proposition 3.4 requires the two following lemmas.. the electronic ournal of combinatorics 203 2013, #P34 10

Lemma 3.5. If A, B, M, N are integers with A B 0 and M N 0, then A B A B. M N N M For these same ranges of parameters, the inequality is strict if and only if A > B, M > N, A M, and B N. Proof. If A = B or M = N then both sides of the inequality are identically equal. If M > A, then M > B and both sides are 0. If N > B, then M > B and, again, both sides are 0. We may now assume A > B, M > N, A M, and B N. Canceling factorials, the desired inequality becomes A N M N > B N M N, where, for non-negative integers n and real numbers α, α n := n 1 i=0 x i is the falling factorial. Since α n > β n if α > β n 1 0, we are done. Lemma 3.6. Suppose that f 1, f 2, g 1, g 2 are polynomials. If 0 f 1 f 2 and 0 g 1 g 2, then f 1 g 1 f 2 g 2. If, additionally, f 1 f 2 and 0 g 2, then f 1 g 1 f 2 g 2. Proof. Denoting the coefficients by f i = 0 f i,y and g i = 0 g i,y for i = 1, 2, the conditions of the lemma state that 0 f 1, f 2, and 0 g 1, g 2, for all 0. Writing their products as f 1 g 1 = l 0 a ly l and f 2 g 2 = l 0 b ly l, the coefficients then satisfy a l =,k 0,+k=l f 1, g 1,k,k 0,+k=l f 2, g 2,k = b l, since f 1, g 1,k f 2, g 2,k for all, k 0. If there is additionally some pair, k such that 0 f 1, < f 2, and 0 < g 2,k, then f 1, g 1,k < f 2, g 2,k, and therefore the stronger conclusion a +k < b +k holds. Proof of Proposition 3.4. By definition, proving that µ l,a yµ l,b y µ l,a 1 yµ l,b+1 y is the same as proving that for each 0 m 2l we have M d=m a b d m d M d=m a 1 b + 1, 3.6 d m d where M := max{0, m l} and M := min{l, m}. Likewise, the stronger condition that µ l,a yµ l,b y µ l,a 1 yµ l,b+1 y is equivalent to additionally proving that there is an m with 0 m 2l for which 3.6 is strict. Applying Pascal s identity a d = a 1 d + a 1 d 1 to the left-side and b+1 m d = b m d + b m d 1 to the right, and then cancelling like terms, 3.6 becomes M d=m a 1 b d 1 m d M d=m a 1 b. 3.7 d m d 1 the electronic ournal of combinatorics 203 2013, #P34 11

Furthermore, after a summation index shift all of the terms but one in 3.7 cancel, leaving only d = M on the left-side and d = M on the right: a 1 b a 1 b. M 1 m M M m M 1 If M = 0, then this inequality is trivially satisfied, and thus so is 3.6. We therefore need only consider the case that M = m l. This means that m l, so in this case M = l, and 3.6 is finally equivalent to the inequality a 1 m l 1 b a 1 l l b m l 1 We now apply Lemma 3.5 with A = a 1, B = b, M = l, N = m l 1 to complete the proof. The inequality easily follows. It is also easy to see that 3.6 is strict in the cases where a b + 2, a 1 l, and 2l m l + 1 and hence l 1.. Remark 3.7. Proposition 3.4 and its proof can also be interpreted combinatorially. In particular, consider two rows consisting of a and b square cells, respectively. The y m coefficient in l a l b y i y i i=0 is the number of ways of marking exactly m of the cells subect to the restriction that there are at most l marked cells in each row, and the result then states that if a > b, then there are at least as many ways to mark two rows of length a 1 and b + 1 subect to the same restriction. Corollary 3.8. If y 0, λ 1, λ 2 m each have integer coordinates, and λ 1 λ 2 then m m µ l,λ1 iy µ l,λ2 iy.. Proof. Suppose λ 1 λ 2. By definition, there must then be two indices 1 α < β m such that λ 1 α > λ 2 α and λ 1 β < λ 2 β. Define λ 1 by setting λ 1α := λ 1 α 1, λ 1β := λ 1 β + 1, and λ 1i := λ 1 i for all i α, β. Importantly, it is still true that λ 1 maorizes λ 2. Noting that λ 1 α > λ 2 α λ 2 β > λ 1 β, Proposition 3.4 now states that µ l,λ1 αyµ l,λ1 βy µ l,λ 1 αyµ l,λ 1 βy which, combined with Lemma 3.6, implies that m m µ l,λ1 ix µ l,λ 1 ix. 3.8 the electronic ournal of combinatorics 203 2013, #P34 12

If λ 1 = λ 2, then 3.8 gives the statement of the corollary. Otherwise, the above procedure is repeated a finite number of steps until this is the case. Applying this result with the partitions λ 1 = n k and λ 2 = k n will finally complete the proof of Lemma 3.2. Proof of Lemma 3.2. Corollary 3.8 implies µ l,n y k µ l,k y n. Since this partial order requires that all coefficients be dominated, this immediately implies that µ l,n y k µ l,k y n for all y 0. Clearly, if k = n, l = 0, l = n, or x = 0, then µ l,n x k = µ l,k x n. It therefore remains to be shown that the inequality is strict if 1 l n 1, k n 1 and x > 0. Proposition 3.4 implies that µ l,n µ l,0 µ l,n 1 µ l,1. Following the proof method of Proposition 3.4, we introduce the dummy term µ l,0 = 1 and find µ k l,n = µ k 1 l,n µ l,nµ l,0 µ k 1 l,n µ l,n 1µ l,1 µ n l,k. The second relation follows from Lemma 3.6, and the third follows from Corollary 3.8. Since µ k l,n µn l,k and x > 0, we conclude that µ l,nx k < µ l,k x n 4 Inequalities for sums of Bernoulli random variables In this brief section we describe the relationship between our quasi-mean inequalities in Theorem 1.3 and the distributions of sums of Bernoulli random variables. Proof of Theorem 1.4. The inequality is trivial if l = n, so we henceforth assume that l < n. Furthermore, if p i = 1 for some i with 1 i n, then PC i l = 0 and the inequality is again trivially true. We therefore also assume that p i [0, 1 for each i. All of the events {C l} are independent, and their individual probabilities are given by PC l = n p m 1 p n m. m Thus P max } l 1 n = 0 m l n =1 0 m l n p m 1 p n m. 4.1 m Similarly, the events {R i l} are also independent, and their probabilities are given by PR i l = p 1 p, so P max {R i} l = 1 i n 0 m l I [n] I I =m 0 m l I [n] I I =m I p 1 p I n. 4.2 the electronic ournal of combinatorics 203 2013, #P34 13

Dividing 4.1 and 4.2 by n =1 1 p, we see that the desired inequality is equivalent to the left inequality from Theorem 1.3 with x i = p i /1 p i for 1 i n. References [1] D. Bertsekas, Nonlinear Programming, Athena Scientific, Belmont, MA, 1995. [2] G. Hardy, J. Littlewood, and G. Pólya, Inequalities, 2nd. ed., Cambridge University Press, Cambridge, 1952. [3] R. Muirhead, Some methods applicable to identities of symmetric algebraic functions of n letters, Proc. Edinburgh Math. Soc. 21 1902/03, 144 157. [4] R. Stanley, Enumerative Combinatorics, vol. 2, Cambridge University Press, Cambridge, 1999. the electronic ournal of combinatorics 203 2013, #P34 14