THE CAUCHY – SCHWARZ MASTER CLASS - PART 12 ppsx

12 Symmetric Sums The kth elementary symmetric function of the n variables x 1 ,x 2 , ,x n is the polynomial defined the formula e k (x 1 ,x 2 , ,x n )=  1≤i 1 <i 2 <···<i k ≤n x i 1 x i 2 ··· x i k . The first three of these polynomials are simply e 0 (x 1 ,x 2 , ,x n )=1,e 1 (x 1 ,x 2 , ,x n )=x 1 + x 2 + + x n , and e 2 (x 1 ,x 2 , ,x n )=  1≤j<k≤n x j x k , while the nth elementary symmetric function is simply the full product e n (x 1 ,x 2 , ,x n )=x 1 x 2 ···x n . These functions are used in virtually every part of the mathematical sciences, yet they draw much of their importance from the connection they provide between the coefficients of a polynomial and functions of its roots. To be explicit, if the polynomial P (t) is written as the product P (t)=(t − x 1 )(t − x 2 ) ···(t − x n ), then it also has the representation P (t)=t n −e 1 (x)t n−1 +···+(−1) k e k (x)t n−k +···+(−1) n e n (x), (12.1) where for brevity we have written e k (x) in place of e k (x 1 ,x 2 , ,x n ). The Classical Inequalities of Newton and Maclaurin The elementary polynomials have many connections with the theory of inequalities. Two of the most famous of these date back to the great Isaac Newton (1642–1727) and the Scottish prodigy Colin Maclaurin 178 Symmetric Sums 179 (1696–1746). Their namesake inequalities are best expressed in terms of the averages E k (x)=E k (x 1 ,x 2 , ,x n )= e k (x 1 ,x 2 , ,x n )  n k  , which bring us to our first challenge problem. Problem 12.1 (Inequalities of Newton and Maclaurin) Show that for all x ∈ R n one has Newton’s inequalities E k−1 (x) · E k+1 (x) ≤ E 2 k (x) for 0 <k<n (12.2) and check that they imply Maclaurin’s inequalities which assert that E 1/n n (x) ≤ E 1/(n−1) n−1 (x) ≤···≤E 2 (x) 1/2 ≤ E 1 (x) (12.3) for all x =(x 1 ,x 2 , ,x n ) such that x k ≥ 0 for all 1 ≤ k ≤ n. Orientation and the AM-GM Connection If we take n =3andsetx =(x,y, z), then Maclaurin’s inequalities simply say (xyz) 1/3 ≤  xy + xz + yz 3  1/2 ≤ x + y + z 3 , which is a sly refinement of the AM-GM inequality. In the general case, Maclaurin’s inequalities insert a whole line of ever increasing expressions between the geometric mean (x 1 x 2 ···x n ) 1/n and the arithmetic mean (x 1 + x 2 + ···+ x n )/n. From Newton to Maclaurin by Geometry For a vector x ∈ R n with only nonnegative coordinates, the values {E k (x):0≤ k ≤ n} are also nonnegative, so we can take logarithms of Newton’s inequalities to deduce that log E k−1 (x) + log E k+1 (x) 2 ≤ log E k (x) (12.4) for all 1 ≤ k<n. In particular, we see for x ∈ [0, ∞) n that Newton’s inequalities are equivalent to the assertion that the piecewise linear curve determined by the point set {(k, log E k (x)) : 0 ≤ k ≤ n} is concave. If L k denotes the line determined by the points (0, 0) = (0, log E 1 (x)) and (k,log E k (x)), then as Figure 12.1 suggests, the slope of L k+1 is never larger than the slope of L k for any k =1, 2, ,n−1. Since the 180 Symmetric Sums Fig. 12.1. If E k (x) ≥ 0 for all 1 ≤ k ≤ n, then Newton’s inequalities are equivalent to the assertion that the piecewise linear curve determined by the points (k, y k ), 1 ≤ k ≤ n, is concave. Maclaurin’s inequalities capitalize on just one part of this geometry. slope of L k is log E k (x)/k, we find log E k (x)/k ≤ log E k+1 (x)/(k + 1), and this is precisely the kth of Maclaurin’s inequalities. The real challenge is to prove Newton’s inequalities. As one might ex- pect for a result that is both ancient and fundamental, there are many possible approaches. Most of these depend on calculus in one way or an- other, but Newton never published a proof of his namesake inequalities, so we do not know if his argument relied on his “method of fluxions.” Polynomials and Their Derivatives Even if Newton took a different path, it does make sense to ask what the derivative P  (t) might tell us about the about the special polynomials E k (x 1 ,x 2 , ,x n ), 1 ≤ k ≤ n. If we write the identity (12.1) in the form P (t)=(t − x 1 )(t − x 2 ) ···(t − x n ) = n  k=0 (−1) k  n k  E k (x 1 ,x 2 , ,x n )t n−k , (12.5) then its derivative is almost a perfect clone. More precisely, we have Q(t)= 1 n P  (t)= n−1  k=0 (−1) k  n k  n − k n E k (x 1 ,x 2 , ,x n )t n−k−1 = n−1  k=0 (−1) k  n − 1 k  E k (x 1 ,x 2 , ,x n )t n−k−1 , where in the second line we used the familiar identity  n k  n − k n = n! k!(n − k)! n − k n = (n − 1)! k!(n − k −1)! =  n − 1 k  . Symmetric Sums 181 If the values x k , k =1, 2, ,n are elements of the interval [a, b], then the polynomial P (t)hasn real roots in [a, b], and Rolle’s theorem tells us that the derivative P  (x)musthaven − 1 real roots in [a, b]. If we denote these roots by {y 1 ,y 2 , ,y n−1 }, then we also have the identity Q(t)= 1 n P  (t)=(t − y 1 )(t − y 2 ) ···(t − y n−1 ) = n−1  k=0 (−1) k  n − 1 k  E k (y 1 ,y 2 , ,y n−1 )t n−k−1 . If we now equate the coefficients in our two formulas for Q(t), we find that for all 0 ≤ k ≤ n − 1 we have the truly remarkable identity E k (x 1 ,x 2 , ,x n )=E k (y 1 ,y 2 , ,y n−1 ). (12.6) Why is It So Remarkable? The left-hand side of the identity (12.6) is a function of the n vector x =(x 1 ,x 2 , ,x n ) while the right side is a function of the n −1 vector y =(y 1 ,y 2 , ,y n−1 ). Thus, if we can prove a relation such as 0 ≤ F (E 0 (y),E 1 (y), ,E n−1 (y)) for all y ∈ [a, b] n−1 , then it follows that we also have the relation 0 ≤ F (E 0 (x),E 1 (x), ,E n−1 (x)) for all x ∈ [a, b] n . That is, any inequality — or identity — which provides a relation be- tween the n −1 quantities E 0 (y),E 1 (y), ,E n−1 (y) and which is valid for all values of y ∈ [a, b] n−1 extends automatically to a corresponding relation for the n−1 quantities E 0 (x),E 1 (x), ,E n−1 (x) which is valid for all values of x ∈ [a, b] n . This presents a rare but valuable situation where to prove a relation for functions of n variables it suffices to prove an analogous relation for functions of just n − 1 variables. This observation can be used in an ad hoc way to produce many special identities which otherwise would be completely baffling, and it can also be used systematically to provide seamless induction proofs for results such as Newton’s inequalities. Induction on the Number of Variables Consider now the induction hypothesis H n which asserts that E j−1 (x 1 ,x 2 , ,x n )E j+1 (x 1 ,x 2 , ,x n ) ≤ E 2 j (x 1 ,x 2 , ,x n ) (12.7) for all x ∈ R n and all 1 <j<n.Forn = 1 this assertion is empty, so 182 Symmetric Sums our induction argument begins with H 2 , in which case we just need to prove one inequality, E 0 (x 1 ,x 2 )E 2 (x 1 ,x 2 ) ≤ E 2 1 (x 1 ,x 2 )orx 1 x 2 ≤  x 1 + x 2 2  2 . (12.8) As we have seen a dozen times before, this holds for all real x 1 and x 2 because of the trivial bound (x 1 − x 2 ) 2 > 0. Logically, we could now address the general induction step, but we first need a clear understanding of the underlying pattern. Thus, we consider the hypothesis H 3 which consists of the two assertions: E 0 (x 1 ,x 2 ,x 3 )E 2 (x 1 ,x 2 ,x 3 ) ≤ E 2 1 (x 1 ,x 2 ,x 3 ), (12.9) E 1 (x 1 ,x 2 ,x 3 )E 3 (x 1 ,x 2 ,x 3 ) ≤ E 2 2 (x 1 ,x 2 ,x 3 ). (12.10) Now the “remarkable identity” (12.6) springs into action. The assertion (12.9) says for three variables what the inequality (12.8) says for two, therefore (12.6) tells us that our first inequality (12.9) is true. We have obtained half of the hypothesis H 3 virtually for free. To complete the proof of H 3 , we now only need prove the second bound (12.10). To make the task clear we first rewrite the bound (12.10) in longhand as  x 1 + x 2 + x 3 3  {x 1 x 2 x 3 }≤  x 1 x 2 + x 1 x 3 + x 2 x 3 3  2 . (12.11) This bound is trivial if x 1 x 2 x 3 = 0, so there is no loss of generality if we assume x 1 x 2 x 3 = 0. We can then divide our bound by (x 1 x 2 x 3 ) 2 to get 1 3  1 x 1 x 2 + 1 x 1 x 3 + 1 x 2 x 3  ≤ 1 9  1 x 1 + 1 x 2 + 1 x 3  2 which may be expanded and simplified to 1 x 1 x 2 + 1 x 1 x 3 + 1 x 2 x 3 ≤ 1 x 2 1 + 1 x 2 2 + 1 x 2 3 . At this stage of our Master Class, this inequality is almost obvious. For a thematic proof, one can apply Cauchy’s inequality to the pair of vectors (1/x 1 , 1/x 3 , 1/x 2 )and(1/x 2 , 1/x 1 , 1/x 3 ), or, a bit more generally, one can sum the three AM-GM bounds 1 x j x k ≤ 1 2  1 x 2 j + 1 x 2 k  1 ≤ j<k≤ 3. Thus, the proof of H 3 is complete, and, moreover, we have found a pattern that should guide us through the general induction step. Symmetric Sums 183 A Pattern Confirmed The general hypothesis H n consists of n−1 inequalities which may be viewed in two groups. First, for x =(x 1 ,x 2 , ,x n )wehavethen −2 inequalities which involve only E j (x) with 0 ≤ j<n, E k−1 (x)E k+1 (x) ≤ E 2 k (x)for1≤ k<n−1, (12.12) then we have one final inequality which involves E n (x), E n−2 (x)E n (x) ≤ E 2 n−1 (x). (12.13) In parallel with the analysis of H 3 , we now see that all of the inequalities in the first group (12.12) follow from the induction hypothesis H n and the identity (12.6). All of the inequalities of H n have come to us for free, except for one. If we write the bound (12.13) in longhand and use ˆx j as a symbol to suggest that x j is omitted, then we see that it remains for us to prove that we have the relation 2 n(n − 1)   1≤j<k≤n x 1 ···ˆx j ···ˆx k ···x n  x 1 x 2 ···x n ≤  1 n n  j=1 x 1 x 2 ···ˆx j ···x n  2 . (12.14) In parallel with our earlier experience, we note that there is no loss of generality in assuming x 1 x 2 ···x n = 0. After division by (x 1 x 2 ···x n ) 2 and some simplification, we see that the bound (12.14) is equivalent to 1  n 2   1≤j<k≤n 1 x j x k ≤  1 n n  j=1 1 x j  2 . (12.15) We could now stick with the pattern that worked for H 3 , but there is a more graceful way to finish which is almost staring us in the face. If we adopt the language of symmetric functions, the target bound (12.15) may be written more systematically as E 0 (1/x 1 , 1/x 2 , ,1/x n )E 2 (1/x 1 , 1/x 2 , ,1/x n ) ≤ E 2 1 (1/x 1 , 1/x 2 , ,1/x n ), and one now sees that this inequality is covered by the first bound of the group (12.12). Thus, the proof of Newton’s inequalities is complete. 184 Symmetric Sums Equality in the Bounds of Newton or Maclaurin From Figure 12.1, we see that we have equality in the kth Maclaurin bound y k+1 /(k +1)≤ y k /k if and only if the dotted and the dashed lines have the same slope. By the concavity of the piecewise linear curve through the points {(j, y j ):0≤ j ≤ n}, this is possible if and only if the three points (k − 1,y k−1 ), (k, y k ), and (k +1,y k+1 ) all lie on a straight line. This is equivalent to the assertion y k =(y k−1 + y k+1 )/2, so, by geometry, we find that equality holds in the kth Maclaurin bound if and only if it holds in the kth Newton bound. It takes only a moment to check that equality holds in each of Newton’s bounds when x 1 = x 2 = ··· = x n , and there are several ways to prove that this is the only circumstance where equality is possible. For us, perhaps the easiest way to prove this assertion is by making some small changes to our induction argument. In fact, the diligent reader will surely want to confirm that our induction argument can be repeated almost word for word while including induction hypothesis (12.7) the condition for strict inequality. Passage to Muirhead David Hilbert once said, “The art of doing mathematics consists in finding that special case which contains all the germs of generality.” The next challenge problem is surely more modest than the examples that Hilbert had in mind, but in this chapter and the next we will see that it amply illustrates Hilbert’s point. Problem 12.2 (A Symmetric Appetizer) Show that for nonnegative x, y,andz one has the bound x 2 y 3 + x 2 z 3 + y 2 x 3 + y 2 z 3 + z 2 x 3 + z 2 y 3 ≤ xy 4 + xz 4 + yx 4 + yz 4 + zx 4 + zy 4 , (12.16) and take inspiration from your discoveries to generalize this result as widely as you can. Making Connections We have already met several problems where the AM-GM inequal- ity helped us to understand the relationship between two homogeneous polynomials, and if we hope to use a similar idea here we need to show that each summand on the left can be written as a weighted geometric mean of the summands on the right. After some experimentation, one Symmetric Sums 185 is sure to observe that for any nonnegative a and b we have the product representation a 2 b 3 =(ab 4 ) 2 3 (a 4 b) 1 3 . The weighted AM-GM inequality (2.9) then gives us the bound a 2 b 3 =(ab 4 ) 2 3 (a 4 b) 1 3 ≤ 2 3 ab 4 + 1 3 a 4 b, (12.17) and now we just need to see how this may be applied. If we replace (a, b) in turn by the ordered pairs (x, y) and (y,x), then the sum of the resulting bounds gives us x 2 y 3 +y 2 x 3 ≤ xy 4 +x 4 y and, in exactly the same way, we can get two analogous inequalities by summing the bound (12.17) for the two pairs (x, z) and (z,x), and the two pairs (y, z) and (z, y). Finally, the sum of the resulting three bounds then gives us our target inequality (12.16). Passage to an Appropriate Generalization This argument can be applied almost without modification to any symmetric sum of two-term products x a y b , but one may feel some un- certainty about sums that contain triple products such as x a y b z c . Such sums may have many terms, and complexity can get the best of us unless we develop a systematic approach. Fortunately, geometry points the way. From Figure 12.2 one sees at a glance that (2, 3) = 2 3 (1, 4)+ 1 3 (4, 1), and, by exponentiation, we see that this recaptures us our decomposition a 2 b 3 =(ab 4 ) 2 3 (a 4 b) 1 3 . Geometry makes quick work of such two-term decompositions, but the real benefit of the geometric point of view is that it suggests useful representation for products of three or more variables. The key is to find the right analog of Figure 12.2. In abstract terms, the solution of the first challenge problem piv- oted on the observation that (2, 3) is in the convex hull of (1, 4) and its permutation (4, 1). Now, more generally, given any pair of n-vectors α =(α 1 ,α 2 , ,α n )andβ =(β 1 ,β 2 , ,β n ), we can consider an anal- ogous situation where α is contained in the convex hull H(β) of the set of points (β τ(1) ,β τ(2) , ,β τ(n) ) which are determined by letting τ run over the set S n of all n! permutations of {1, 2, ,n}. This suggestion points us to a far reaching generalization of our first challenge problem. The result is due to another Scot, Robert Franklin Muirhead (1860–1941). It has been known since 1903, and, at first, it may look complicated. Nevertheless, with experience one finds that it has both simplicity and a timeless grace. 186 Symmetric Sums Fig. 12.2. If the point (α 1 ,α 2 ) is in the convex hull of (β 1 ,β 2 )and(β 2 ,β 1 )then x α 1 y α 2 is bounded by a linear combination of x β 1 y β 2 and x β 2 y β 1 . This leads to some engaging inequalities when applied to symmetric sums of products, and there are exceptionally revealing generalizations of these bounds. Problem 12.3 (Muirhead’s inequality) Given that α ∈ H(β) where α =(α 1 ,α 2 , ,α n ) and β =(β 1 ,β 2 , ,β n ), show that for all positive x 1 ,x 2 , ,x n one has the bound  σ∈S n x α 1 σ(1) x α 2 σ(2) ···x α n σ(n) ≤  σ∈S n x β 1 σ(1) x β 2 σ(2) ···x β n σ(n) (12.18) A Quick Orientation To familiarize this notation, one might first check that Muirhead’s inequality does indeed contain the bound given by our second challenge problem (page 184). In that case, S 3 is the set of six permutations of the set {1, 2, 3}, and we have (x 1 ,x 2 ,x 3 )=(x, y, z). We also have (α 1 ,α 2 ,α 3 )=(2, 3, 0) and (β 1 ,β 2 ,β 3 )=(1, 4, 0), and since (2, 3, 0) = 2 3 (1, 4, 0)+ 1 3 (4, 1, 0) we find that α ∈ H(β). Finally, one has the α-sum  σ∈S 3 x α 1 σ(1) x α 2 σ(2) x α 3 σ(3) = x 2 y 3 + x 2 z 3 + y 2 x 3 + y 2 z 3 + z 2 x 3 + z 2 y 3 , while the β-sum is given by  σ∈S 3 x β 1 σ(1) x β 2 σ(2) x β 3 σ(3) = xy 4 + xz 4 + yx 4 + yz 4 + zx 4 + zy 4 , so Muirhead’s inequality (12.18) does indeed give us a generalization of our first challenge bound (12.16). Finally, before we address the proof, we should note that there is no constraint on the sign of the coordinates of α and β in Muirhead’s inequality. Thus, for example, if we take α =(1/2, 1/2, 0) and take Symmetric Sums 187 Fig. 12.3. The geometry of the condition α ∈ H(β) is trivial in dimension 2, and this figure shows how it may be visualized in dimension 3. In higher dimensions, geometric intuition is still suggestive, but algebra serves as our unfailing guide. β =(−1, 2, 0), then Muirhead’s inequality tells us that for positive x, y, and z one has 2  √ xy + √ xz + √ yz  ≤ x 2 y + x 2 z + y 2 x + y 2 z + z 2 x + z 2 y . (12.19) This instructive bound can be proved in many ways; for example, both Cauchy’s inequality and the AM-GM bound provide easy derivations. Nevertheless, it is Muirhead’s inequality which makes the bound most immediate and which embeds the bound in the richest context. Proof of Muirhead’s Inequality We were led to conjecture Muirhead’s inequality by the solution of our first challenge problem, so we naturally hope to prove it by leaning on our earlier argument. First, just to make the hypothesis α ∈ H(β) concrete, we note that it is equivalent to the assertion that (α 1 ,α 2 , ,α n )=  τ∈S n p τ (β τ(1) ,β τ(2) , ,β τ(n) ) where p τ ≥ 0and  τ∈S n p τ =1. Now, if we use the jth coordinate of this identity to express x α j σ(j) as a product, then we can take the product over all j to obtain the identity x α 1 σ(1) x α 2 σ(2) ···x α n σ(n) =  τ∈S n  x β τ(1) σ(1) x β τ(2) σ(2) ···x β τ(n) σ(n)  p τ . From this point the AM-GM inequality and arithmetic do the rest of [...]... an )1/n ≤ 2 n(n − 1) √ aj ak (12. 22) 1≤j . E 2 2 (x 1 ,x 2 ,x 3 ). (12. 10) Now the “remarkable identity” (12. 6) springs into action. The assertion (12. 9) says for three variables what the inequality (12. 8) says for two, therefore (12. 6) tells us. first group (12. 12) follow from the induction hypothesis H n and the identity (12. 6). All of the inequalities of H n have come to us for free, except for one. If we write the bound (12. 13) in longhand. simply the full product e n (x 1 ,x 2 , ,x n )=x 1 x 2 ···x n . These functions are used in virtually every part of the mathematical sciences, yet they draw much of their importance from the connection they