THE CAUCHY – SCHWARZ MASTER CLASS - PART 12 ppsx

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 fluxio

Symmetric Sums

The kth elementary symmetric function of the n variables x1, x2, , x n

is the polynomial defined the formula

e k (x1, x2, , x n) = 

1≤i1<i2< ···<i k ≤n

x i1x i2· · · x i k

The first three of these polynomials are simply

e0(x1, x2, , x n) = 1, e1(x1, x2, , x n) = x1+ x2+ + xn ,

and e2(x1, x2, , x n) = 

1≤j<k≤n

x j x k ,

while the nth elementary symmetric function is simply the full product

e n (x1, x2, , x n ) = x1x2· · · 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 − x1)(t − x2)· · · (t − x n), then it also has the representation

P (t) = t n −e1(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 (x1, x2, , 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) = Ek(x1, x2, , x n) = e k (x1, x2n , , x 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)≤ E2

k(x) for 0 < k < n (12.2)

and check that they imply Maclaurin’s inequalities which assert that

E n 1/n(x)≤ E 1/(n −1)

n−1 (x)≤ · · · ≤ E2(x)1/2 ≤ E1(x) (12.3)

for all x = (x1, x2, , x n ) such that x k ≥ 0 for all 1 ≤ k ≤ n.

Orientation and the AM-GM Connection

If we take n = 3 and set x = (x, y, z), then Maclaurin’s inequalities

simply say

(xyz) 1/3 ≤



xy + xz + yz

3

1/2

≤ x + y + z

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 (x1x2· · · x n) 1/n and the arithmetic mean

(x1+ x2+· · · + 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)

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 Lk denotes the line determined by the points (0, 0) = (0, log E1(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 Lk 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 Lk is log Ek(x)/k, we find log Ek (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 (x1, x2, , x n), 1≤ k ≤ n If we write the identity (12.1) in the form

P (t) = (t − x1)(t − x2)· · · (t − x n)

=

n



k=0

(−1) k



n k



E k (x1, x2, , 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 (x1, x2, , x n )t

n−k−1

=

n−1



k=0

(−1) k



n − 1 k



E k (x1, x2, , x n )t n −k−1 ,

where in the second line we used the familiar identity



n

k



n − k

n!

k!(n − k)!

n − k

(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) has n real roots in [a, b], and Rolle’s theorem tells

us that the derivative P  (x) must have n − 1 real roots in [a, b] If we

denote these roots by{y1, y2, , y n−1 }, then we also have the identity Q(t) = 1

n P

 (t) = (t − y1)(t − y2)· · · (t − y n −1)

=

n−1

k=0

(−1) k



n − 1 k



E k (y1, y2, , 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 (x1, x2, , x n ) = E k (y1, y2, , 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 = (x1, x2, , x n ) while the right side is a function of the n − 1 vector

y = (y1, y2, , y n−1) Thus, if we can prove a relation such as

0≤ F (E0(y), E1(y), , E n−1(y)) for all y∈ [a, b] n −1 ,

then it follows that we also have the relation

0≤ F (E0(x), E1(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 E0(y), E1(y), , En−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 E0(x), E1(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 Hn which asserts that

E j −1 (x1, x2, , x n)Ej+1(x1, x2, , x n)≤ E2

j (x1, x2, , x n) (12.7)

for all x ∈ R n and all 1 < j < n For n = 1 this assertion is empty, so

182 Symmetric Sums

our induction argument begins with H2, in which case we just need to prove one inequality,

E0(x1, x2)E2(x1, x2)≤ E2

1(x1, x2) or x1x2≤



x1+ x2

2

2

. (12.8)

As we have seen a dozen times before, this holds for all real x1 and x2 because of the trivial bound (x1− x2)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 H3which consists of the two assertions:

E0(x1, x2, x3)E2(x1, x2, x3)≤ E2

1(x1, x2, x3), (12.9)

E1(x1, x2, x3)E3(x1, x2, x3)≤ E2

2(x1, x2, x3). (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 H3 virtually for free

To complete the proof of H3, 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



x1+ x2+ x3

3



{x1x2x3} ≤



x1x2+ x1x3+ x2x3

3

2

. (12.11)

This bound is trivial if x1x2x3= 0, so there is no loss of generality if we

assume x1x2x3= 0 We can then divide our bound by (x1x2x3)2 to get

1

3



1

x1x2 +

1

x1x3 +

1

x2x3



≤ 1

9

 1

x1 +

1

x2 +

1

x3

2

which may be expanded and simplified to

1

x1x2

x1x3

x2x3 ≤ 1

x2 + 1

x2 + 1

x2.

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/x1, 1/x3, 1/x2) and (1/x2, 1/x1, 1/x3), or, a bit more generally, one can sum the three AM-GM bounds

1

x j x k ≤ 1

2

 1

x2

j

+ 1

x2



1≤ j < k ≤ 3.

Thus, the proof of H3 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 Hn consists of n −1 inequalities which may be

viewed in two groups First, for x = (x1, x2, , x n) we have the n − 2

inequalities which involve only E j(x) with 0≤ j < n,

E k−1 (x)Ek+1(x) ≤ E2(x) for 1≤ k < n − 1, (12.12)

then we have one final inequality which involves En(x),

E n−2 (x)E n(x)≤ E2

In parallel with the analysis of H3, we now see that all of the inequalities

in the first group (12.12) follow from the induction hypothesis Hn 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 xj 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

x1· · · ˆx j · · · ˆx k · · · x n



x1x2· · · x n

≤

 1

n

n



j=1

x1x2· · · ˆx j · · · x n

2

In parallel with our earlier experience, we note that there is no loss of

generality in assuming x1x2· · · x n = 0 After division by (x1x2· · · 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

We could now stick with the pattern that worked for H3, 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

E0(1/x1, 1/x2, , 1/x n )E2(1/x1, 1/x2, , 1/x n)

≤ E2

1(1/x1, 1/x2, , 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 yk+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, yk), and (k + 1, yk+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 x1 = x2 =· · · = 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, and z one has the bound

x2y3+ x2z3+ y2x3+ y2z3+ z2x3+ z2y3

≤ xy4+ xz4+ yx4+ yz4+ zx4+ zy4, (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 a2b3 = (ab4)2(a4b)1 The weighted AM-GM inequality (2.9) then gives us the bound

a2b3= (ab4)2(a4b)1 ≤2

3ab

4+1

3a

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 x2y3+ y2x3≤ xy4+ x4y 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) = 23(1, 4)+13(4, 1), and, by exponentiation, we see that this recaptures us our decomposition a2b3 = (ab4)2(a4b)1 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 = (α1, α2, , α n) and β = (β1, β2, , 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 setS 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 (α1, α2) is in the convex hull of (β1, β2) and (β2, β1) then

x α1y α2 is bounded by a linear combination of x β1y β2 and x β2y β1 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 = (α1, α2, , α n ) and β = (β1, β2, , β n ),

show that for all positive x1, x2, , x n one has the bound



σ∈S n

x α1

σ(1) x α2

σ(2) · · · x α n

σ(n) ≤ 

σ∈S n

x β σ(1)1 x β σ(2)2 · · · x β

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, S3 is the set of six permutations of the set{1, 2, 3}, and we have (x1, x2, x3) = (x, y, z) We also have

(α1, α2, α3) = (2, 3, 0) and (β1, β2, β3) = (1, 4, 0),

and since (2, 3, 0) =23(1, 4, 0) +13(4, 1, 0) we find that α ∈ H(β) Finally,

one has the α-sum



σ∈S3

x α1

σ(1) x α2

σ(2) x α3

σ(3) = x2y3+ x2z3+ y2x3+ y2z3+ z2x3+ z2y3,

while the β-sum is given by



σ∈S3

x β1

σ(1) x β2

σ(2) x β3

σ(3) = xy4+ xz4+ yx4+ yz4+ zx4+ zy4,

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

≤ x2

y +

x2

z +

y2

x +

y2

z +

z2

x +

z2

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

(α1, α2, , α n) = 

τ∈S n

p τ (β τ (1) , β τ (2) , , β τ (n))

where p τ ≥ 0 and 

τ ∈S n

p τ = 1.

Now, if we use the jth coordinate of this identity to express x α σ(j) j as a

product, then we can take the product over all j to obtain the identity

x α1

σ(1) x α2

σ(2) · · · x α n

σ(n)= 

τ ∈S n

x β σ(1) τ(1) x β σ(2) τ(2) · · · x β τ(n)

σ(n)

pτ

.

From this point the AM-GM inequality and arithmetic do the rest of