Introduction An algebraic inequality Generalizationzlu/talks/2007-uci-mathclub/ucimathclub.pdf · Introduction An algebraic inequality Generalization History of the Erd¨os-Mordell

IntroductionAn algebraic inequality

Generalization

Erdos-Mordell inequality and beyond

Zhiqin Lu

The Math ClubUniversity of California, Irvine

November 28, 2007

Zhiqin Lu Erdos-Mordell inequality and beyond


Generalization

History of the Erdos-Mordell inequalityMordell’s proof

In 1935, the following problem proposal appeared in the“Advanced Problems” section of the American MathematicalMonthly:

Advanced section, Problem 3740American Math. Monthly, 42, 1935.

3740. Proposed by Paul Erdos, The University of Manchester,England.



Generalization


In 1935, the following problem proposal appeared in the“Advanced Problems” section of the American MathematicalMonthly:

Advanced section, Problem 3740American Math. Monthly, 42, 1935.

3740. Proposed by Paul Erdos, The University of Manchester,England.



Generalization


From a point P inside a given triangle ABC the perpendicularsPPA,PPB,PPC are drawn to its sides. Prove that

PA + PB + PC ≥ 2(PPA + PPB + PPC).



Generalization


From MathWorld

This inequality was proposed by Erdos (1935), and solved byMordell and Barrow (1937) two years later. Elementary proofswere subsequently found by Kazarinoff in 1945 (Kazarinoff1962, p. 78) and Bankoff (1958).Oppenheim (1961) and Mordell (1962) also showed that

PA× PB × PC ≥ (PPB + PPC)(PPC + PPA)(PPA + PPB).

(These results are all in Amer. Math. Monthly.)There are many different proofs, simplifications, andgeneralizations of the result.



Generalization


Proof given by Mordell, 1937

The Mordell’s proof was given two years later and wasregarded as a “simple proof”. However, a good knowledge intrigonometry and algebra is needed.



Generalization


c

y x

We let ∠C = γ, ∠A = α, and ∠B = β. Law ofsine:

PAPB = c sin γ.

Law of cosine

PAP2B = y2 + x2 − 2yx cos(π − γ).

Thus we have

c2 sin2 γ = y2 + x2 + 2yx cos γ.



Generalization


c

y x


PAPB = c sin γ.

Law of cosine

PAP2B = y2 + x2 − 2yx cos(π − γ).

Thus we have




Generalization


c

y x


PAPB = c sin γ.

Law of cosine

PAP2B = y2 + x2 − 2yx cos(π − γ).

Thus we have




Generalization


c

y x


PAPB = c sin γ.

Law of cosine

PAP2B = y2 + x2 − 2yx cos(π − γ).

Thus we have




Generalization


Lagrange’s method of complete square

Sinceα+ β + γ = π,

we have

cos γ = − cos(α+ β) = − cosα cosβ + sinα sinβ.

Then we have (The Lagrange’s complete square method)

y2 + x2 + 2yx cos γ = (y cosα− x cosβ)2 + (y sinα+ x sinβ)2



Generalization


Lagrange’s method of complete square

Sinceα+ β + γ = π,

we have

cos γ = − cos(α+ β) = − cosα cosβ + sinα sinβ.

Then we have (The Lagrange’s complete square method)

y2 + x2 + 2yx cos γ = (y cosα− x cosβ)2 + (y sinα+ x sinβ)2



Generalization


Remeber, by the laws of sine and cosine, we have

c2 sin2 γ = (y cosα− x cosβ)2 + (y sinα+ x sinβ)2.

Thus we havec sin γ ≥ y sinα+ x sinβ.

Or, in a more symmetric way

c ≥ ysinαsin γ

+ xsinβsin γ



Generalization


Using the same method, we have

b ≥ zsinαsinβ

+ xsin γsinβ

a ≥ ysin γsinα

+ zsinβsinα

togethe with

c ≥ ysinαsin γ

+ xsinβsin γ

we geta + b + c ≥ 2(x + y + z),

as desired.



Generalization


In summary, we have used the following tools:The laws of sine and cosine;The trigonometric addition formula;The Lagrange’s complete square;Most imporantly, the fact a2 + b2 ≥ 2ab.



Generalization





Generalization





Generalization





Generalization

An optimal inequalityThe Lagrange’s method revisitedErdos-Mordell inequality revisited

To repeat, let a,b be real numbers. Then we have

a2 + b2 ≥ 2ab.

However, for three real numbers, the inequality

a2 + b2 + c2 ≥ 2(ab + bc + ca)

is not correct. The correct inequality is

a2 + b2 + c2 ≥ ab + bc + ca,

which can be proved by using

(a− b)2 + (b − c)2 + (c − a)2 ≥ 0.



Generalization



a2 + b2 ≥ 2ab.


a2 + b2 + c2 ≥ 2(ab + bc + ca)


a2 + b2 + c2 ≥ ab + bc + ca,


(a− b)2 + (b − c)2 + (c − a)2 ≥ 0.



Generalization



a2 + b2 ≥ 2ab.


a2 + b2 + c2 ≥ 2(ab + bc + ca)


a2 + b2 + c2 ≥ ab + bc + ca,


(a− b)2 + (b − c)2 + (c − a)2 ≥ 0.



Generalization


We can prove that the above inequality is optimal in the sensethat if

a2 + b2 + c2 ≥ s(ab + bc + ca)

is true for any s, then s ≤ 1. For details, see

Zhiqin LuAn optimal inequalityto appear in The Math Gazette.

The URL is

http://math.uci.edu/ zlu/publications/non-research-papers/20060311.pdf



Generalization


We can prove that the above inequality is optimal in the sensethat if

a2 + b2 + c2 ≥ s(ab + bc + ca)

is true for any s, then s ≤ 1. For details, see

Zhiqin LuAn optimal inequalityto appear in The Math Gazette.

The URL is

http://math.uci.edu/ zlu/publications/non-research-papers/20060311.pdf



Generalization


Lagrange’s method revisited

We have the following algebraic result:

TheoremLet α+ β + γ = π. Then

a + b + c ≥ 2√

bc cosα+ 2√

ca cosβ + 2√

ab cos γ

In particular, if α = β = γ = π/3, we get the result we knewbefore

a + b + c ≥√

bc +√

ca +√

ab.



Generalization


The one line proof

As before, we have

cos γ = − cos(α+ β) = − cosα cosβ + sinα sinβ

Using the Lagrange’s complete square, we have

a + b + c − 2√

bc cosα− 2√

ca cosβ − 2√

ab cos γ

= (√

c −√

a cosβ −√

b cosα)2 + (√

a sinβ −√

b sinα)2 ≥ 0.



Generalization


The one line proof

As before, we have

cos γ = − cos(α+ β) = − cosα cosβ + sinα sinβ

Using the Lagrange’s complete square, we have

a + b + c − 2√

bc cosα− 2√

ca cosβ − 2√

ab cos γ

= (√

c −√

a cosβ −√

b cosα)2 + (√

a sinβ −√

b sinα)2 ≥ 0.



Generalization


Let ∠BPC = α, ∠CPA = β, and ∠APB = γ.Then we have

α+ β + γ = 2π

We can prove that

z ≤√

ab cos(γ/2)

x ≤√

bc cos(α/2)

y ≤√

ca cos(β/2)

Note that α/2 + β/2 + γ/2 = π, theinequality a + b + c ≥ 2(x + y + z) follows.



Generalization



α+ β + γ = 2π

We can prove that

z ≤√

ab cos(γ/2)

x ≤√

bc cos(α/2)

y ≤√

ca cos(β/2)




Generalization



α+ β + γ = 2π

We can prove that

z ≤√

ab cos(γ/2)

x ≤√

bc cos(α/2)

y ≤√

ca cos(β/2)




Generalization

A variation of the inequalityMatrix and commutatorTwo conjectures

We rewrite the above theorem

a + b + c ≥ 2√

bc cosα+ 2√

ca cosβ + 2√

ab cos γ

into the following form

a2 + b2 + c2 − 2bc cosα− 2ca cosβ − 2ab cos γ ≥ 0

Or

(a +b + c)2 ≥ 2ab(1+ cos γ)+2bc(1+ cosα)+2ca(1+ cosβ).



Generalization


We rewrite the above theorem

a + b + c ≥ 2√

bc cosα+ 2√

ca cosβ + 2√

ab cos γ

into the following form

a2 + b2 + c2 − 2bc cosα− 2ca cosβ − 2ab cos γ ≥ 0

Or

(a +b + c)2 ≥ 2ab(1+ cos γ)+2bc(1+ cosα)+2ca(1+ cosβ).



Generalization


By the double angle formula, we have

(a + b + c)2 ≥ 4ab cos2 γ

2+ 4bc cos2 α

2+ 4ca cos2 β

2

which can be re-written as

(a2 + b2 + c2)2 ≥ 4a2b2 cos2 γ

2+ 4b2c2 cos2 α

2+ 4c2a2 cos2 β

2



Generalization


By the double angle formula, we have

(a + b + c)2 ≥ 4ab cos2 γ

2+ 4bc cos2 α

2+ 4ca cos2 β

2

which can be re-written as

(a2 + b2 + c2)2 ≥ 4a2b2 cos2 γ

2+ 4b2c2 cos2 α

2+ 4c2a2 cos2 β

2



Generalization


The key observation

We let ~A, ~B, ~C be vectors in R3 with length a,b, c, respectively.Assume that the angle between ~A, ~B, ~C areπ/2 + α/2, π/2 + β/2, π/2 + γ/2, respectivley. Then we have

(|~A|2 + |~B|2 + |~C|2)2 ≥ 4(|~A× ~B|2 + |~B × ~C|2 + |~C × ~A|2)



Generalization


Review of the cross product

Let ~A, ~B be two vectors. By definition ~A× ~B is a vector. Let thecomponents of ~A, ~B be (x1, x2, x3) and (y1, y2, y3), respectively.Then

~A× ~B = ~C,

where

~C = (x2y3 − x3y2, x1y3 − x3y1, x1y2 − x2y1).

We also have|~C| = |~A| · |~B| sinα,

where α is the angle between ~A and ~B.



Generalization


Review of the cross product

Let ~A, ~B be two vectors. By definition ~A× ~B is a vector. Let thecomponents of ~A, ~B be (x1, x2, x3) and (y1, y2, y3), respectively.Then

~A× ~B = ~C,

where

~C = (x2y3 − x3y2, x1y3 − x3y1, x1y2 − x2y1).

We also have|~C| = |~A| · |~B| sinα,

where α is the angle between ~A and ~B.



Generalization


The definition of the cross product is somewhat mysterious. Wecan make the following matrix interpretation. Recall that we candefine the product of two n × n matrices. For example(

a bc d

)·(

x yz w

)=

(ax + bz ay + bwcx + dz cy + dw

)In general, for two matrices, we have AB 6= BA. Thus we definethe commutator of the matrices as

[A,B] = AB − BA

which measures the non-commutativity of the matrices.



Generalization



a bc d

)·(

x yz w

)=



[A,B] = AB − BA




Generalization



a bc d

)·(

x yz w

)=



[A,B] = AB − BA




Generalization



a bc d

)·(

x yz w

)=



[A,B] = AB − BA




Generalization


An matrix definition of the cross product

We make the following identification:

(x1, x2, x3)→

0 x3 −x2−x3 0 x1x2 −x1 0

(y1, y2, y3)→

0 y3 −y2−y3 0 y1y2 −y1 0



Generalization


Then we have

~A× ~B →

0 x3 −x2−x3 0 x1x2 −x1 0

,

0 y3 −y2−y3 0 y1y2 −y1 0

In mathematics, this phenomena is called isomorphism.



Generalization


Then we have

~A× ~B →

0 x3 −x2−x3 0 x1x2 −x1 0

,

0 y3 −y2−y3 0 y1y2 −y1 0

In mathematics, this phenomena is called isomorphism.



Generalization


Interpretation of isomorphism



Generalization


Using the notion of commutator, we have the following result

TheoremLet A,B,C be 3× 3 skew-symmetric matrices with zerodiagonal parts. Then we have

(||A||2 + ||B||2 + ||C||2)2 ≥ 8(||[A,B]||2 + ||[B,C]||2 + ||[C,A]||2).

The above inequality implies the Erdos-Mordell inequality.



Generalization


Using the notion of commutator, we have the following result

TheoremLet A,B,C be 3× 3 skew-symmetric matrices with zerodiagonal parts. Then we have

(||A||2 + ||B||2 + ||C||2)2 ≥ 8(||[A,B]||2 + ||[B,C]||2 + ||[C,A]||2).

The above inequality implies the Erdos-Mordell inequality.



Generalization


Conjecture (Normal scalar curvature conjecture)Let A1, · · · ,Am be n × n symmetric matrices. Then we have

(∑||Ai ||2)2 ≥ 2

∑i<j

||[Ai ,Aj ]||2.

Conjecture (Bottcher-Wenzel Conjecture)Let A,B be two n × n matrices. Then

2||[A,B]||2 ≤ (||A||2 + ||B||2)2.

In summer of 2007, I proved both conjectures.



Generalization



(∑||Ai ||2)2 ≥ 2

∑i<j

||[Ai ,Aj ]||2.


2||[A,B]||2 ≤ (||A||2 + ||B||2)2.




Generalization



(∑||Ai ||2)2 ≥ 2

∑i<j

||[Ai ,Aj ]||2.


2||[A,B]||2 ≤ (||A||2 + ||B||2)2.




Generalization


I made the following conjecture:

Conjecture (Zhiqin Lu)Let A,B be the bounded trace class operators in a separableHilbert space. Then we have

2||[A,B]||2 ≤ (||A||2 + ||B||2)2,

where the normal is defined as

||A|| =√

Tr(A∗A).



Generalization


Thank you!

Q.E.D.(quod erat demonstrandum)


Documents

Introduction An algebraic inequality Generalizationzlu/talks/2007-uci-mathclub/ucimathclub.pdf · Introduction An algebraic inequality Generalization History of the Erd¨os-Mordell