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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
IntroductionAn algebraic inequality
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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).
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
History of the Erdos-Mordell inequalityMordell’s proof
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
An optimal inequalityThe Lagrange’s method revisitedErdos-Mordell inequality revisited
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
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
An optimal inequalityThe Lagrange’s method revisitedErdos-Mordell inequality revisited
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
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
An optimal inequalityThe Lagrange’s method revisitedErdos-Mordell inequality revisited
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
An optimal inequalityThe Lagrange’s method revisitedErdos-Mordell inequality revisited
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
An optimal inequalityThe Lagrange’s method revisitedErdos-Mordell inequality revisited
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
An optimal inequalityThe Lagrange’s method revisitedErdos-Mordell inequality revisited
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
An optimal inequalityThe Lagrange’s method revisitedErdos-Mordell inequality revisited
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
An optimal inequalityThe Lagrange’s method revisitedErdos-Mordell inequality revisited
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
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).
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
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).
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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)
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
Interpretation of isomorphism
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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.
Zhiqin Lu Erdos-Mordell inequality and beyond
IntroductionAn algebraic inequality
Generalization
A variation of the inequalityMatrix and commutatorTwo conjectures
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).
Zhiqin Lu Erdos-Mordell inequality and beyond