aarjaneiro/planetmath huggingface dataset:
derivation of geometric mean as the limit of the power mean

derivation of geometric mean as the limit of the power mean

Fix

x

1

,

x

2

,

…

,

x

n

∈

ℝ

+

. Then let

μ

⁢

(

r

)

:=

(

x

1

r

+

⋯

+

x

n

r

n

)

1

/

r

.

For

r

≠

0

, by

definition

μ

⁢

(

r

)

is the

r

th

power mean

of the

x

i

. It is also clear that

μ

⁢

(

r

)

is a

differentiable function

for

r

≠

0

. What is

lim

r

→

0

⁡

μ

⁢

(

r

)

?

We will first calculate

lim

r

→

0

⁡

log

⁡

μ

⁢

(

r

)

using l’Hôpital’s rule (

http://planetmath.org/LHpitalsRule

).

lim

r

→

0

⁡

log

⁡

μ

⁢

(

r

)

=

lim

r

→

0

⁡

log

⁡

(

x

1

r

+

⋯

+

x

n

r

n

)

r

=

lim

r

→

0

⁡

(

x

1

r

⁢

log

⁡

x

1

+

⋯

+

x

n

r

⁢

log

⁡

x

n

n

)

(

x

1

r

+

⋯

+

x

n

r

n

)

=

lim

r

→

0

⁡

x

1

r

⁢

log

⁡

x

1

+

⋯

+

x

n

r

⁢

log

⁡

x

n

x

1

r

+

⋯

+

x

n

r

=

log

⁡

x

1

+

⋯

+

log

⁡

x

n

n

=

log

⁡

x

1

⁢

⋯

⁢

x

n

n

.

It follows immediately that

lim

r

→

0

⁡

(

x

1

r

+

⋯

+

x

n

r

n

)

1

/

r

=

x

1

⁢

⋯

⁢

x

n

n

.

Title

derivation of

geometric mean

as the limit of the power mean

Canonical

name

DerivationOfGeometricMeanAsTheLimitOfThePowerMean

Date of creation

2013-03-22 14:17:13

Last modified on

2013-03-22 14:17:13

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

8

Author

Mathprof (13753)

Entry type

Derivation

Classification

msc 26D15

Related topic

LHpitalsRule

Related topic

PowerMean

Related topic

WeightedPowerMean

Related topic

ArithmeticGeometricMeansInequality

Related topic

ArithmeticMean

Related topic

GeometricMean

Related topic

DerivationOfZerothWeightedPowerMean

Generated on Fri Feb 9 20:01:29 2018 by

LaTeXML
general means inequality

general means inequality

The

power means inequality

is a generalization of

arithmetic-geometric means inequality

.

If

0

≠

r

∈

ℝ

, the

r

-mean (or

r

-th

power mean

) of the nonnegative
numbers

a

1

,

…

,

a

n

is defined as

M

r

⁢

(

a

1

,

a

2

,

…

,

a

n

)

=

(

1

n

⁢

∑

k

=

1

n

a

k

r

)

1

/

r

Given

real numbers

x

,

y

such that

x

⁢

y

≠

0

and

x

<

y

, we have

M

x

≤

M

y

and the equality holds if and only if

a

1

=

…

=

a

n

.

Additionally, if we define

M

0

to be the

geometric mean

(

a

1

⁢

a

2

⁢

…

⁢

a

n

)

1

/

n

, we have
that the

inequality

above holds for arbitrary real numbers

x

<

y

.

The mentioned inequality is a special case of this one, since

M

1

is the

arithmetic mean

,

M

0

is the geometric mean and

M

-

1

is the

harmonic mean

.

This inequality can be further generalized using

weighted power means

.

Title

general means inequality

Canonical

name

GeneralMeansInequality

Date of creation

2013-03-22 12:39:49

Last modified on

2013-03-22 12:39:49

Owner

drini (3)

Last modified by

drini (3)

Numerical id

6

Author

drini (3)

Entry type

Theorem

Classification

msc 26D15

Synonym

power means inequality

Related topic

ArithmeticGeometricMeansInequality

Related topic

ArithmeticMean

Related topic

GeometricMean

Related topic

HarmonicMean

Related topic

PowerMean

Related topic

ProofOfArithmeticGeometricHarmonicMeansI

Related topic

RootMeanSquare3

Related topic

DerivationOfZerothWeightedPowerMean

Related topic

ProofOfArithmeticGeometricHarmonicMeansInequality

Related topic

ComparisonOfPythagor

Generated on Fri Feb 9 20:01:31 2018 by

LaTeXML
Hardy’s inequality

Hardy’s inequality

Suppose

p

>

1

and

{

a

n

}

is a sequence of nonnegative

real numbers

.
Let

A

n

=

∑

i

=

1

n

a

i

. Then

∑

n

≥

1

(

A

n

n

)

p

<

(

p

p

-

1

)

p

⁢

∑

n

≥

1

a

n

p

,

unless all the

a

n

are zero. The constant is best possible.

This theorem has an

integral

analogue:
Suppose that

p

>

1

and

f

≥

0

on

(

0

,

∞

)

.
Let

F

⁢

(

x

)

=

∫

0

x

f

⁢

(

t

)

⁢

𝑑

t

. Then

∫

0

∞

(

F

x

)

p

⁢

𝑑

x

<

(

p

p

-

1

)

p

⁢

∫

0

∞

f

p

⁢

(

x

)

⁢

𝑑

x

,

unless

f

≡

0

. The constant is best possible.

References

1

G.H. Hardy, J.E. Littlewood and G.Pólya,

Inequalities

, Cambridge University Press,
Cambridge, 2nd edition, 1952, pp. 239-240.

Title

Hardy’s inequality

Canonical name

HardysInequality

Date of creation

2013-03-22 17:04:32

Last modified on

2013-03-22 17:04:32

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

8

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 26D15

Generated on Fri Feb 9 20:01:33 2018 by

LaTeXML
Hermite-Hadamard integral inequality

Hermite-Hadamard integral inequality

Let

f

:

[

a

,

b

]

→

ℝ

be a

convex function

.
Then

f

⁢

(

a

+

b

2

)

≤

1

b

-

a

⁢

∫

a

b

f

⁢

(

t

)

⁢

𝑑

t

≤

f

⁢

(

a

)

+

f

⁢

(

b

)

2

.

Title

Hermite-Hadamard integral inequality

Canonical name

HermiteHadamardIntegralInequality

Date of creation

2013-03-22 16:56:30

Last modified on

2013-03-22 16:56:30

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

8

Author

Mathprof (13753)

Entry type

Theorem

Classification

msc 26D15

Classification

msc 26D10

Generated on Fri Feb 9 20:01:34 2018 by

LaTeXML
MacLaurin’s inequality

MacLaurin’s inequality

Let

a

1

,

a

2

,

…

,

a

n

be positive real numbers , and define the sums

S

k

as follows :

S

k

=

∑

1

≤

i

1

<

i

2

<

⋯

<

i

k

≤

n

a

i

1

⁢

a

i

2

⁢

⋯

⁢

a

i

k

(

n

k

)

Then the following chain of

inequalities

is true :

S

1

≥

S

2

≥

S

3

3

≥

⋯

≥

S

n

n

Note

:

S

k

are called the

averages

of the

elementary

symmetric

sums

This inequality is in fact important because it shows that the

arithmetic-geometric mean inequality

is nothing but a

consequence

of a chain of stronger inequalities

Title

MacLaurin’s inequality

Canonical

name

MacLaurinsInequality

Date of creation

2013-03-22 13:19:28

Last modified on

2013-03-22 13:19:28

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

7

Author

Mathprof (13753)

Entry type

Definition

Classification

msc 26D15

Generated on Fri Feb 9 20:01:36 2018 by

LaTeXML
Minkowski inequality

Minkowski inequality

If

p

≥

1

and

a

k

,

b

k

are

real numbers

for

k

=

1

,

…

, then

(

∑

k

=

1

n

|

a

k

+

b

k

|

p

)

1

/

p

≤

(

∑

k

=

1

n

|

a

k

|

p

)

1

/

p

+

(

∑

k

=

1

n

|

b

k

|

p

)

1

/

p

The

Minkowski inequality

is in fact valid for all

L

p

norms with

p

≥

1

on arbitrary

measure spaces

. This covers the case of

ℝ

n

listed here as well as spaces of sequences and

spaces of functions

, and also complex

L

p

spaces.

Title

Minkowski inequality

Canonical name

MinkowskiInequality

Date of creation

2013-03-22 11:46:24

Last modified on

2013-03-22 11:46:24

Owner

drini (3)

Last modified by

drini (3)

Numerical id

13

Author

drini (3)

Entry type

Theorem

Classification

msc 26D15

Related topic

LebesgueMeasure

Related topic

MeasurableSpace

Generated on Fri Feb 9 20:01:38 2018 by

LaTeXML
Muirhead’s theorem

Muirhead’s theorem

Let

0

≤

s

1

≤

⋯

≤

s

n

and

0

≤

t

1

≤

…

≤

t

n

be

real numbers

such that

∑

i

=

1

n

s

i

=

∑

i

=

1

n

t

i

and

∑

i

=

1

k

s

i

≤

∑

i

=

1

k

t

i

(

k

=

1

,

…

,

n

-

1

)

Then for any nonnegative numbers

x

1

,

…

,

x

n

,

∑

σ

x

1

s

σ

⁢

(

1

)

⁢

…

⁢

x

n

s

σ

⁢

(

n

)

≥

∑

σ

x

1

t

σ

⁢

(

1

)

⁢

…

⁢

x

n

t

σ

⁢

(

n

)

where the sums run over all

permutations

σ

of

{

1

,

2

,

…

,

n

}

.

Title

Muirhead’s

theorem

Canonical name

MuirheadsTheorem

Date of creation

2013-03-22 13:15:25

Last modified on

2013-03-22 13:15:25

Owner

Koro (127)

Last modified by

Koro (127)

Numerical id

7

Author

Koro (127)

Entry type

Theorem

Classification

msc 26D15

Synonym

Muirhead’s

inequality

Related topic

CharacterizationsOfMajorization

Generated on Fri Feb 9 20:01:40 2018 by

LaTeXML
PTAH inequality result

PTAH inequality result

There is an additional classic

inequality

to be added to the list (A) through (F), namely the maximum-entropy inequality (in logarithmic form)

-

∑

i

=

1

n

p

i

⁢

log

⁡

p

i

≤

log

⁡

n

.

Also, inequality (C) should be labeled “the Kullback–Leibler inequality”.

Then, the

labeling

could be modified: the first time the

PTAH inequality

appears it could be labeled (P);
then the maximum-entropy inequality could be labeled (G), followed by Hölder’s inequality (already labeled (H));
and the statement could be “the inequalities (A) through (G) are

equivalent

, each is a special case of (H),
and (H) is a special case of (P). However, it appears that none of the reverse

implications

hold.”

Title

PTAH inequality result

Canonical name

PTAHInequalityResult

Date of creation

2013-03-22 18:08:33

Last modified on

2013-03-22 18:08:33

Owner

cappymate (20671)

Last modified by

cappymate (20671)

Numerical id

4

Author

cappymate (20671)

Entry type

Result

Classification

msc 26D15

Generated on Fri Feb 9 20:01:45 2018 by

LaTeXML
power mean

power mean

The

r

-th

power mean

of the numbers

x

1

,

x

2

,

…

,

x

n

is defined as:

M

r

⁢

(

x

1

,

x

2

,

…

,

x

n

)

=

(

x

1

r

+

x

2

r

+

⋯

+

x

n

r

n

)

1

/

r

.

The

arithmetic mean

is a special case when

r

=

1

.
The power mean is a

continuous function

of

r

, and taking limit when

r

→

0

gives us the

geometric mean

:

M

0

⁢

(

x

1

,

x

2

,

…

,

x

n

)

=

x

1

⁢

x

2

⁢

⋯

⁢

x

n

n

.

Also, when

r

=

-

1

we get

M

-

1

⁢

(

x

1

,

x

2

,

…

,

x

n

)

=

n

1

x

1

+

1

x

2

+

⋯

+

1

x

n

the

harmonic mean

.

A

generalization

of power means are

weighted power means

.

Title

power mean

Canonical

name

PowerMean

Date of creation

2013-03-22 11:47:17

Last modified on

2013-03-22 11:47:17

Owner

drini (3)

Last modified by

drini (3)

Numerical id

14

Author

drini (3)

Entry type

Definition

Classification

msc 26D15

Classification

msc 16D10

Classification

msc 00-01

Classification

msc 34-00

Classification

msc 35-00

Related topic

WeightedPowerMean

Related topic

ArithmeticGeometricMeansInequality

Related topic

ArithmeticMean

Related topic

GeometricMean

Related topic

HarmonicMean

Related topic

GeneralMeansInequality

Related topic

RootMeanSquare3

Related topic

ProofOfGeneralMeansInequality

Related topic

DerivationOfZerothWeightedPowerMean

Related topic

DerivationOfHarmonicMeanAsTheLimitOfThePowerMean

Generated on Fri Feb 9 20:01:47 2018 by

LaTeXML
proof of arithmetic-geometric means inequality

proof of arithmetic-geometric means inequality

A short geometrical proof can be given for the case

n

=

2

of the

arithmetic-geometric means inequality

.

Let

a

and

b

be two non

negative numbers

.
Draw the line

A

⁢

B

such that

A

⁢

P

has length

a

, and

P

⁢

B

has length

b

, as in the following picture, and draw a

semicircle

with

diameter

A

⁢

B

. Let

O

be the center of the circle.

Now raise

perpendiculars

P

⁢

Q

and

O

⁢

T

to

A

⁢

B

. Notice that

O

⁢

T

is a radius, and so

O

⁢

T

=

A

⁢

B

2

=

a

+

b

2

Also notice that

P

⁢

Q

≤

O

⁢

T

for any point

P

, and equality is obtained only when

P

=

O

, that is, when

a

=

b

.

Notice also that

P

⁢

Q

is a height over the

hypotenuse

on

right triangle

△

⁢

A

⁢

Q

⁢

B

. We have then triangle

similarities

△

⁢

A

⁢

Q

⁢

B

∼

△

⁢

A

⁢

P

⁢

Q

∼

△

⁢

Q

⁢

P

⁢

B

, and thus

A

⁢

P

P

⁢

Q

=

P

⁢

Q

P

⁢

B

which implies

P

⁢

Q

=

A

⁢

P

⋅

P

⁢

B

=

a

⁢

b

. Since

P

⁢

Q

≤

O

⁢

T

, we conclude

a

⁢

b

≤

a

+

b

2

.

This special case can also be proved using

rearrangement inequality

.
Let

a

,

b

non negative numbers, and assume

a

≤

b

. Let

x

1

=

a

,

x

2

=

b

, and then

x

1

≤

x

2

.
Now suppose

y

1

and

y

2

are such that one of them is

x

1

and the other is

x

2

. Rearrangement inequality states that

x

1

⁢

y

1

+

x

2

⁢

y

2

is

maximum

when

y

1

≤

y

2

and

x

1

≤

x

2

.
So, we have

x

1

⁢

x

2

+

x

2

⁢

x

1

≤

x

1

2

+

x

2

2

and substituting back

a

,

b

gives

2

⁢

a

⁢

b

≤

(

a

)

2

+

(

b

)

2

=

a

+

b

where it follows the desired result.

One more proof can be given as follows.
Let

x

=

a

,

y

=

b

. Then

(

x

-

y

)

2

≥

0

, and equality holds only when

x

=

y

. Then,

x

2

-

2

⁢

x

⁢

y

+

y

2

≥

0

becomes

x

2

+

y

2

≥

2

⁢

x

⁢

y

and substituting back

a

,

b

gives the desired result as in the previous proof.

Title

proof of arithmetic-geometric means inequality

Canonical

name

ProofOfArithmeticgeometricMeansInequality

Date of creation

2013-03-22 14:49:14

Last modified on

2013-03-22 14:49:14

Owner

mathcam (2727)

Last modified by

mathcam (2727)

Numerical id

10

Author

mathcam (2727)

Entry type

Proof

Classification

msc 26D15

Generated on Fri Feb 9 20:01:49 2018 by

LaTeXML
proof of arithmetic-geometric-harmonic means inequality

proof of arithmetic-geometric-harmonic means inequality

For the

Arithmetic

Geometric Inequality

, I claim it is enough to prove that if

∏

i

=

1

n

x

i

=

1

with

x

i

≥

0

then

∑

i

=

1

n

x

i

≥

n

. The arithmetic geometric inequality for

y

1

,

…

,

y

n

will follow by taking

x

i

=

y

i

∏

k

=

1

n

y

k

n

. The geometric

harmonic

inequality

follows from the arithmetic geometric by taking

x

i

=

1

y

i

.

So, we show that if

∏

i

=

1

n

x

i

=

1

with

x

i

≥

0

then

∑

i

=

1

n

x

i

≥

n

by

induction

on

n

.

Clear for

n

=

1

.

Induction Step: By reordering indices we may assume the

x

i

are

increasing

, so

x

n

≥

1

≥

x

1

. Assuming the statement is true for

n

-

1

, we have

x

2

+

⋯

+

x

n

-

1

+

x

1

⁢

x

n

≥

n

-

1

. Then,

∑

i

=

1

n

x

i

≥

n

-

1

+

x

n

+

x

1

-

x

1

⁢

x

n

by adding

x

1

+

x

n

to both sides and subtracting

x

1

⁢

x

n

. And so,

∑

i

=

1

n

x

i

≥

n

+

(

x

n

-

1

)

+

(

x

1

-

x

1

⁢

x

n

)

=

n

+

(

x

n

-

1

)

-

x

1

⁢

(

x

n

-

1

)

=

n

+

(

x

n

-

1

)

⁢

(

1

-

x

1

)

≥

n

The last line follows since

x

n

≥

1

≥

x

1

.

Title

proof of arithmetic-geometric-harmonic means inequality

Canonical name

ProofOfArithmeticgeometricharmonicMeansInequality

Date of creation

2013-03-22 15:09:37

Last modified on

2013-03-22 15:09:37

Owner

Mathprof (13753)

Last modified by

Mathprof (13753)

Numerical id

10

Author

Mathprof (13753)

Entry type

Proof

Classification

msc 26D15

Generated on Fri Feb 9 20:01:51 2018 by

LaTeXML
proof of Chebyshev’s inequality

proof of Chebyshev’s inequality

Let

x

1

,

x

2

,

…

,

x

n

and

y

1

,

y

2

,

…

,

y

n

be

real numbers

such
that

x

1

≤

x

2

≤

⋯

≤

x

n

. Write the

product

(

x

1

+

x

2

+

⋯

+

x

n

)

⁢

(

y

1

+

y

2

+

⋯

+

y

n

)

as

(

x

1

⁢

y

1

+

x

2

⁢

y

2

+

⋯

+

x

n

⁢

y

n

)

(1)

+

(

x

1

⁢

y

2

+

x

2

⁢

y

3

+

⋯

+

x

n

-

1

⁢

y

n

+

x

n

⁢

y

1

)

+

(

x

1

⁢

y

3

+

x

2

⁢

y

4

+

⋯

+

x

n

-

2

⁢

y

n

+

x

n

-

1

⁢

y

1

+

x

n

⁢

y

2

)

+

⋯

+

(

x

1

⁢

y

n

+

x

2

⁢

y

1

+

x

3

⁢

y

2

+

⋯

+

x

n

⁢

y

n

-

1

)

.

•

If

y

1

≤

y

2

≤

⋯

≤

y

n

, each of the

n

terms in parentheses
is less than or equal to

x

1

⁢

y

1

+

x

2

⁢

y

2

+

⋯

+

x

n

⁢

y

n

, according to
the

rearrangement inequality

. From this, it follows that

(

x

1

+

x

2

+

⋯

+

x

n

)

⁢

(

y

1

+

y

2

+

⋯

+

y

n

)

≤

n

⁢

(

x

1

⁢

y

1

+

x

2

⁢

y

2

+

⋯

+

x

n

⁢

y

n

)

or (dividing by

n

2

)

(

x

1

+

x

2

+

⋯

+

x

n

n

)

⁢

(

y

1

+

y

2

+

⋯

+

y

n

n

)

≤

x

1

⁢

y

1

+

x

2

⁢

y

2

+

⋯

+

x

n

⁢

y

n

n

.

•

If

y

1

≥

y

2

≥

⋯

≥

y

n

, the same reasoning gives

(

x

1

+

x

2

+

⋯

+

x

n

n

)

⁢

(

y

1

+

y

2

+

⋯

+

y

n

n

)

≥

x

1

⁢

y

1

+

x

2

⁢

y

2

+

⋯

+

x

n

⁢

y

n

n

.

It is clear that equality holds if

x

1

=

x

2

=

⋯

=

x

n

or

y

1

=

y

2

=

⋯

=

y

n

. To see that this condition is also necessary,
suppose that not all

y

i

’s are equal, so that

y

1

≠

y

n

.
Then the second term in parentheses of (

1

) can only be
equal to

x

1

⁢

y

1

+

x

2

⁢

y

2

+

⋯

+

x

n

⁢

y

n

if

x

n

-

1

=

x

n

, the third
term only if

x

n

-

2

=

x

n

-

1

, and so on, until the last term which
can only be equal to

x

1

⁢

y

1

+

x

2

⁢

y

2

+

⋯

+

x

n

⁢

y

n

if

x

1

=

x

2

. This
implies that

x

1

=

x

2

=

⋯

=

x

n

. Therefore, Chebyshev’s

inequality

is an equality if and only if

x

1

=

x

2

=

⋯

=

x

n

or

y

1

=

y

2

=

⋯

=

y

n

.

Title

proof of Chebyshev’s inequality

Canonical name

ProofOfChebyshevsInequality

Date of creation

2013-03-22 13:08:38

Last modified on

2013-03-22 13:08:38

Owner

pbruin (1001)

Last modified by

pbruin (1001)

Numerical id

4

Author

pbruin (1001)

Entry type

Proof

Classification

msc 26D15

Generated on Fri Feb 9 20:01:59 2018 by

LaTeXML
proof of Jensen’s inequality

proof of Jensen’s inequality

We prove an

equivalent

, more convenient formulation: Let

X

be some

random variable

, and let

f

⁢

(

x

)

be a

convex function

(defined at least on a

segment

containing the range of

X

). Then the

expected value

of

f

⁢

(

X

)

is at least the value of

f

at the mean of

X

:

𝔼

⁡

[

f

⁢

(

X

)

]

≥

f

⁢

(

𝔼

⁡

[

X

]

)

.

Indeed, let

c

=

𝔼

⁡

[

X

]

. Since

f

⁢

(

x

)

is convex, there exists a supporting line for

f

⁢

(

x

)

at

c

:

φ

⁢

(

x

)

=

α

⁢

(

x

-

c

)

+

f

⁢

(

c

)

for some

α

, and

φ

⁢

(

x

)

≤

f

⁢

(

x

)

. Then

𝔼

⁡

[

f

⁢

(

X

)

]

≥

𝔼

⁡

[

φ

⁢

(

X

)

]

=

𝔼

⁡

[

α

⁢

(

X

-

c

)

+

f

⁢

(

c

)

]

=

f

⁢

(

c

)

as claimed.

Title

proof of Jensen’s

inequality

Canonical name

ProofOfJensensInequality

Date of creation

2013-03-22 12:45:15

Last modified on

2013-03-22 12:45:15

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

6

Author

Andrea Ambrosio (7332)

Entry type

Proof

Classification

msc 26D15

Classification

msc 39B62

Generated on Fri Feb 9 20:02:05 2018 by

LaTeXML
proof of Minkowski inequality

proof of Minkowski inequality

For

p

=

1

the result follows immediately from the

triangle inequality

, so we may assume

p

>

1

.

We have

|

a

k

+

b

k

|

p

=

|

a

k

+

b

k

|

⁢

|

a

k

+

b

k

|

p

-

1

≤

(

|

a

k

|

+

|

b

k

|

)

⁢

|

a

k

+

b

k

|

p

-

1

by the triangle inequality. Therefore we have

|

a

k

+

b

k

|

p

≤

|

a

k

|

⁢

|

a

k

+

b

k

|

p

-

1

+

|

b

k

|

⁢

|

a

k

+

b

k

|

p

-

1

Set

q

=

p

p

-

1

. Then

1

p

+

1

q

=

1

, so by the Hölder

inequality

we have

∑

k

=

0

n

|

a

k

|

⁢

|

a

k

+

b

k

|

p

-

1

≤

(

∑

k

=

0

n

|

a

k

|

p

)

1

p

⁢

(

∑

k

=

0

n

|

a

k

+

b

k

|

(

p

-

1

)

⁢

q

)

1

q

∑

k

=

0

n

|

b

k

|

⁢

|

a

k

+

b

k

|

p

-

1

≤

(

∑

k

=

0

n

|

b

k

|

p

)

1

p

⁢

(

∑

k

=

0

n

|

a

k

+

b

k

|

(

p

-

1

)

⁢

q

)

1

q

Adding these two inequalities, dividing by the factor common to the right sides of both, and observing that

(

p

-

1

)

⁢

q

=

p

by definition, we have

(

∑

k

=

0

n

|

a

k

+

b

k

|

p

)

1

-

1

q

≤

∑

k

=

0

n

(

|

a

k

|

+

|

b

k

|

)

⁢

|

a

k

+

b

k

|

p

-

1

(

∑

k

=

0

n

|

a

k

+

b

k

|

p

)

1

q

≤

(

∑

k

=

0

n

|

a

k

|

p

)

1

p

+

(

∑

k

=

0

n

|

b

k

|

p

)

1

p

Finally, observe that

1

-

1

q

=

1

p

, and the result follows as required. The proof for the

integral

version is analogous.

Title

proof of Minkowski inequality

Canonical name

ProofOfMinkowskiInequality

Date of creation

2013-03-22 12:42:14

Last modified on

2013-03-22 12:42:14

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

10

Author

Andrea Ambrosio (7332)

Entry type

Proof

Classification

msc 26D15

Related topic

HolderInequality

Generated on Fri Feb 9 20:02:08 2018 by

LaTeXML
proof of rearrangement inequality

proof of rearrangement inequality

We first prove the

rearrangement inequality

for the case

n

=

2

. Let

x

1

,

x

2

,

y

1

,

y

2

be

real numbers

such that

x

1

≤

x

2

and

y

1

≤

y

2

. Then

(

x

2

-

x

1

)

⁢

(

y

2

-

y

1

)

≥

0

,

and therefore

x

1

⁢

y

1

+

x

2

⁢

y

2

≥

x

1

⁢

y

2

+

x

2

⁢

y

1

.

Equality holds iff

x

1

=

x

2

or

y

1

=

y

2

.

For the general case, let

x

1

,

x

2

,

…

,

x

n

and

y

1

,

y

2

,

…

,

y

n

be real numbers such that

x

1

≤

x

2

≤

⋯

≤

x

n

. Suppose that

(

z

1

,

z

2

,

…

,

z

n

)

is a

permutation

(rearrangement) of

{

y

1

,

y

2

,

…

,

y

n

}

such that the sum

x

1

⁢

z

1

+

x

2

⁢

z

2

+

⋯

+

x

n

⁢

z

n

is maximized. If there exists a pair

i

<

j

with

z

i

>

z

j

, then

x

i

⁢

z

j

+

x

j

⁢

z

i

≥

x

i

⁢

z

i

+

x

j

⁢

z

j

(the

n

=

2

case); equality holds iff

x

i

=

x

j

. Therefore,

x

1

⁢

z

1

+

x

2

⁢

z

2

+

⋯

+

x

n

⁢

z

n

is not

maximal

unless

z

1

≤

z

2

≤

⋯

≤

z

n

or

x

i

=

x

j

for all pairs

i

<

j

such that

z

i

>

z

j

. In the latter case, we can consecutively
interchange these pairs until

z

1

≤

z

2

≤

⋯

≤

z

n

(this is
possible because the number of pairs

i

<

j

with

z

i

>

z

j

decreases
with each step). So

x

1

⁢

z

1

+

x

2

⁢

z

2

+

⋯

+

x

n

⁢

z

n

is maximized if

z

1

≤

z

2

≤

⋯

≤

z

n

.

To show that

x

1

⁢

z

1

+

x

2

⁢

z

2

+

⋯

+

x

n

⁢

z

n

is minimal for a
permutation

(

z

1

,

z

2

,

…

,

z

n

)

of

{

y

1

,

y

2

,

…

,

y

n

}

if

z

1

≥

z

2

≥

⋯

≥

z

n

, observe that

-

(

x

1

⁢

z

1

+

x

2

⁢

z

2

+

⋯

+

x

n

⁢

z

n

)

=

x

1

⁢

(

-

z

1

)

+

x

2

⁢

(

-

z

2

)

+

⋯

+

x

n

⁢

(

-

z

n

)

is maximized if

-

z

1

≤

-

z

2

≤

⋯

≤

-

z

n

. This implies that

x

1

⁢

z

1

+

x

2

⁢

z

2

+

⋯

+

x

n

⁢

z

n

is minimized if

z

1

≥

z

2

≥

⋯

≥

z

n

.

Title

proof of rearrangement inequality

Canonical name

ProofOfRearrangementInequality

Date of creation

2013-03-22 13:08:41

Last modified on

2013-03-22 13:08:41

Owner

pbruin (1001)

Last modified by

pbruin (1001)

Numerical id

4

Author

pbruin (1001)

Entry type

Proof

Classification

msc 26D15

Related topic

ChebyshevsInequality

Generated on Fri Feb 9 20:02:17 2018 by

LaTeXML
proof of Schur’s inequality

proof of Schur’s inequality

By Schur’s theorem, a

unitary matrix

U

and an

upper triangular matrix

T

exist such that

A

=

U

⁢

T

⁢

U

H

,

T

being

diagonal

if and only if

A

is normal.
Then

A

H

⁢

A

=

U

⁢

T

H

⁢

U

H

⁢

U

⁢

T

⁢

U

H

=

U

⁢

T

H

⁢

T

⁢

U

H

, which means

A

H

⁢

A

and

T

H

⁢

T

are

similar

; so they have the same trace. We have:

∥

A

∥

F

2

=

Tr

⁡

(

A

H

⁢

A

)

=

Tr

⁡

(

T

H

⁢

T

)

=

∑

i

=

1

n

|

λ

i

|

2

+

∑

i

<

j

|

t

i

⁢

j

|

2

=

=

Tr

⁡

(

D

H

⁢

D

)

+

∑

i

<

j

|

t

i

⁢

j

|

2

≥

Tr

⁡

(

D

H

⁢

D

)

=

∥

D

∥

F

2

.

If and only if

A

is normal,

T

=

D

and therefore equality holds.

□

Title

proof of Schur’s inequality

Canonical name

ProofOfSchursInequality

Date of creation

2013-03-22 15:35:25

Last modified on

2013-03-22 15:35:25

Owner

Andrea Ambrosio (7332)

Last modified by

Andrea Ambrosio (7332)

Numerical id

7

Author

Andrea Ambrosio (7332)

Entry type

Proof

Classification

msc 26D15

Classification

msc 15A42

Generated on Fri Feb 9 20:02:19 2018 by

LaTeXML