proof of general means inequality

proof of general means inequality

Let

w

1

,

w

2

, …,

w

n

be positive real numbers such that

w

1

+

w

2

+

⋯

+

w

n

=

1

. For any

real number

r

≠

0

, the

weighted power mean

of degree

r

of

n

positive real numbers

x

1

,

x

2

, …,

x

n

(with respect to the weights

w

1

, …,

w

n

) is defined as

M

w

r

⁢

(

x

1

,

x

2

,

…

,

x

n

)

=

(

w

1

⁢

x

1

r

+

w

2

⁢

x

2

r

+

⋯

+

w

n

⁢

x

n

r

)

1

/

r

.

The

definition

is extended to the case

r

=

0

by taking the limit

r

→

0

; this yields the weighted geometric mean

M

w

0

⁢

(

x

1

,

x

2

,

…

,

x

n

)

=

x

1

w

1

⁢

x

2

w

2

⁢

…

⁢

x

n

w

n

(see

derivation of zeroth weighted power mean

). We will prove the
weighted power means

inequality

, which states that for any two real
numbers

r

<

s

, the weighted power means of orders

r

and

s

of

n

positive real numbers

x

1

,

x

2

, …,

x

n

satisfy the
inequality

M

w

r

⁢

(

x

1

,

x

2

,

…

,

x

n

)

≤

M

w

s

⁢

(

x

1

,

x

2

,

…

,

x

n

)

with equality if and only if all the

x

i

are equal.

First, let us suppose that

r

and

s

are nonzero. We
distinguish three cases for the signs of

r

and

s

:

r

<

s

<

0

,

r

<

0

<

s

, and

0

<

r

<

s

. Let us consider the last case, i.e. assume

r

and

s

are both

positive

; the others are similar. We write

t

=

s

r

and

y

i

=

x

i

r

for

1

≤

i

≤

n

; this implies

y

i

t

=

x

i

s

. Consider the function

f

:

(

0

,

∞

)

→

(

0

,

∞

)

x

↦

x

t

.

Since

t

>

1

, the

second derivative

of

f

satisfies

f

′′

⁢

(

x

)

=

t

⁢

(

t

-

1

)

⁢

x

t

-

2

>

0

for all

x

>

0

, so

f

is a

strictly
convex function

. Therefore, according to Jensen’s inequality,

(

w

1

⁢

y

1

+

w

2

⁢

y

2

+

⋯

+

w

n

⁢

y

n

)

t

=

f

⁢

(

w

1

⁢

y

1

+

w

2

⁢

y

2

+

⋯

+

w

n

⁢

y

n

)

≤

w

1

⁢

f

⁢

(

y

1

)

+

w

2

⁢

f

⁢

(

y

2

)

+

⋯

+

w

n

⁢

f

⁢

(

y

n

)

=

w

1

⁢

y

1

t

+

w

2

⁢

y

2

t

+

⋯

+

w

n

⁢

y

n

t

,

with equality if and only if

y

1

=

y

2

=

⋯

=

y

n

. By substituting

t

=

s

r

and

y

i

=

x

i

r

back into this inequality, we get

(

w

1

⁢

x

1

r

+

w

2

⁢

x

2

r

+

⋯

+

w

n

⁢

x

n

r

)

s

/

r

≤

w

1

⁢

x

1

s

+

w

2

⁢

x

2

s

+

⋯

+

w

n

⁢

x

n

s

with equality if and only if

x

1

=

x

2

=

⋯

=

x

n

. Since

s

is
positive, the function

x

↦

x

1

/

s

is

strictly increasing

, so
raising both sides to the power

1

/

s

preserves

the inequality:

(

w

1

⁢

x

1

r

+

w

2

⁢

x

2

r

+

⋯

+

w

n

⁢

x

n

r

)

1

/

r

≤

(

w

1

⁢

x

1

s

+

w

2

⁢

x

2

s

+

⋯

+

w

n

⁢

x

n

s

)

1

/

s

,

which is the inequality we had to prove. Equality holds if and only
if all the

x

i

are equal.

If

r

=

0

, the inequality is still correct:

M

w

0

is defined as

lim

r

→

0

⁡

M

w

r

, and since

M

w

r

≤

M

w

s

for all

r

<

s

with

r

≠

0

, the same holds for the limit

r

→

0

. The same

argument

shows that the inequality also holds for

s

=

0

, i.e. that

M

w

r

≤

M

w

0

for all

r

<

0

. We conclude that for all real numbers

r

and

s

such that

r

<

s

,

M

w

r

⁢

(

x

1

,

x

2

,

…

,

x

n

)

≤

M

w

s

⁢

(

x

1

,

x

2

,

…

,

x

n

)

.

Title

proof of general means inequality

Canonical

name

ProofOfGeneralMeansInequality

Date of creation

2013-03-22 13:10:26

Last modified on

2013-03-22 13:10:26

Owner

pbruin (1001)

Last modified by

pbruin (1001)

Numerical id

5

Author

pbruin (1001)

Entry type

Proof

Classification

msc 26D15

Related topic

ArithmeticMean

Related topic

GeometricMean

Related topic

HarmonicMean

Related topic

RootMeanSquare3

Related topic

PowerMean

Related topic

WeightedPowerMean

Related topic

ArithmeticGeometricMeansInequality

Related topic

JensensInequality

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