proof of arithmetic-geometric means inequality using Lagrange multipliers

proof of arithmetic-geometric means inequality using Lagrange multipliers

As an interesting example of the

Lagrange multiplier method

,
we employ it to prove the

arithmetic-geometric means inequality

:

x

1

⁢

⋯

⁢

x

n

n

≤

x

1

+

⋯

+

x

n

n

,

x

i

≥

0

,

with equality if and only if all the

x

i

are equal.

To begin with, define

f

:

ℝ

≥

0

n

→

ℝ

≥

0

by

f

⁢

(

x

)

=

(

x

1

⁢

⋯

⁢

x

n

)

1

/

n

.

Fix

a

>

0

(the

arithmetic mean

),
and consider the set

M

=

{

x

∈

ℝ

n

:

x

i

≥

0

⁢

for all

i

,

⁢

∑

i

x

i

=

n

⁢

a

}

.
This is clearly a closed and

bounded set

in

ℝ

n

, and hence is

compact

.
Therefore

f

on

M

has both a

maximum

and

minimum

.

M

is almost a manifold (a

differentiable

surface), except that it has a

boundary

consisting
of points with

x

i

=

0

for some

i

.
Clearly

f

attains the minimum zero on this boundary,
and on the “

interior

” of

M

,
that is

M

′

=

{

x

∈

ℝ

n

:

x

i

>

0

⁢

for all

i

,

⁢

∑

i

x

i

=

n

⁢

a

}

,

f

is

positive

. Now

M

′

is a manifold, and the maximum of

f

on

M

is evidently a

local maximum

on

M

′

. Hence
the Lagrange multiplier method may be applied.

We have the constraint

1

1

Although

g

does not constrain that

x

i

>

0

,
this does not really matter — for those who are worried about such
things, a

rigorous

way to escape this annoyance is to simply
define

f

⁢

(

x

)

=

0

to be zero whenever

x

i

≤

0

for some

i

.
Clearly the new

f

cannot have a maximum at such points,
so we may simply ignore the points with

x

i

≤

0

when solving the system.
And for the same reason, the fact that

f

fails to be differentiable at the boundary
points is immaterial.

0

=

g

⁢

(

x

)

=

∑

i

x

i

-

n

⁢

a

under which we are to maximize

f

.

We compute:

f

⁢

(

x

)

n

⁢

x

i

=

∂

⁡

f

∂

⁡

x

i

=

λ

⁢

∂

⁡

g

∂

⁡

x

i

=

λ

.

If we take the

product

,

i

=

1

,

…

,

n

, of the extreme left- and right- hand sides of this

equation

,
we obtain

1

n

n

=

f

⁢

(

x

)

n

n

n

⁢

x

1

⁢

⋯

⁢

x

n

=

λ

n

from which it follows that

λ

=

1

/

n

(

λ

=

-

1

/

n

is obviously inadmissible).
Then substituting back, we get

f

⁢

(

x

)

=

x

i

, which implies that the

x

i

are all equal to

a

,
The value of

f

at this point is

a

.

Since we have obtained a unique solution to the

Lagrange multiplier

system,
this unique solution must be the unique maximum point of

f

on

M

.
So

f

⁢

(

x

)

≤

a

amongst all numbers

x

i

whose arithmetic mean is

a

,
with equality occurring if and only if all the

x

i

are equal to

a

.

The case

a

=

0

, which was not included in the above

analysis

, is

trivial

.

Proof of concavity

The question of whether the point obtained from solving Lagrange system
is actually a maximum for

f

was taken care of by our preliminary

compactness

argument

. Another popular way to determine whether a given

stationary point

is a local maximum or minimum is by studying the

Hessian

of the

function

at that point.
For the sake of illustrating the relevant techniques,
we will prove again that

x

i

=

a

is a local maximum for

f

,
this time by showing that the Hessian is

negative definite

.

Actually, it turns out to be not much
harder to show that

f

is weakly

concave

everywhere

on

ℝ

≥

0

n

, not just on

M

.
A plausibility argument for this fact
is that the graph of

t

↦

t

n

is concave, and since

f

also involves an

n

th root,
it should be concave too. We will prove concavity of

f

by
showing that the Hessian of

f

is negative semi-definite.

Since

M

is a

convex

2

2

If

M

is not convex, then the arguments that follow will not work.
In general, a prescription to determine whether a

critical point

is a

local minimum

or maximum can be found
in

tests for local extrema in Lagrange multiplier method

.

set, the

restriction

of

f

to

M

will also be weakly concave;
then we can conclude that
the critical point

x

i

=

a

on

M

must be a

global

minimum on

M

.

We begin by calculating

second-order

derivatives

:

∂

2

⁡

f

∂

⁡

x

j

⁢

∂

⁡

x

i

=

∂

∂

⁡

x

j

⁢

(

f

⁢

(

x

)

n

⁢

x

i

)

=

1

n

⁢

x

i

⁢

∂

⁡

f

∂

⁡

x

j

+

f

⁢

(

x

)

⁢

∂

∂

⁡

x

j

⁢

1

n

⁢

x

i

=

f

⁢

(

x

)

n

2

⁢

x

i

⁢

x

j

-

δ

i

⁢

j

⁢

f

⁢

(

x

)

n

⁢

x

i

2

.

(The symbol

δ

i

⁢

j

is the

Kronecker delta

.)
Thus the Hessian matrix is:

D

2

⁡

f

⁢

(

x

)

=

f

⁢

(

x

)

n

2

⁢

[

1

x

i

⁢

x

j

]

i

⁢

j

-

f

⁢

(

x

)

n

2

⁢

[

n

x

1

2

⋱

n

x

n

2

]

Now if we had actual numbers to substitute for the

x

i

,
then we can go to a

computer

and ask if the matrix is negative definite or not.
But we want to prove global concavity, so we have to employ some clever

algebra

, and work directly from the

definition

.

Let

v

=

(

v

1

,

…

,

v

n

)

be a test vector.
Negative semi-definiteness means

v

⁢

D

2

⁡

f

⁢

(

x

)

⁢

v

T

≤

0

for all

v

.
So let us write out:

v

⁢

D

2

⁡

f

⁢

(

x

)

⁢

v

T

=

f

⁢

(

x

)

n

2

⁢

(

∑

i

,

j

v

i

x

i

⁢

v

j

x

j

-

n

⁢

∑

i

v

i

2

x

i

2

)

=

f

⁢

(

x

)

n

2

⁢

(

(

∑

i

v

i

x

i

)

2

-

n

⁢

∑

i

(

v

i

x

i

)

2

)

.

We would be done if we can prove

(

∑

i

w

i

)

2

≤

n

⁢

∑

i

w

i

2

,

w

i

=

v

i

x

i

.

(1)

But look, this is just the

Cauchy-Schwarz inequality

,
concerning the

dot product

of the vectors

(

w

1

,

…

,

w

n

)

and

(

1

,

…

,

1

)

!
So

D

2

⁡

f

⁢

(

x

)

is negative semi-definite everywhere,
i.e.

f

is weakly concave.

Moreover, the case of equality in the Cauchy-Schwarz inequality (

1

)
only occurs when

(

w

1

,

…

,

w

n

)

is

parallel

to

(

1

,

…

,

1

)

,
i.e.

v

i

=

μ

⁢

x

i

for some scalar

μ

.
Notice that

(

1

,

…

,

1

)

is the

normal vector

to
the

tangent space

of

M

,
so

(

1

,

…

,

1

)

⋅

v

=

μ

⁢

∑

i

x

i

≠

0

unless

μ

=

0

.
This means, equality never holds in (

1

)
if

v

≠

0

is

restricted to the tangent space

of

M

.
In turn, this means that

f

is

strictly

concave
on the convex set

M

, and
so

x

i

=

a

is a strict

global minimum

of

f

on

M

.

Proof of inequality by symmetry argument

Observe that the maximum point

x

i

=

a

is pleasantly

symmetric

.
We can show that the solution has to be symmetric,
by exploiting the

symmetry

of the function

f

and the set

M

.

We have already calculated that

f

is strictly concave on

M

;
this was

independent

of the Lagrange multiplier method.
Since

f

is strictly concave on a closed convex set,
it has a unique maximum there; call it

x

.

Pick any indices

i

and

j

, ranging from

1

to

n

, and form
the

linear transformation

T

, which switches the

i

th and

j

th

coordinates

of its argument.

By definition,

f

⁢

(

x

)

≥

f

⁢

(

ξ

)

for all

ξ

∈

M

.
But

f

∘

T

=

f

, so this means

f

⁢

(

T

⁢

x

)

≥

f

⁢

(

T

⁢

ξ

)

for all

ξ

∈

M

.
But

M

is mapped to itself by the

transformation

T

,
so this is the same as saying that

f

⁢

(

T

⁢

x

)

≥

f

⁢

(

ξ

)

for all

ξ

∈

M

.
But the

global maximum

is unique, so

T

⁢

x

=

x

,
that is,

x

i

=

x

j

.

Note that the argument in this last proof
is extremely general — it takes the form:
if

x

is the unique maximum of

f

:

M

→

ℝ

,
and

T

is any symmetry of

M

such that

f

∘

T

=

f

,
then

T

⁢

x

=

x

. This kind of symmetry argument was used to great effect
by Jakob Steiner in his famous attempted proof of the

isoperimetric inequality

:
among all

closed curves

of a given

arc length

, the circle maximizes
the area enclosed. However, Steiner did not realize that
the

assumption

that there is a unique maximizing curve
has to be proved. Actually, that is the hardest part of the proof —
if the assumption is known, then one may easily calculate using the

Lagrange multiplier method

that the maximizing curve must be a circle!

Title

proof of arithmetic-geometric means inequality using Lagrange multipliers

Canonical

name

ProofOfArithmeticgeometricMeansInequalityUsingLagrangeMultipliers

Date of creation

2013-03-22 15:25:53

Last modified on

2013-03-22 15:25:53

Owner

stevecheng (10074)

Last modified by

stevecheng (10074)

Numerical id

16

Author

stevecheng (10074)

Entry type

Example

Classification

msc 49-00

Classification

msc 26B12

Related topic

ArithmeticGeometricMeansInequality

Related topic

UsingJensensInequalityToProveTheArithmeticGeometricHarmonicMeansInequality

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