proof of arithmetic-geometric-harmonic means inequality

proof of arithmetic-geometric-harmonic means inequality

Let

M

be

max

⁡

{

x

1

,

x

2

,

x

3

,

…

,

x

n

}

and let

m

be

min

⁡

{

x

1

,

x

2

,

x

3

,

…

,

x

n

}

.

Then

M

=

M

+

M

+

M

+

⋯

+

M

n

≥

x

1

+

x

2

+

x

3

+

⋯

+

x

n

n

m

=

n

n

m

=

n

1

m

+

1

m

+

1

m

+

⋯

+

1

m

≤

n

1

x

1

+

1

x

2

+

1

x

3

+

⋯

+

1

x

n

where all the

summations

have

n

terms.
So we have proved in this way the two

inequalities

at the extremes.

Now we shall prove the inequality between

arithmetic mean

and

geometric mean

.

1

Case

n

=

2

We do first the case

n

=

2

.

(

x

1

-

x

2

)

2

≥

0

x

1

-

2

⁢

x

1

⁢

x

2

+

x

2

≥

0

x

1

+

x

2

≥

2

⁢

x

1

⁢

x

2

x

1

+

x

2

2

≥

x

1

⁢

x

2

2

Case

n

=

2

k

Now we prove the inequality for any power of

2

(that is,

n

=

2

k

for some

integer

k

) by using mathematical induction.

x

1

+

x

2

+

⋯

+

x

2

k

+

x

2

k

+

1

+

⋯

+

x

2

k

+

1

2

k

+

1

=

(

x

1

+

x

2

+

⋯

+

x

2

k

2

k

)

+

(

x

2

k

+

1

+

x

2

k

+

2

+

⋯

+

x

2

k

+

1

2

k

)

2

and using the case

n

=

2

on the last

expression

we can state the following inequality

x

1

+

x

2

+

⋯

+

x

2

k

+

x

2

k

+

1

+

⋯

+

x

2

k

+

1

2

k

+

1

≥

(

x

1

+

x

2

+

⋯

+

x

2

k

2

k

)

⁢

(

x

2

k

+

1

+

x

2

k

+

2

+

⋯

+

x

2

k

+

1

2

k

)

≥

x

1

⁢

x

2

⁢

⋯

⁢

x

2

k

2

k

⁢

x

2

k

+

1

⁢

x

2

k

+

2

⁢

⋯

⁢

x

2

k

+

1

2

k

where the last inequality was obtained by applying the

induction hypothesis

with

n

=

2

k

. Finally, we see that the last expression is equal to

x

1

⁢

x

2

⁢

x

3

⁢

⋯

⁢

x

2

k

+

1

2

k

+

1

and so we have proved the truth of the inequality when the number of terms is a

power of two

.

3

Inequality for

n

numbers implies inequality for

n

-

1

Finally, we prove that if the inequality holds for any

n

, it must also hold for

n

-

1

, and this

proposition

, combined with the preceding proof for powers of

2

, is enough to prove the inequality for any

positive integer

.

Suppose that

x

1

+

x

2

+

⋯

+

x

n

n

≥

x

1

⁢

x

2

⁢

⋯

⁢

x

n

n

is known for a given value of

n

(we just proved that it is true for powers of two, as example).
Then we can replace

x

n

with the

average

of the first

n

-

1

numbers. So

x

1

+

x

2

+

⋯

+

x

n

-

1

+

(

x

1

+

x

2

+

⋯

+

x

n

-

1

n

-

1

)

n

=

(

n

-

1

)

⁢

x

1

+

(

n

-

1

)

⁢

x

2

+

⋯

+

(

n

-

1

)

⁢

x

n

-

1

+

x

1

+

x

2

+

⋯

+

x

n

-

1

n

⁢

(

n

-

1

)

=

n

⁢

x

1

+

n

⁢

x

2

+

⋯

+

n

⁢

x

n

-

1

n

⁢

(

n

-

1

)

=

x

1

+

x

2

+

⋯

+

x

n

-

1

(

n

-

1

)

On the other hand

x

1

⁢

x

2

⁢

⋯

⁢

x

n

-

1

⁢

(

x

1

+

x

2

+

⋯

+

x

n

-

1

n

-

1

)

n

=

x

1

⁢

x

2

⁢

⋯

⁢

x

n

-

1

n

⁢

x

1

+

x

2

+

⋯

+

x

n

-

1

n

-

1

n

which, by

hypothesis

(the inequality holding for

n

numbers) and the observations made above, leads to:

(

x

1

+

x

2

+

⋯

+

x

n

-

1

n

-

1

)

n

≥

(

x

1

⁢

x

2

⁢

⋯

⁢

x

n

)

⁢

(

x

1

+

x

2

+

⋯

+

x

n

-

1

n

-

1

)

and so

(

x

1

+

x

2

+

⋯

+

x

n

-

1

n

-

1

)

n

-

1

≥

x

1

⁢

x

2

⁢

⋯

⁢

x

n

from where we get that

x

1

+

x

2

+

⋯

+

x

n

-

1

n

-

1

≥

x

1

⁢

x

2

⁢

⋯

⁢

x

n

n

-

1

.

So far we have proved the inequality between the arithmetic mean and the geometric mean. The geometric-harmonic inequality is easier. Let

t

i

be

1

/

x

i

.

From

t

1

+

t

2

+

⋯

+

t

n

n

≥

t

1

⁢

t

2

⁢

t

3

⁢

⋯

⁢

t

n

n

we obtain

1

x

1

+

1

x

2

+

1

x

3

+

⋯

+

1

x

n

n

≥

1

x

1

⁢

1

x

2

⁢

1

x

3

⁢

⋯

⁢

1

x

n

n

and therefore

x

1

⁢

x

2

⁢

x

3

⁢

⋯

⁢

x

n

n

≥

n

1

x

1

+

1

x

2

+

1

x

3

+

⋯

+

1

x

n

and so, our proof is completed.

Title

proof of arithmetic-geometric-harmonic means inequality

Canonical

name

ProofOfArithmeticgeometricharmonicMeansInequality

Date of creation

2013-03-22 12:41:25

Last modified on

2013-03-22 12:41:25

Owner

drini (3)

Last modified by

drini (3)

Numerical id

6

Author

drini (3)

Entry type

Proof

Classification

msc 26D15

Related topic

ArithmeticMean

Related topic

GeometricMean

Related topic

HarmonicMean

Related topic

GeneralMeansInequality

Related topic

WeightedPowerMean

Related topic

PowerMean

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