The Proofs of the Arithmetic-Geometric Mean Inequality Through Both the Product and Binomial Inequalities 


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS
Vol. 11, No. 4, 2018, 1100-1107
ISSN 1307-5543 – www.ejpam.com
Published by New York Business Global

The Proofs of the Arithmetic-Geometric Mean
Inequality Through Both the Product and Binomial

Inequalities

Benedict Barnes1,∗, E. Harris1, N. F. Darquah2, G. Hughes3

1 Department of Mathematics, Kwame Nkrumah University of Science and Technology,
Kumasi, Ghana
2 Department of Computer Science, Kwame Nkrumah University of Science and Technology,
Kumasi, Ghana
3 Department of Economics, Central University, Accra, Ghana

Abstract. In this paper, we show new ways of proving the arithmetic-geometric mean AGM
inequality through the first product and the second product inequalities. In addition, we prove the
AGM inequality through the binomial inequalities. These methods are alternative ways of proving
AGM inequalities.

2010 Mathematics Subject Classifications: 44B51 44B52

Key Words and Phrases: Arithmetic-geometric inequality, first product inequality, second
product inequality and binomial inequalities

1. Introduction

The importance of inequalities cannot be underestimated as they play central role in
mathematical analysis. In the 21st century, the AGM inequality has received much atten-
tion and has been applied in the areas of statistics and engineering. The AGM inequality
was first introduced by Lagrange (as cited in [1]). Since then the AGM inequality has
used to establish the relationships between the areas and perimeters of geometrical plane
figures, the so-called isoperimetric inequalities, for example, see authors in [2, 3]. However,
a substantial progress has been made to increase the understanding of the AGM inequality
by the researchers across the globe. In [4], the author proved the AGM inequality using
the heuristic method. The author in [5] observed that the mutatis mutandis’ method for
proving the AGM inequality was similar to the result obtained by Jacobsthal and Rado, for

∗Corresponding author.
DOI: https://doi.org/10.29020/nybg.ejpam.v11i4.3300

Email addresses: ewiekwamina@gmail.com bbarnes.cos@knust.edu.gh (B. Barnes)

http://www.ejpam.com 1100 c© 2018 EJPAM All rights reserved.



B. Barnes et al. / Eur. J. Pure Appl. Math, 11 (4) (2018), 1100-1107 1101

example, see [6]. In [7], the author proved the AGM inequality through Taylor’s theorem
about x = 1

2 by setting the function f(x) equals to the Heinz mean. Thus,

f(x) =
axb1−x + a1−xbx

2
, ∀ 0 ≤ x ≤ 1 and f ∈ C2[0, 1].

In a similar development, another refinement of the AGM inequality was given by the
authors in [8]. They obtained their result through Taylor’s theorem about x = 1

2 , by
setting

f(x) =
ax + a1−x

2
, ∀ 0 ≤ x ≤ 1 and f ∈ Ck(0,∞).

Notwithstanding, in [9], the authors proved the AGM inequality with the use of second
derivative test by setting

f(x) =
(x− a)2

a(x+ max{x, a})
+ lnx, ∀ a > 0 and f ∈ C2(0,∞).

Some researchers have applied AGM inequality to solve matrix algebra. For example,
see authors in [10, 11]. The author in [12] extended the AGM inequality to include the
harmonic mean called AGHM inequality.

In this paper, the AGM inequality is proved through the first product inequality in
a closed interval [0, 2], then through the second product inequality in a half open ended
interval [2,∞) and finally, through the binomial inequalities of rational numbers.

Definition 1. Let A be a linear vector space defined over the real number field R. A
scalar-valued function p : A×A→ R that associates with each pair a1, a2 of vectors in A
a scalar, denoted (a1, a2), is called an inner product on A if and only if

(i) (a1, a2) > 0 whenever a 6= 0, and (a1, a1) = 0 if and only if a1 = 0

(ii) (a1, a2) = (a2, a1), ∀ a1, a2 ∈ A

(iii) (αa1 + βa2, a3) = α(a1, a3) + β(a2, a3), ∀ α, β ∈ R, and a1, a2, a3 ∈ V . See [13]

Definition 2. Let A be a linear space over R. A norm on A is a real-valued function
‖ ·‖ : A→ [0,∞) such that for any a1, a2 ∈ A and α ∈ R the following conditions are met:

‖a‖ ≥ 0, and ‖a‖ = 0, iff a = 0

‖αa‖ = |α|‖ua‖, ∀ a ∈ A and α ∈ R

‖a1 ± a2‖ ≤ ‖a1‖+ ‖a2‖, ∀ a1, a2 ∈ A,

See [14].



B. Barnes et al. / Eur. J. Pure Appl. Math, 11 (4) (2018), 1100-1107 1102

Definition 3 (First and Second Product Inequalities). Let a1 and a2 be any two positive
real numbers, then

(i)‖a1‖‖a2‖ ≤ ‖a1‖+ ‖a2‖, ∀a1, a2 ∈ [0, 2]. (1)

(ii)‖a1‖+ ‖a2‖ ≤ ‖a1‖‖a2‖, ∀a1, a2 ∈ [2,∞). (2)

See [15].

1.1. The Proof of the AGM Inequality through the First Product In-
equality

In this section, we obtain the AGM inequality through both the first and second
product inequalities by induction as follows. Multiplying both sides of inequality (1) by
(1− p) yields

(1− p)
(
‖a1‖‖a2‖

)
≤ (1− p)

{
‖a1‖+ ‖a2‖

}
, ∀ p ∈ [0, 1]. (3)

But, we see that:(
‖a1‖‖a2‖

)(1−p)
≤ (1− p)

(
‖a1‖‖a2‖

)
. (4)

Substituting inequality (4) into inequality (3) yields(
‖a1‖‖a2‖

)(1−p)
≤ (1− p)

(
‖a1‖+ ‖a2‖

)
. (5)

Setting p = 1
2 into inequality (5) yields(

‖a1‖‖a2‖
) 1

2 ≤ 1

2

(
‖a1‖+ ‖a2‖

)
⇒
( 2∏

i=1

ai

) 1
2 ≤ 1

2

2∑
i=1

ai.

We can see that for any three positive real numbers n = 3, the following inequality
holds.

‖a1‖‖a2‖‖a3‖ ≤ ‖a1‖+ ‖a2‖+ ‖a3‖, ∀a1, a2, a3 ∈ [0, 2]

⇒ (1− p)
(
‖a1‖‖a2‖‖a3‖

)
= (1− p)

(
‖a1‖+ ‖a2‖+ ‖a3‖

)
⇒
(
‖a1‖‖a2‖‖a3‖

)(1−p)
≤ (1− p)

(
‖a1‖+ ‖a2‖+ ‖a3‖

)
.

Setting p = 2
3 into the above inequality, we obtain(

‖a1‖‖a2‖‖a3‖
) 1

3 ≤ 1

3

(
‖a1‖+ ‖a2‖+ ‖a3‖

)



B. Barnes et al. / Eur. J. Pure Appl. Math, 11 (4) (2018), 1100-1107 1103

⇒
( 3∏

i=1

ai

) 1
3 ≤ 1

3

3∑
i=1

ai.

For any number of positive real numbers n, the following inequalities are observed:

‖a1‖‖a2‖ . . . ‖an‖ ≤ ‖a1‖+ ‖a2‖+ . . .+ ‖an‖ ∀a1, a2, . . . , an ∈ [0, 2]

⇒ (1− p)
(
‖a1‖‖a2‖, . . . , ‖an‖

)
≤ (1− p)

(
‖a1‖+ ‖a2‖+ . . .+ ‖an‖

)
⇒
(
‖a1‖‖a2‖, . . . , ‖an‖

)(1−p)
≤ (1− p)

(
‖a1‖+ ‖a2‖+ . . .+ ‖an‖

)
.

Setting p = (n−1)
n into the above inequality yields(
‖a1‖‖a2‖, . . . , ‖an‖

) 1
n ≤ 1

n

(
‖a1‖+ ‖a2‖+ . . .+ ‖an‖

)
⇒
( n∏

i=1

ai

) 1
n ≤ 1

n

n∑
i=1

ai ∀ a1, a2, . . . , an ∈ [0, 2].

1.2. The Proof of the AGM Inequality through the Second Product In-
equality

In a similar development, we prove the AGM inequality through the second product
inequality. The AGM inequality is obtained by induction. Multiplying both sides of
inequality (2) by p yields

p
(
‖a1‖+ ‖a2‖

)
≤ p
{
‖a1‖‖a2‖

}
, ∀ p ∈ [0, 1]. (6)

But, we see that:(
‖a1‖‖a2‖

)p
≤ p
(
‖a1‖‖a2‖

)
. (7)

Substituting inequality (7) into inequality (6), we get(
‖a1‖‖a2‖

)p
≤ p
(
‖a1‖+ ‖a2‖

)
. (8)

Setting p = 1
2 into inequality (8) yields(

‖a1‖‖a2‖
) 1

2 ≤ 1

2

(
‖a1‖+ ‖a2‖

)
⇒
( 2∏

i=1

ai

) 1
2 ≤ 1

2

2∑
i=1

ai.

Again, we observed that:

‖a1‖+ ‖a2‖+ ‖a3‖ ≤ ‖a1‖‖a2‖‖a3‖ ∀a1, a2, a3 ∈ [2,∞)



B. Barnes et al. / Eur. J. Pure Appl. Math, 11 (4) (2018), 1100-1107 1104

⇒ p
(
‖a1‖+ ‖a2‖+ ‖a3‖

)
≤ p

(
‖a1‖‖a2‖‖a3‖

)
. (9)

We observed that:(
‖a1‖‖a2‖‖a3‖

)p
≤ p
(
‖a1‖‖a2‖‖a3‖

)
. (10)

Substituting inequality (10) into inequality (9) yields(
‖a1‖‖a2‖‖a3‖

)p
≤ p
(
‖a1‖+ ‖a2‖+ ‖a3‖

)
.

Setting p = 1
3 into the above inequality, we obtain(

‖a1‖‖a2‖‖a3‖
) 1

3 ≤ 1

3

(
‖a1‖+ ‖a2‖+ ‖a3‖

)
⇒
( 3∏

i=1

ai

) 1
3 ≤ 1

3

3∑
i=1

ai.

We observed for any number of positive real numbers n, we have:

‖a1‖+ ‖a2‖+ . . .+ ‖an‖ ≤ ‖a1‖‖a2‖ . . . ‖an‖ ∀a1, a2, . . . , an ∈ [2,∞)

⇒ p
(
‖a1‖+ ‖a2‖+ . . .+ ‖an‖

)
≤ p

(
‖a1‖‖a2‖ . . . ‖an‖

)
. (11)

But we see that:(
‖a1‖‖a2‖ . . . ‖an‖

)p
≤ p
(
‖a1‖‖a2‖ . . . ‖an‖

)
. (12)

Substituting inequality (12) into inequality (11) yields(
‖a1‖‖a2‖ . . . ‖an‖

)p
≤ p
(
‖a1‖+ ‖a2‖+ . . .+ ‖an‖

)
.

Setting p = 1
n in the above equation yields

( n∏
i=1

ai

) 1
n ≤ 1

n

n∑
i=1

ai ∀ a1, a2, . . . , an ∈ [2,∞).

1.3. The Proof of the AGM Inequality through the Binomial Inequalities

In this section, the AGM inequality is proved through new binomial inequalities of
rational numbers. We can see that n = 2, the following inequality holds.

(
√
a1 +

√
a2)

2 ≥ 0

⇒
(
‖a1a2‖

) 1
2 ≤ 1

2

(
‖a1‖+ ‖a2‖

)



B. Barnes et al. / Eur. J. Pure Appl. Math, 11 (4) (2018), 1100-1107 1105

⇒ ‖a1‖
1
2 ‖a2‖

1
2 ≤ 1

2

(
‖a1‖+ ‖a2‖

)
⇒
( 2∏

i=1

ai

) 1
2 ≤ 1

2

2∑
i=1

ai.

Also, let a1, a2 and a3 be three positive real numbers, then

(
√
a1 +

√
a2 +

√
a3)

2 ≥ 0

⇒ −2
(√

a1a2 +
√
a1a3 +

√
a2a3

)
≤ (a1 + a2 + a3)

⇒ −2

3

(√
a1a2 +

√
a1a3 +

√
a2a3

)
=

(a1 + a2 + a3)

3

⇒ −4

3

{1

2

(√
a1a2 +

√
a3

)}
≤ (a1 + a2 + a3)

3

⇒ ‖−4

3

{1

2

(√
a1a2 +

√
a3

)}
‖ ≤ ‖(a1 + a2 + a3)

3
‖

⇒ 4

3

∥∥∥{1

2

(√
a1a2 +

√
a3

)}∥∥∥ ≤ 1

3
(‖a1‖+ ‖a2‖+ ‖a3‖)

⇒ 4

3

(
‖a1‖‖a2‖‖a3‖

) 1
4 ≤ 1

3

(
‖a1‖+ ‖a2‖+ ‖a3‖

)
. (13)

We see that:(
‖a1‖‖a2‖‖a3‖

) 1
3 ≤ 4

3

(
‖a1‖‖a2‖‖a3‖

) 1
4
. (14)

Substituting inequality (13) into inequality (14) yields

⇒
( 3∏

i=1

ai

) 1
3 ≤ 1

3

3∑
i=1

ai.

Similarly, we can see that:

(
√
a1 +

√
a2 +

√
a3 +

√
a4)

2 ≥ 0

⇒ (a1 + a2 + a3 + a4) ≥ −2
(√

a1a2 +
√
a3a4 +

√
a1a3 +

√
a1a4 +

√
a2a3 +

√
a2a4

)
⇒ −1

2

(√
a1a2 +

√
a3a4 +

√
a1a3 +

√
a1a4 +

√
a2a3 +

√
a2a4

)
≤ (a1 + a2 + a3 + a4)

4

⇒ −1

2

(√
a1a2 +

√
a3a4

)
≤ (a1 + a2 + a3 + a4)

4

⇒ ‖− 1

2

(√
a1a2 +

√
a3a4

)
‖ = ‖(a1 + a2 + a3 + a4)

4
‖

⇒ 1

2
‖
(√

a1a2 +
√
a3a4

)
‖ ≤ 1

4
(‖a1‖+ ‖a2‖+ ‖a3‖+ ‖a4‖)

⇒
(
‖a1‖‖a2‖‖a3‖‖a4‖

) 1
4 ≤ 1

4
(‖a1‖+ ‖a2‖+ ‖a3‖+ ‖a4‖)



REFERENCES 1106

⇒
( 4∏

i=1

ai

) 1
4 ≤ 1

4

4∑
i=1

ai.

By the principle of induction, we see that for any n number of real numbers, we have:( n∏
i=1

ai

) 1
n ≤ 1

n

n∑
i=1

ai.

This completes the prove.

2. Conclusion

In a nutsell, we have provided the new ways of proving the AGM inequality through
the product and binomial inequalities.

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