Classroom
In this section of Resonance, we invite readers to pose questions likely to be raised in a
classroom situation. We may suggest strategies for dealing with them, or invite responses,
or both. “Classroom” is equally a forum for raising broader issues and sharing personal
experiences and viewpoints on matters related to teaching and learning science.
Prithwijit De
The Arithmetic Mean – Geometric Mean – Harmonic
Mean: Inequalities and a Spectrum of Applications
Homi Bhabha Centre for Science
Education
Tata Institute of Fundamental
Research
Mumbai
The Arithmetic Mean – Geometric Mean – Harmonic Mean
inequality, AM–GM–HM inequality in short, is one of the
fundamental inequalities in Algebra, and it is used extensively in olympiad mathematics to solve many problems. The
aim of this article is to acquaint students with the inequality,
its proof and various applications.
Before we state the AM–GM–HM inequality, let us define the
Arithmetic Mean (AM), Geometric Mean (GM) and Harmonic
Mean (HM). Let a1 , a2 , . . . , an be n positive real numbers. The
AM (An ), GM (Gn ) and HM (Hn ) of these n numbers are defined
as:
a1 + a2 + · · · + an
,
n
√
Gn = n a1 a2 · · · an ,
n
.
Hn =
1
1
1
+
+ ··· +
a1 a2
an
An =
Email:
[email protected]
AM–GM–HM inequality
is one of the fundamental
inequalities in Algebra.
(1)
(2)
(3)
The AM–GM–HM inequality states that
Keywords
Arithmetic mean, geometric
An ≥ Gn ≥ Hn ,
RESONANCE | December 2016
(4)
mean, harmonic mean, Nesbit’s
inequality, Euler’s inequality.
1119
CLASSROOM
The AM–GM–HM
inequality states that
An ≥ Gn ≥ Hn , and
equality occurs if and
only if
a1 = a2 = · · · = an .
and equality occurs if and only if a1 = a2 = · · · = an .
The result is easy to prove when n = 2. In this case we need to
show,
√
a1 + a2
2
≥ a1 a2 ≥
.
(5)
1
1
2
+
a1 a2
Observe that
√
√
( a1 − a2 )2
a1 + a2 √
− a1 a2 =
≥ 0.
2
2
(6)
This proves that A2 ≥ G2 . To prove G2 ≥ H2 observe that writing
b1 = 1/a1 and b2 = 1/a2 reduces the relevant inequality to
b1 + b2
≥ b1 b2
2
To prove Gn ≥ Hn for
any n is same as proving
An ≥ Gn for the
reciprocals of the given
real numbers.
(7)
which we have already proved. This proves the AM–GM–HM
inequality for n = 2. Note that to prove Gn ≥ Hn for any n is
same as proving An ≥ Gn for the reciprocals of the given real
numbers. Thus to prove the AM–GM–HM inequality it suffices
to prove that An ≥ Gn for all n. For n = 3 the inequality can be
proved by using elementary algebra. We have to prove
a + b + c √3
≥ abc .
3
(8)
This is same as proving,
(a + b + c)3 ≥ 27abc.
(9)
But,
(a + b + c)3 = a3 + b3 + c3 + 3 (a + b) (b + c)(c + a).
(10)
Therefore,
(a + b + c)3 − 27abc = (a3 + b3 + c3 − 3abc)
+ 3{(a + b) (b + c) (c + a)
− 8abc}.
1120
(11)
RESONANCE | December 2016
CLASSROOM
We can write,
a3 + b3 + c3 − 3abc = (a + b)3 + c3 − 3ab (a + b + c)
= (a + b + c) {(a + b)2
− (a + b) c + c2 }
− 3ab (a + b + c) ,
(12)
and further simplify it to,
a3 + b3 + c3 − 3abc = (a + b + c) (a2 + b2 + c2 − ab − bc − ca). (13)
But,
1
a2 + b2 + c2 − ab − bc − ca = {(a − b)2 + (b − c)2 + (c − a)2 }. (14)
2
Thus,
1
a3 + b3 + c3 − 3abc = (a + b + c){(a − b)2 + (b − c)2 + (c − a)2 } ≥ 0.
2
(15)
Using the AM–GM–HM inequality for n = 2 we see that,
√
√
√
(a + b )(b + c) (c + a) ≥ (2 ab) (2 bc) (2 ca) = 8abc. (16)
Therefore,
(a + b + c)3 − 27abc = (a3 + b3 + c3 − 3abc)
+ 3{(a + b) (b + c) (c + a)
− 8abc } ≥ 0 ,
(17)
and we have proved A3 ≥ G3 . Observe that A3 = G3 if and only if
equality occurs in (15) and (16) which is same as saying a = b =
c. One may continue in the same spirit to prove the inequality for
higher values of n but would realize during the process that the
algebra becomes cumbersome and would almost certainly give
up the chase. The case n = 4 can be dealt with quite elegantly by
a clever reduction to the case n = 2. How? Observe that,
a1 + a2 + a3 + a4
=
4
RESONANCE | December 2016
a1 + a2 a3 + a4
+
2
2 .
2
(18)
1121
CLASSROOM
a1 + a2
a3 + a4
√
and b2 =
. Then b1 ≥ a1 a2 ,
Now set b1 =
2
2
√
b2 ≥ a3 a4 and
a1 + a2 a3 + a4
+
√
b1 + b2
2
2
=
≥ b1 b2 = 4 a1 a2 a3 a4 .
2
2
(19)
Hence,
a1 + a2 + a3 + a4 √4
≥ a1 a2 a3 a4 ,
(20)
4
with equality if and only if a1 = a2 = a3 = a4 . This argument
can be easily extended to n = 8 and more generally to any n of
the form 2k for some positive integer k. For instance, when n = 8,
apply A4 ≥ G4 to the set of numbers
a + a a + a a + a a + a 1
2
3
4
5
6
7
8
,
,
,
2
2
2
2
followed by A2 ≥ G2 to the sets of numbers {a1 , a2 }, {a3 , a4 },
{a5 , a6 }, {a7 , a8 } . Thus for n = 2k ,
a1 + a2 + · · · + a2k
√
k
≥ 2 a1 a2 · · · a2k .
k
2
(21)
If n is not a power of two then we can find a unique k such that
2k < n < 2k+1 and we observe that
(2k+1 − n) times
An =
≥
a1 + a2 + · · · + an + (An + · · · + An )
2k+1
2k+1
a1 a2 · · · an .(An )2k+1 −n .
(22)
This can be simplified further to obtain,
Ann ≥ a1 a2 · · · an ,
(23)
and this is same as An ≥ Gn . The idea of this proof is due to
the great French mathematician Augustin-Louis Cauchy (1789–
1857). There is another way to prove An ≥ Gn for all positive
integers n. Let,
ai
xi = √n
,
(24)
a1 a2 · · · an
1122
RESONANCE | December 2016
CLASSROOM
for i = 1, 2, . . . , n. Then x1 x2 · · · xn = 1 and to prove An ≥ Gn we
need to show that,
x1 + x2 + · · · + xn ≥ n ,
(25)
AM–GM–HM inequality
is used extensively in
olympiad mathematics to
solve many problems.
subject to x1 x2 · · · xn = 1. When n = 2 it boils down to showing
x1 + x2 ≥ 2 if x1 x2 = 1. This follows easily because
√
√
√
√
√
x1 + x2 − 2 = ( x1 )2 + ( x2 )2 − 2 x1 x2 = ( x1 − x2 )2 ≥ 0.
(26)
Assume that we have shown that for m ≥ 3 ,
x1 + x2 + · · · + xm−1 ≥ m − 1 ,
(27)
if x1 x2 · · · xm−1 = 1. Now x1 x2 · · · xm = 1. Without loss of generality let x1 ≤ 1 ≤ x2 . Thus,
(x2 − 1) (1 − x1 ) ≥ 0 ,
(28)
x1 + x2 ≥ 1 + x1 x2 .
(29)
implying
Observe that x1 x2 · · · xm = 1 = y1 y2 · · · ym−1 where y1 = x1 x2 and
yi = xi+1 for 2 ≤ i ≤ m − 1. Therefore,
y1 + y2 + · · · + ym−1 ≥ m − 1.
(30)
But y2 + y3 + · · · + ym−1 = x3 + x4 + · · · + xm and 1 + y1 = 1 + x1 x2 ≤
x1 + x2 . Thus,
(31)
x1 + x2 + · · · + xm ≥ m ,
and by induction it follows that x1 + x2 + · · · + xn ≥ n for every
positive integer n ≥ 2 if x1 x2 · · · xn = 1.
Applications
Having presented two proofs of the AM–GM–HM inequality, let
us look at some applications.
1. Let a1 , a2 , . . . , an be positive real numbers. Let b1 , b2 , . . . , bn
be a permutation of a1 , a2 , . . . , an . Then,
RESONANCE | December 2016
1123
CLASSROOM
an
a1 a2
+
+ ··· +
≥ n.
b1 b2
bn
In order to prove this just note that a1 a2 · · · an = b1 b2 · · · bn .
Thus,
a1 a2
an
+
+ ··· +
a1 a2 . . . an
b1 b2
bn
≥ n
= 1.
n
b1 b2 . . . bn
(32)
2. (Nesbit’s Inequality)
Let a, b, c be positive real numbers. Then,
b
c
3
a
+
+
≥ ,
b+c c+a a+b 2
and equality occurs if and only if a = b = c.
b
c
a
+
+
. Then,
Let P =
b+c c+a a+b
1
1
1
+
+
.
P + 3 = (a + b + c)
b+c c+a a+b
(33)
Now A3 ≥ H3 implies,
a+b b+c c+a
+
+
2
2
2 ≥
3
3
.
2
2
2
+
+
a+b b+c c+a
(34)
3
9
and hence P ≥ . Equality occurs if and
2
2
b+c
a+b
only if A3 = H3 which occurs if and only if
=
=
2
2
c+a
, i.e., a = b = c.
2
Therefore P + 3 ≥
3. Let a, b, c be real numbers greater than or equal to 1. Then,
a3 + 2
b3 + 2
c3 + 2
+
+
≥ 9.
b2 − b + 1 c2 − c + 1 a2 − a + 1
To see this, observe that a3 +2−3 (a2 −a+1) = a3 −3a2 +3a−1 =
a3 + 2
≥ 3. Since every term in the
(a − 1)3 ≥ 0, whence 2
a −a+1
1124
RESONANCE | December 2016
CLASSROOM
left hand side of the inequality is positive we have,
a3 + 2
b3 + 2
c3 + 2
+
+
b2 − b + 1 c2 − c + 1 a2 − a + 1
1/3
(a3 + 2) (b3 + 2) (c3 + 2)
≥3
= 9. (35)
(a2 − a + 1) (b2 − b + 1) (c2 − c + 1)
1 1 1
4. Let a, b, c be positive real numbers such that + + = 1.
a b c
Then, we have the inequality,
8
1
1
+
≤ .
(a − 1) (b − 1) (c − 1) (a + 1) (b + 1) (c + 1) 4
For
this, observe that a > 1, b > 1, c > 1 and a − 1 =
2a
1 1
≥ √ (by AM–GM inequality). Similarly, we
a +
b c
bc
2c
2b
get b − 1 ≥ √ and c − 1 ≥ √ . Multiplying these and
ca
ab
taking the reciprocal we obtain,
1
1
≤ .
(36)
(a − 1) (b − 1) (c − 1) 8
√
2
2 2
a+1
= 1+
≥ √
Next observe that
whence, a+1 ≥
a−1
a−1
a−1 √
√
2 2 (a − 1). Similarly we obtain b + 1 ≥ 2 2 (b − 1) and
√
c + 1 ≥ 2 2 (c − 1). Multiplying these yields
√
(a + 1) (b + 1) (c + 1) ≥ 16 2(a − 1) (b − 1) (c − 1) ≥
√
16 2.8 = 64.
Therefore,
1
8
≤ .
(a + 1) (b + 1) (c + 1) 8
(37)
By adding (36) and (37) we get,
1
8
1
+
≤ .
(a − 1) (b − 1) (c − 1) (a + 1) (b + 1) (c + 1) 4
RESONANCE | December 2016
1125
CLASSROOM
5. (Euler’s Inequality)
Let R be the circumradius and r be the inradius of a triangle.
Then R ≥ 2r with equality if and only if the triangle is equilateral.
This is perhaps the most elegant geometric inequality that one
comes across in school level mathematics. There is a geometric proof of this result. Here, we shall present an algebraic
proof.
Let ABC be the given triangle. Denote by a, b, and c the
lengths of the sides BC, CA, and AB respectively. Let s =
a+b+c
, where s is the semi-perimeter, and let Δ denote the
2
area of ABC. Observe that s − a, s − b and s − c are positive
abc
Δ
. Therefore,
quantities. Recall that r = and R =
s
4Δ
abc
R abcs
=
=
.
2
r
4 (s − a) (s − b) (s − c)
4Δ
(38)
Now observe that,
a = 2s − b − c = (s − b) + (s − c) ≥ 2
b = 2s − c − a = (s − c) + (s − a) ≥ 2
(s − b) (s − c), (39)
(s − c) (s − a), (40)
and,
c = 2s − a − b = (s − a) + (s − b) ≥ 2 (s − a) (s − b). (41)
Multiplying these three inequalities leads to,
whence,
abc ≥ 8(s − a) (s − b) (s − c)
(42)
abc
R
=
≥ 2.
r
4(s − a) (s − b) (s − c)
(43)
Equality occurs if and only if s − a = s − b = s − c, that
is, a = b = c, which is same as saying that the triangle is
equilateral.
1126
RESONANCE | December 2016
CLASSROOM
Solving Some Equations Using the Inequalities
In all the examples discussed so far, we applied the AM–GM–
HM inequality to show that one algebraic expression is at least
as large as another algebraic expression under certain conditions
on the variables involved. It can also be used to solve algebraic
equations. Here are some examples.
AM–GM–HM inequality
can be used to solve
algebraic equations.
1. Suppose we wish to find all quadruples a, b, c, d of positive
real numbers that are solutions to the system of equations
a + b + c +d = 4,
1
1
1
1
+ 12 + 12 + 12 (1 + 3abcd ) = 16.
12
a
b
c
d
By A4 ≥ G4 we get,
1=
a + b + c + d √4
≥ abcd ,
4
(44)
1
1
1
1
+ 12 + 12 + 12
12
4
a
b
c
d ≥
(abcd)−12 ,
(45)
4
Multiplying both sides of (45) by (1 + 3abcd ) and simplifying
we obtain,
1
1
1
4 (1 + 3abcd )
1
+
+
+
. (46)
(1 + 3abcd) ≥
a12 b12 c12 d12
(abcd )3
4 (1 + 3t )
. This implies 4t3 −
t3
3t − 1 ≥ 0. But 4t3 − 3t − 1 = (t − 1) (2t + 1)2 . Therefore
(t − 1) (2t + 1)2 ≥ 0. Hence t ≥ 1. But earlier we obtained
t ≤ 1. Therefore t = 1.
Let t = abcd. Then 16 ≥
So now we have,
a + b + c + d √4
= abcd ,
4
a+b+c+d
= 1.
that is, A4 = G4 . Hence a = b = c = d =
4
It is easy to see that (a, b, c, d) = (1, 1, 1, 1) indeed satisfies the
given system of equations.
RESONANCE | December 2016
1127
CLASSROOM
2. Let us solve the simultaneous system of equations for real values of x, y and z.
x=
2y2
2z2
2x2
;
y
=
;
z
=
.
1 + y2
1 + z2
1 + x2
Observe that x = y = z = 0 is a solution. If x, y, z are different
from zero then they must be positive (Why?). Now observe
that for any positive real number t,
1
1
1
t+
≥ t. = 1.
(47)
2
t
t
Therefore,
x=
y=
y
1
1
y+
2
y
≤ y,
z
≤ z,
1
1
z+
2
z
(48)
(49)
and,
z=
x
1
1
x+
2
x
≤ x.
(50)
Thus x ≤ y ≤ z ≤ x. Hence x = y = z and the common value is
1. Thus the solutions are (x, y, z) = (0, 0, 0), (1, 1, 1). It is easy
to see that these triples indeed satisfy the given system.
Optimization Problems
Some optimization questions can also be solved using the AM–
GM–HM inequalities.
1. Among all triangles with a fixed perimeter, let us find which
triangle has the largest area. Let us also determine the maximum area.
1128
RESONANCE | December 2016
CLASSROOM
Let p be the fixed perimeter. Let a, b and c be the sides of a
triangle whose perimeter is p. Let s = p/2. The area of the
triangle is
Δ (a, b, c) = s (s − a) (s − b) (s − c) .
(51)
Some optimization
questions can also be
solved using the
AM–GM–HM
inequalities.
Note that the area is a function of the side-lengths a, b and c.
Now,
(s − a) + (s − b) + (s − c) 3
≥ (s − a) (s − b) (s − c) , (52)
3
with equality if and only if s − a = s − b = s − c, that is,
a = b = c. Therefore,
Δ (a, b, c) =
s (s − a) (s − b) (s − c) ≤
Thus the maximum value of the area is
when the triangle is equilateral.
s2
p2
√ =
√ .
3 3 12 3
(53)
p2
√ and it is attained
12 3
2. Let a sector with central angle θ be removed from a circle of
unit radius and rolled into a right circular cone. What is the
value of θ for which the volume of the cone thus formed is
maximum?
Let r and h be respectively the radius of the base and the height
of the cone. Then,
2πr = θ ,
(54)
and,
θ
πr r2 + h2 = .
2
Solving for h to obtain,
θ 2
h = 1 − r2 = 1 −
.
2π
Therefore the volume of the cone is,
θ 2
2
2
θ
πr h π
= ·
1−
.
V=
3
3 2π
2π
RESONANCE | December 2016
(55)
(56)
(57)
1129
CLASSROOM
AM–GM–HM
inequalities can be
applied to a system of
quadratic equations.
Let t =
θ 2
. Observe that 0 < t < 1 and,
2π
t
t
+ + (1 − t) t t
3
2 2
≥
· · (1 − t)
3
2 2
whence,
(58)
√
t 1−t ≤
2
(59)
√ .
3 3
t
2
Equality is attained if and only if = 1 − t, that is, t = and
2
3
2
this leads to θ = 2π
. Therefore the volume of the cone is
3
2
= 294◦
maximised when the central angle is equal to 2π
3
approximately.
Here is an application of the AM–GM–HM inequalities to a system of quadratic equations.
Suppose that the three equations ax2 −2bx+c = 0, bx2 −2cx+a = 0
and cx2 − 2ax + b = 0 all have only positive roots. Show that
a = b = c.
Let α1 , β1 be the roots of ax2 − 2bx + c = 0, α2 , β2 be the roots of
bx2 − 2cx + a = 0 and α3 , β3 be the roots of cx2 − 2ax + b = 0.
Then,
α 1 + β1 =
2b
c
; α1 β1 = ,
a
a
(60)
α2 + β2 =
2c
a
; α2 β 2 = ,
b
b
(61)
α3 + β 3 =
2a
b
; α3 β 3 = .
c
c
(62)
The roots are positive. By A2 ≥ G2 we obtain,
1130
RESONANCE | December 2016
CLASSROOM
b α1 + β1
=
≥ α1 β1 =
a
2
c α2 + β2
=
≥ α2 β2 =
b
2
a α3 + β 3
=
≥ α3 β3 =
c
2
c
,
a
(63)
a
,
b
(64)
b
,
c
(65)
which are equivalent to b2 ≥ ca, c2 ≥ ab and a2 ≥ bc. Suppose
on the contrary, either a, b, c are distinct or any two of them are
equal but are different from the third. If a, b, c are distinct then
as the inequalities are symmetric with respect to a, b, c, we may
assume without loss of generality that a > b > c. In that case
c2 < ab and we have a contradiction. If a = b > c then also
c2 < ab and we have a contradiction. By symmetry a > b = c and
c = a > b will lead to contradictions too. Therefore a = b = c.
Weighted AM–GM–HM Inequalities
Let us point out to the reader that for the AM–GM inequality to
hold, the quantities a1 , a2 , . . . , an may assumed to be non-negative
but they have to be positive for the AM–HM inequality or the
GM–HM inequality to be valid. The AM–GM–HM inequality
can be generailzed to the weighted AM–GM–HM inequality by
assigning weights to the quantities. We state the inequality without proof.
Let a1 , a2 , . . . , an and w1 , w2 , . . . , wn be positive real numbers.
Then,
n
wi ai
i=1
n
i=1
wi
≥
(aw1 1
·
aw2 2
1
n
wi
· · · awn n ) i=1
n
≥
wi
i=1
n wi
i=1
.
ai
Here is an application.
Let a, b, c be positive real numbers. Prove that
RESONANCE | December 2016
1131
CLASSROOM
1
1
1
+
+
≤
10a + 11b + 11c 11a + 10b + 11c 11a + 11b + 10c
1
1
1
+
+
.
32a 32b 32c
By the weighted AM–HM inequality we have,
32
10/a + 11/b + 11/c
≤
,
10a + 11b + 11c
32
11/a + 10/b + 11/c
32
≤
,
11a + 10b + 11c
32
11/a + 11/b + 10/c
32
≤
.
11a + 11b + 10c
32
Upon adding the above inequalities and rearranging terms we obtain,
1
1
1
+
+
≤
10a + 11b + 11c 11a + 10b + 11c 11a + 11b + 10c
1
1
1
+
+
.
32a 32b 32c
We conclude this article by leaving some problems for the student.
Problems
1. Let a1 , a2 , . . . , an be n positive real numbers whose product
is 1. Prove that
(1 + a1 ) (1 + a2 ) · · · (1 + an ) ≥ 2n .
2. Let a, b, c be real numbers greater than or equal to 1. Prove
that
a (b2 + 3) b (c2 + 3) c (a2 + 3)
+
+
≥ 3.
3c2 + 1
3a2 + 1
3b2 + 1
3. Let 0 = a0 < a1 < · · · < an < an+1 = 1 be such that
a1 + a2 + · · · + an = 1. Prove that
a1
a2
an
1
+
+ ··· +
≥ .
a2 − a0 a3 − a1
an+1 − an−1 an
1132
RESONANCE | December 2016
CLASSROOM
4. Let x1 , x2 , . . . , xn be positive real numbers with x1 + x2 +
· · · + xn = 1. Then show that
n
n
xi
.
≥
2n − 1
i=1 2 − xi
5. If a, b, c ∈ (0, 1) satisfy a + b + c = 2, prove that
abc
≥ 8.
(1 − a) (1 − b) (1 − c)
6. Let a, b, c be positive real numbers. Prove that
a
b
c
+
+
≤
2a + b + c 2b + c + a 2c + a + b
b
c
a
+
+
.
2b + 2c 2c + 2a 2a + 2b
7. Let a, b, c be positive real numbers such that ab + bc + ca =
3. Prove that
a3 + b3
b3 + c3
2
2
+ (3b + 2) 2
+
(3a + 2) 2
a + ab + b2
b + bc + c2
c3 + a3
≥ 10abc.
(3c2 + 2) 2
c + ca + a2
8. Let a, b, c be positive real numbers such that abc = 1.
Prove that
√
√
√
√
√
√
( b + c)4
( c + a)4
( a + b)4
+
+
≥ 24.
a+b
b+c
c+a
9. Let ha , hb and hc be the altitudes and r the inradius of a
triangle. Prove that
ha − 2r hb − 2r hc − 2r 3
+
+
≥ .
ha + 2r hb + 2r hc + 2r 5
10. Let a, b, c be positive numbers and abc = 8. Prove that
1
1
1
a4 + b4 b4 + c4 c4 + a4
+
+
≥ 64 5 + 5 + 5 + 6
c3
a3
b3
a
b
c
Suggested Reading
[1] B J Venkatachala, Inequalities - An Approach Through Problems, Hindustan
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RESONANCE | December 2016
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