A.M., G.M., H.M. Inequalities

# A.M., G.M., H.M. Inequalities | Mathematics (Maths) Class 11 - Commerce PDF Download

``` Page 1

H. A.M. ? G.M. ? H.M. INEQUALITIES
Theorem : If A, G, H are respectively AM, GM, HM between a & b both being unequal & positive then,
(i) G² = AH (ii)  A > G > H (G > 0) .  Note that  A, G, H  constitute a GP.
Let a
1
, a
2
, a
3
,....... a
n
be n positive real numbers, then we define their
A.M. =
n
a ...... a a a
n 3 2 1
? ? ? ?
, G.M. = (a
1
a
2
a
3
...........a
n
)
1/n
and H.M. =
n 2 1
a
1
.....
a
1
a
1
n
? ? ?
It can be shown that A.M. ? G.M. ? H.M. and equality holds at either places iff
a
1
= a
2
= a
3
= ............... = a
n
Ex. If a, b, c, d be four distinct positive quantities in G.P., then show that
(a) a + d > b + c (b)
?
?
?
?
?
?
? ? ? ?
1
ac
1
bd
1
2
ab
1
cd
1
.
Sol. Since a, b, c d are in G.P., therefore
(a)  using A.M. > G.M., we have for the first three terms
a c
2
?
> b i.e. a + c > 2b ....(1)
2
d b ?
> c i.e. b + d > 2c ....(2)
From results (1) and (2), we have a + c + b + d > 2b + 2c i.e. a + d > b + c
which is the desired result.
Page 2

H. A.M. ? G.M. ? H.M. INEQUALITIES
Theorem : If A, G, H are respectively AM, GM, HM between a & b both being unequal & positive then,
(i) G² = AH (ii)  A > G > H (G > 0) .  Note that  A, G, H  constitute a GP.
Let a
1
, a
2
, a
3
,....... a
n
be n positive real numbers, then we define their
A.M. =
n
a ...... a a a
n 3 2 1
? ? ? ?
, G.M. = (a
1
a
2
a
3
...........a
n
)
1/n
and H.M. =
n 2 1
a
1
.....
a
1
a
1
n
? ? ?
It can be shown that A.M. ? G.M. ? H.M. and equality holds at either places iff
a
1
= a
2
= a
3
= ............... = a
n
Ex. If a, b, c, d be four distinct positive quantities in G.P., then show that
(a) a + d > b + c (b)
?
?
?
?
?
?
? ? ? ?
1
ac
1
bd
1
2
ab
1
cd
1
.
Sol. Since a, b, c d are in G.P., therefore
(a)  using A.M. > G.M., we have for the first three terms
a c
2
?
> b i.e. a + c > 2b ....(1)
2
d b ?
> c i.e. b + d > 2c ....(2)
From results (1) and (2), we have a + c + b + d > 2b + 2c i.e. a + d > b + c
which is the desired result.
(b) using G.M. > H.M., we have for the first three terms b >
c a
ac 2
?
i.e. ab + bc > 2ac....(3)
and for the last three terms c >
2bd
b d ?
i.e. bc + cd > 2bd ....(4)
From results (3) and (4), we have ab + cd + 2bc > 2ac+ 2bd i.e. ab + cd > 2(ac + bd - bc)
i.e.
1 1 1 1 1
2
? ?
? ? ? ?
? ?
? ?
[dividing both sides by abcd] which is the desired result.
Ex. If a, b, c are in H.P. and they are distinct and positive then prove that a
n
+ c
n
> 2b
n
Sol. Let a
n
and c
n
be two numbers then
2
c a
n n
?
> (a
n
c
n
)
1/2
? a
n
+ c
n
> 2 (ac)
n/2
.....(i)
Also G.M. > H.M. i.e.
ac
> b (ac)
n/2
> b
n
.....(ii) hence from (i) and (ii) a
n
+ c
n
> 2b
n
Ex. Show that n
n

n 2
2
1 n
?
?
?
?
?
? ?
> (n !)
3
.
Sol. Consider the unequal positive numbers 1
3
, 2
3
, ........n
3
.
Since A.M. > G.M., therefore
n
n ...... 2 1
3 3 3
? ? ?
> (1
3
. 2
3
....n
3
)
1/n
, i.e.,
4
) 1 n ( n
2
?
> {(n !)
3
}
1/n
.
Raising both sides to power n, we have  n
n
n 2
2
1 n
?
?
?
?
?
? ?
> (n !)
3
.
Ex. If a, b, c are positive real numbers such that a + b + c = 1, then find the minimum value of
ac
1
bc
1
ab
1
? ?
.
Sol.
abc
1
abc
c b a
ac
1
bc
1
ab
1
?
? ?
? ? ?
Also,
3
c b a ? ?
? (abc)
1/3
? abc ?
27
1
?
abc
1
? 27. Hence
ac
1
bc
1
ab
1
? ?
? 27
Ex. In the equation x
4
+px
3
+qx
2
+rx+5=0 has four positive real roots, then find the minimum value of pr.
Sol. Let ? ? ?? ? ? ?? ?? be the four positive real roots of the given equation. Then
? ? ? ? ? ? ? ? ? ? ? ? ? = –p, ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ?? = q, ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = –r, ? ? ? ? = 5.
Using A.M. ? G.M. ? ? ?
5
4
.
4
4 3 3 3 3
4
? ? ? ?? ? ? ? ? ? ? ? ?? ?
? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
4
r
4
p
? 5 ? pr ? 80 ? minimum value of pr = 80.
Ex.65 If

S = a+b+ c then prove that
S
S a ?
+
S
S b ?
+
S
S c ?

>
9
2
where  a

,

b & c  are distinct positive reals.
Sol.
(S a) S b) (S c)
3
? ? ? ? ?

>

? ?
( ) ( ) ( )
/
S a S b S c ? ? ?
1 3
........(i)
Page 3

H. A.M. ? G.M. ? H.M. INEQUALITIES
Theorem : If A, G, H are respectively AM, GM, HM between a & b both being unequal & positive then,
(i) G² = AH (ii)  A > G > H (G > 0) .  Note that  A, G, H  constitute a GP.
Let a
1
, a
2
, a
3
,....... a
n
be n positive real numbers, then we define their
A.M. =
n
a ...... a a a
n 3 2 1
? ? ? ?
, G.M. = (a
1
a
2
a
3
...........a
n
)
1/n
and H.M. =
n 2 1
a
1
.....
a
1
a
1
n
? ? ?
It can be shown that A.M. ? G.M. ? H.M. and equality holds at either places iff
a
1
= a
2
= a
3
= ............... = a
n
Ex. If a, b, c, d be four distinct positive quantities in G.P., then show that
(a) a + d > b + c (b)
?
?
?
?
?
?
? ? ? ?
1
ac
1
bd
1
2
ab
1
cd
1
.
Sol. Since a, b, c d are in G.P., therefore
(a)  using A.M. > G.M., we have for the first three terms
a c
2
?
> b i.e. a + c > 2b ....(1)
2
d b ?
> c i.e. b + d > 2c ....(2)
From results (1) and (2), we have a + c + b + d > 2b + 2c i.e. a + d > b + c
which is the desired result.
(b) using G.M. > H.M., we have for the first three terms b >
c a
ac 2
?
i.e. ab + bc > 2ac....(3)
and for the last three terms c >
2bd
b d ?
i.e. bc + cd > 2bd ....(4)
From results (3) and (4), we have ab + cd + 2bc > 2ac+ 2bd i.e. ab + cd > 2(ac + bd - bc)
i.e.
1 1 1 1 1
2
? ?
? ? ? ?
? ?
? ?
[dividing both sides by abcd] which is the desired result.
Ex. If a, b, c are in H.P. and they are distinct and positive then prove that a
n
+ c
n
> 2b
n
Sol. Let a
n
and c
n
be two numbers then
2
c a
n n
?
> (a
n
c
n
)
1/2
? a
n
+ c
n
> 2 (ac)
n/2
.....(i)
Also G.M. > H.M. i.e.
ac
> b (ac)
n/2
> b
n
.....(ii) hence from (i) and (ii) a
n
+ c
n
> 2b
n
Ex. Show that n
n

n 2
2
1 n
?
?
?
?
?
? ?
> (n !)
3
.
Sol. Consider the unequal positive numbers 1
3
, 2
3
, ........n
3
.
Since A.M. > G.M., therefore
n
n ...... 2 1
3 3 3
? ? ?
> (1
3
. 2
3
....n
3
)
1/n
, i.e.,
4
) 1 n ( n
2
?
> {(n !)
3
}
1/n
.
Raising both sides to power n, we have  n
n
n 2
2
1 n
?
?
?
?
?
? ?
> (n !)
3
.
Ex. If a, b, c are positive real numbers such that a + b + c = 1, then find the minimum value of
ac
1
bc
1
ab
1
? ?
.
Sol.
abc
1
abc
c b a
ac
1
bc
1
ab
1
?
? ?
? ? ?
Also,
3
c b a ? ?
? (abc)
1/3
? abc ?
27
1
?
abc
1
? 27. Hence
ac
1
bc
1
ab
1
? ?
? 27
Ex. In the equation x
4
+px
3
+qx
2
+rx+5=0 has four positive real roots, then find the minimum value of pr.
Sol. Let ? ? ?? ? ? ?? ?? be the four positive real roots of the given equation. Then
? ? ? ? ? ? ? ? ? ? ? ? ? = –p, ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ?? = q, ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = –r, ? ? ? ? = 5.
Using A.M. ? G.M. ? ? ?
5
4
.
4
4 3 3 3 3
4
? ? ? ?? ? ? ? ? ? ? ? ?? ?
? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
4
r
4
p
? 5 ? pr ? 80 ? minimum value of pr = 80.
Ex.65 If

S = a+b+ c then prove that
S
S a ?
+
S
S b ?
+
S
S c ?

>
9
2
where  a

,

b & c  are distinct positive reals.
Sol.
(S a) S b) (S c)
3
? ? ? ? ?

>

? ?
( ) ( ) ( )
/
S a S b S c ? ? ?
1 3
........(i)
&

1
3

1 1 1
S a S b S c ?
?
?
?
?
?
?
?
?
?
?

>

1 1 1
1 3
S a S b S c ?
?
?
?
?
?
?
?
?
?
?
/
........(ii)
Multiple (i) & (ii) =
1
9
2 (a + b + c)

1 1 1
S a S b S c ?
?
?
?
?
?
?
?
?
?
?
>

1
=
2
9

S S S
S a S b S c
? ?
? ?
? ?
? ? ?
? ?

>

1 or
S
S a ?
+
S
S b ?
+
S
S c ?

>
9
2
Ex.66 Let  a > 1  and  n ? N . Prove the inequality,   a
n
? 1  ?  n ( ) a a
n n ? ?
?
1
2
1
2
.
Sol. Put   a = ?
2
T P T ? ?? ?
2n
? 1  ? ? ?    n ( ?
n + 1
? ? ?
n ? 1
)   ? ? ?   ?
2n
? 1  ? ? ?

n ?
n ? 1
( ?
2
? 1)   ?
?
?
2
2
1
1
n
?
?
?   n . ?
n ? 1
? ? ?1 + ?
2
+ ?
4
+ ...... + a
2 (n ? 1)
?   n  ?
n ? 1
?
1
2 4 2 1
. . . . . . . . . .
( )
? ? ?
n
n
?
. Which is True as  A.M. ?

G.M.
Ex. If

s

be the sum of

n

positive unequal quantities  a, b, c  then prove the inequality ;
s
s a
s
s b
s
s c ? ? ?
? ?
+ ...... >
n
n
2
1 ?
(n ? 2).
Sol. AM  >  GM  ?  [(s ? a) + (s ? b) + (s ? c) + ...... ]  >  n [(s ? a) + (s ? b) + (s ? c) + ...... ]
1/n
?
1 1 1
s a s b s c ?
?
?
?
?
?
?
?
?
?
?
?
. . . . . .
>  n
1
1
( ) ( ) ( )
/
s a s b s c
n
? ? ?
?
?
?
?
?
?
Multiplying   [(s ? a) + (s ? b) + ...... ]
1 1 1
s a s b s c ?
?
?
?
?
?
?
?
?
?
?
?
. . . . . .
>  n
2
or
( ) ( ) ( )
. . . . . .
s a
s
s b
s
s c
s
?
?
?
?
?
?
?
?
?
?
?
?

s
s a
s
s b
s
s c ?
?
?
?
?
?
?
?
?
?
?
?
. . . . . .
>  n
2
?
s
s a
s
s b
s
s c ?
?
?
?
?
?
?
?
?
?
?
?
. . . . . .

n s s
s
? ?
?
?
?
?
?  >  n
2
or
s a
s
s b
s
s c
s
?
?
?
?
?
+ ......  >
n
n
2
1 ?
Ex. If  a, b, c, d  are all positive and the sum of any three is greater than twice the fourth, then show that,
a b c d

>

(b + c + d ? 2a)

(c + d + a ? 2b)

(d + a + b ? 2c)

(a + b + c ? 2d).
Sol. Use    A.M.  >  G.M.
a + b + c ? 2d  =  m
1
,  b + c + d ? 2a  =  m
2
,  c + d + a ? 2b  =  m
3
,  d + a + b ? 2c  =  m
4
Now m
1
+ m
2
+ m
3
= 3

c ?    c  =
m m m
1 2 3
3
? ?
>  (m
1
m
2
m
3
)
1/3
m
2
+ m
3
+ m
4

=  3

d ?     d  =
m m m
2 3 4
3
? ?
>  (m
2
m
3
m
4
)
1/3
m
3
+ m
4
+ m
1

=  3

a ?     a  =
m m m
3 4 1
3
? ?
>  (m
3
m
4
m
1
)
1/3
m
4
+ m
1
+ m
2

=  3

b ?     b  =
m m m
4 1 2
3
? ?
>  (m
4
m
1
m
2
)
1/3
Hence   a b c d  >  m
1
m
2
m
3
m
4
?   Result
Page 4

H. A.M. ? G.M. ? H.M. INEQUALITIES
Theorem : If A, G, H are respectively AM, GM, HM between a & b both being unequal & positive then,
(i) G² = AH (ii)  A > G > H (G > 0) .  Note that  A, G, H  constitute a GP.
Let a
1
, a
2
, a
3
,....... a
n
be n positive real numbers, then we define their
A.M. =
n
a ...... a a a
n 3 2 1
? ? ? ?
, G.M. = (a
1
a
2
a
3
...........a
n
)
1/n
and H.M. =
n 2 1
a
1
.....
a
1
a
1
n
? ? ?
It can be shown that A.M. ? G.M. ? H.M. and equality holds at either places iff
a
1
= a
2
= a
3
= ............... = a
n
Ex. If a, b, c, d be four distinct positive quantities in G.P., then show that
(a) a + d > b + c (b)
?
?
?
?
?
?
? ? ? ?
1
ac
1
bd
1
2
ab
1
cd
1
.
Sol. Since a, b, c d are in G.P., therefore
(a)  using A.M. > G.M., we have for the first three terms
a c
2
?
> b i.e. a + c > 2b ....(1)
2
d b ?
> c i.e. b + d > 2c ....(2)
From results (1) and (2), we have a + c + b + d > 2b + 2c i.e. a + d > b + c
which is the desired result.
(b) using G.M. > H.M., we have for the first three terms b >
c a
ac 2
?
i.e. ab + bc > 2ac....(3)
and for the last three terms c >
2bd
b d ?
i.e. bc + cd > 2bd ....(4)
From results (3) and (4), we have ab + cd + 2bc > 2ac+ 2bd i.e. ab + cd > 2(ac + bd - bc)
i.e.
1 1 1 1 1
2
? ?
? ? ? ?
? ?
? ?
[dividing both sides by abcd] which is the desired result.
Ex. If a, b, c are in H.P. and they are distinct and positive then prove that a
n
+ c
n
> 2b
n
Sol. Let a
n
and c
n
be two numbers then
2
c a
n n
?
> (a
n
c
n
)
1/2
? a
n
+ c
n
> 2 (ac)
n/2
.....(i)
Also G.M. > H.M. i.e.
ac
> b (ac)
n/2
> b
n
.....(ii) hence from (i) and (ii) a
n
+ c
n
> 2b
n
Ex. Show that n
n

n 2
2
1 n
?
?
?
?
?
? ?
> (n !)
3
.
Sol. Consider the unequal positive numbers 1
3
, 2
3
, ........n
3
.
Since A.M. > G.M., therefore
n
n ...... 2 1
3 3 3
? ? ?
> (1
3
. 2
3
....n
3
)
1/n
, i.e.,
4
) 1 n ( n
2
?
> {(n !)
3
}
1/n
.
Raising both sides to power n, we have  n
n
n 2
2
1 n
?
?
?
?
?
? ?
> (n !)
3
.
Ex. If a, b, c are positive real numbers such that a + b + c = 1, then find the minimum value of
ac
1
bc
1
ab
1
? ?
.
Sol.
abc
1
abc
c b a
ac
1
bc
1
ab
1
?
? ?
? ? ?
Also,
3
c b a ? ?
? (abc)
1/3
? abc ?
27
1
?
abc
1
? 27. Hence
ac
1
bc
1
ab
1
? ?
? 27
Ex. In the equation x
4
+px
3
+qx
2
+rx+5=0 has four positive real roots, then find the minimum value of pr.
Sol. Let ? ? ?? ? ? ?? ?? be the four positive real roots of the given equation. Then
? ? ? ? ? ? ? ? ? ? ? ? ? = –p, ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ?? = q, ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = –r, ? ? ? ? = 5.
Using A.M. ? G.M. ? ? ?
5
4
.
4
4 3 3 3 3
4
? ? ? ?? ? ? ? ? ? ? ? ?? ?
? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
4
r
4
p
? 5 ? pr ? 80 ? minimum value of pr = 80.
Ex.65 If

S = a+b+ c then prove that
S
S a ?
+
S
S b ?
+
S
S c ?

>
9
2
where  a

,

b & c  are distinct positive reals.
Sol.
(S a) S b) (S c)
3
? ? ? ? ?

>

? ?
( ) ( ) ( )
/
S a S b S c ? ? ?
1 3
........(i)
&

1
3

1 1 1
S a S b S c ?
?
?
?
?
?
?
?
?
?
?

>

1 1 1
1 3
S a S b S c ?
?
?
?
?
?
?
?
?
?
?
/
........(ii)
Multiple (i) & (ii) =
1
9
2 (a + b + c)

1 1 1
S a S b S c ?
?
?
?
?
?
?
?
?
?
?
>

1
=
2
9

S S S
S a S b S c
? ?
? ?
? ?
? ? ?
? ?

>

1 or
S
S a ?
+
S
S b ?
+
S
S c ?

>
9
2
Ex.66 Let  a > 1  and  n ? N . Prove the inequality,   a
n
? 1  ?  n ( ) a a
n n ? ?
?
1
2
1
2
.
Sol. Put   a = ?
2
T P T ? ?? ?
2n
? 1  ? ? ?    n ( ?
n + 1
? ? ?
n ? 1
)   ? ? ?   ?
2n
? 1  ? ? ?

n ?
n ? 1
( ?
2
? 1)   ?
?
?
2
2
1
1
n
?
?
?   n . ?
n ? 1
? ? ?1 + ?
2
+ ?
4
+ ...... + a
2 (n ? 1)
?   n  ?
n ? 1
?
1
2 4 2 1
. . . . . . . . . .
( )
? ? ?
n
n
?
. Which is True as  A.M. ?

G.M.
Ex. If

s

be the sum of

n

positive unequal quantities  a, b, c  then prove the inequality ;
s
s a
s
s b
s
s c ? ? ?
? ?
+ ...... >
n
n
2
1 ?
(n ? 2).
Sol. AM  >  GM  ?  [(s ? a) + (s ? b) + (s ? c) + ...... ]  >  n [(s ? a) + (s ? b) + (s ? c) + ...... ]
1/n
?
1 1 1
s a s b s c ?
?
?
?
?
?
?
?
?
?
?
?
. . . . . .
>  n
1
1
( ) ( ) ( )
/
s a s b s c
n
? ? ?
?
?
?
?
?
?
Multiplying   [(s ? a) + (s ? b) + ...... ]
1 1 1
s a s b s c ?
?
?
?
?
?
?
?
?
?
?
?
. . . . . .
>  n
2
or
( ) ( ) ( )
. . . . . .
s a
s
s b
s
s c
s
?
?
?
?
?
?
?
?
?
?
?
?

s
s a
s
s b
s
s c ?
?
?
?
?
?
?
?
?
?
?
?
. . . . . .
>  n
2
?
s
s a
s
s b
s
s c ?
?
?
?
?
?
?
?
?
?
?
?
. . . . . .

n s s
s
? ?
?
?
?
?
?  >  n
2
or
s a
s
s b
s
s c
s
?
?
?
?
?
+ ......  >
n
n
2
1 ?
Ex. If  a, b, c, d  are all positive and the sum of any three is greater than twice the fourth, then show that,
a b c d

>

(b + c + d ? 2a)

(c + d + a ? 2b)

(d + a + b ? 2c)

(a + b + c ? 2d).
Sol. Use    A.M.  >  G.M.
a + b + c ? 2d  =  m
1
,  b + c + d ? 2a  =  m
2
,  c + d + a ? 2b  =  m
3
,  d + a + b ? 2c  =  m
4
Now m
1
+ m
2
+ m
3
= 3

c ?    c  =
m m m
1 2 3
3
? ?
>  (m
1
m
2
m
3
)
1/3
m
2
+ m
3
+ m
4

=  3

d ?     d  =
m m m
2 3 4
3
? ?
>  (m
2
m
3
m
4
)
1/3
m
3
+ m
4
+ m
1

=  3

a ?     a  =
m m m
3 4 1
3
? ?
>  (m
3
m
4
m
1
)
1/3
m
4
+ m
1
+ m
2

=  3

b ?     b  =
m m m
4 1 2
3
? ?
>  (m
4
m
1
m
2
)
1/3
Hence   a b c d  >  m
1
m
2
m
3
m
4
?   Result
Ex. If the polynomial  f (x) = 4x
4
– ax
3
+ bx
2
– cx + 5 where a,b,c ? R has four positive real roots say
r
1
, r
2
, r
3
and r
4
, such that
2
r
1
+
4
r
2
+
5
r
3
+
8
r
4
= 1. Find the value of 'a'.
Sol. Consider 4 positive terms
2
r
1
,
4
r
2
,
5
r
3
,
8
r
4
A.M. =
4
1
?
?
?
?
?
?
?
?
? ? ?
8
r
5
r
4
r
2
r
4 3 2 1
=
4
1
× 1 =
4
1
G.M. =
4 1
4 3 2 1
8
r
·
5
r
·
4
r
·
2
r
?
?
?
?
?
?
?
?
=
4 1
4 3 2 1
8 · 5 · 4 · 2
r · r · r · r
?
?
?
?
?
?
?
?
now, r
1
r
2
r
3
r
4
=
4
5
? ? ?G.M. =
4 1
) 8 · 5 · 4 · 2 ( 4
5
?
?
?
?
?
?
?
?
=
4 1
8
2
1
?
?
?
?
?
?
=
4
1
.  Hence A.M. = G.M. ? ? ? ?All numbers are equal
2
r
1
=
4
r
2
=
5
r
3
=
8
r
4
= k ? ? ?r
1
= 2k;   r
2
= 4k;    r
3
= 5k;   r
4
= 8k
? ? ?
?
1
r = (2 · 4 · 5 · 8)k
4
?
4
5
=  (2 · 4 · 5 · 8)k
4
? k = 1/4
hence r
1
=
2
1
;  r
2
= 1;  r
3
=
4
5
;  r
4
= 2 ?
?
1
r =
4
19
but r
1
+ r
2
+ r
3
+ r
4
=
4
a
?
4
19
=
4
a
? a = 19
Ex. If  x > 0  and  n ? N ,  show that
x
x x x
n
n
1
2 2
? ? ? ? . . . . . .
?
1
1 2 ? n
.
Sol. x
k
+ x
?k
?  2,   k = 1, 2, 3, ...... , n  ?  1 +
k
n
?
?
1
(x
k
+ x
?k
)  ?  2

n + 1  Expand  ? and interpret
Ex. If a
1
, a
2
,......,a
n
are n distinct odd natural numbers not divisible by any prime greater than 5, then
prove that
n 2 1
a
1
...
a
1
a
1
? ? ?
< 2.
Sol. Since each a
i
is an odd number not divisible by a prime greater than 5, a
i
can be written as
a
i
= 3
r
5
s
where r, s are non-negative integers. Thus, for all n ? N.
n 2 1
a
1
...
a
1
a
1
? ? ?
<
?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
? ? ? ....
5
1
5
1
1 ....
3
1
3
1
1
2 2
=
?
?
?
?
?
?
? 3 / 1 1
1

?
?
?
?
?
?
? 5 / 1 1
1
=
?
?
?
?
?
?
2
3

?
?
?
?
?
?
4
5
=
8
15
< 2.
Ex. A
1
, A
2
,...., A
n
are n A.M’s, and H
1
, H
2
,....., H
n
are n H.M’s inserted between a and b. Prove that
2
2
n 1
n 1
G
A
H H
A A
?
?
?
, where A is the arithmetic mean and G is the geometric mean of a and b.
Sol. Since A
1
, A
2
,...., A
n
are n arithmetic means between a and b,
A
1
=
1 n
nb a
A ,
1 n
b na
n
?
?
?
?
?
?  A
1
+ A
n
= a + b    ...(1) Also,
1 n
1 a nb 1 an b
,
H (n 1)ab H (n 1)ab
? ?
? ?
? ?
....(2)
Page 5

H. A.M. ? G.M. ? H.M. INEQUALITIES
Theorem : If A, G, H are respectively AM, GM, HM between a & b both being unequal & positive then,
(i) G² = AH (ii)  A > G > H (G > 0) .  Note that  A, G, H  constitute a GP.
Let a
1
, a
2
, a
3
,....... a
n
be n positive real numbers, then we define their
A.M. =
n
a ...... a a a
n 3 2 1
? ? ? ?
, G.M. = (a
1
a
2
a
3
...........a
n
)
1/n
and H.M. =
n 2 1
a
1
.....
a
1
a
1
n
? ? ?
It can be shown that A.M. ? G.M. ? H.M. and equality holds at either places iff
a
1
= a
2
= a
3
= ............... = a
n
Ex. If a, b, c, d be four distinct positive quantities in G.P., then show that
(a) a + d > b + c (b)
?
?
?
?
?
?
? ? ? ?
1
ac
1
bd
1
2
ab
1
cd
1
.
Sol. Since a, b, c d are in G.P., therefore
(a)  using A.M. > G.M., we have for the first three terms
a c
2
?
> b i.e. a + c > 2b ....(1)
2
d b ?
> c i.e. b + d > 2c ....(2)
From results (1) and (2), we have a + c + b + d > 2b + 2c i.e. a + d > b + c
which is the desired result.
(b) using G.M. > H.M., we have for the first three terms b >
c a
ac 2
?
i.e. ab + bc > 2ac....(3)
and for the last three terms c >
2bd
b d ?
i.e. bc + cd > 2bd ....(4)
From results (3) and (4), we have ab + cd + 2bc > 2ac+ 2bd i.e. ab + cd > 2(ac + bd - bc)
i.e.
1 1 1 1 1
2
? ?
? ? ? ?
? ?
? ?
[dividing both sides by abcd] which is the desired result.
Ex. If a, b, c are in H.P. and they are distinct and positive then prove that a
n
+ c
n
> 2b
n
Sol. Let a
n
and c
n
be two numbers then
2
c a
n n
?
> (a
n
c
n
)
1/2
? a
n
+ c
n
> 2 (ac)
n/2
.....(i)
Also G.M. > H.M. i.e.
ac
> b (ac)
n/2
> b
n
.....(ii) hence from (i) and (ii) a
n
+ c
n
> 2b
n
Ex. Show that n
n

n 2
2
1 n
?
?
?
?
?
? ?
> (n !)
3
.
Sol. Consider the unequal positive numbers 1
3
, 2
3
, ........n
3
.
Since A.M. > G.M., therefore
n
n ...... 2 1
3 3 3
? ? ?
> (1
3
. 2
3
....n
3
)
1/n
, i.e.,
4
) 1 n ( n
2
?
> {(n !)
3
}
1/n
.
Raising both sides to power n, we have  n
n
n 2
2
1 n
?
?
?
?
?
? ?
> (n !)
3
.
Ex. If a, b, c are positive real numbers such that a + b + c = 1, then find the minimum value of
ac
1
bc
1
ab
1
? ?
.
Sol.
abc
1
abc
c b a
ac
1
bc
1
ab
1
?
? ?
? ? ?
Also,
3
c b a ? ?
? (abc)
1/3
? abc ?
27
1
?
abc
1
? 27. Hence
ac
1
bc
1
ab
1
? ?
? 27
Ex. In the equation x
4
+px
3
+qx
2
+rx+5=0 has four positive real roots, then find the minimum value of pr.
Sol. Let ? ? ?? ? ? ?? ?? be the four positive real roots of the given equation. Then
? ? ? ? ? ? ? ? ? ? ? ? ? = –p, ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ?? = q, ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = –r, ? ? ? ? = 5.
Using A.M. ? G.M. ? ? ?
5
4
.
4
4 3 3 3 3
4
? ? ? ?? ? ? ? ? ? ? ? ?? ?
? ?? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
? ? ?
?
?
?
?
?
?
? ?
?
?
?
?
?
?
4
r
4
p
? 5 ? pr ? 80 ? minimum value of pr = 80.
Ex.65 If

S = a+b+ c then prove that
S
S a ?
+
S
S b ?
+
S
S c ?

>
9
2
where  a

,

b & c  are distinct positive reals.
Sol.
(S a) S b) (S c)
3
? ? ? ? ?

>

? ?
( ) ( ) ( )
/
S a S b S c ? ? ?
1 3
........(i)
&

1
3

1 1 1
S a S b S c ?
?
?
?
?
?
?
?
?
?
?

>

1 1 1
1 3
S a S b S c ?
?
?
?
?
?
?
?
?
?
?
/
........(ii)
Multiple (i) & (ii) =
1
9
2 (a + b + c)

1 1 1
S a S b S c ?
?
?
?
?
?
?
?
?
?
?
>

1
=
2
9

S S S
S a S b S c
? ?
? ?
? ?
? ? ?
? ?

>

1 or
S
S a ?
+
S
S b ?
+
S
S c ?

>
9
2
Ex.66 Let  a > 1  and  n ? N . Prove the inequality,   a
n
? 1  ?  n ( ) a a
n n ? ?
?
1
2
1
2
.
Sol. Put   a = ?
2
T P T ? ?? ?
2n
? 1  ? ? ?    n ( ?
n + 1
? ? ?
n ? 1
)   ? ? ?   ?
2n
? 1  ? ? ?

n ?
n ? 1
( ?
2
? 1)   ?
?
?
2
2
1
1
n
?
?
?   n . ?
n ? 1
? ? ?1 + ?
2
+ ?
4
+ ...... + a
2 (n ? 1)
?   n  ?
n ? 1
?
1
2 4 2 1
. . . . . . . . . .
( )
? ? ?
n
n
?
. Which is True as  A.M. ?

G.M.
Ex. If

s

be the sum of

n

positive unequal quantities  a, b, c  then prove the inequality ;
s
s a
s
s b
s
s c ? ? ?
? ?
+ ...... >
n
n
2
1 ?
(n ? 2).
Sol. AM  >  GM  ?  [(s ? a) + (s ? b) + (s ? c) + ...... ]  >  n [(s ? a) + (s ? b) + (s ? c) + ...... ]
1/n
?
1 1 1
s a s b s c ?
?
?
?
?
?
?
?
?
?
?
?
. . . . . .
>  n
1
1
( ) ( ) ( )
/
s a s b s c
n
? ? ?
?
?
?
?
?
?
Multiplying   [(s ? a) + (s ? b) + ...... ]
1 1 1
s a s b s c ?
?
?
?
?
?
?
?
?
?
?
?
. . . . . .
>  n
2
or
( ) ( ) ( )
. . . . . .
s a
s
s b
s
s c
s
?
?
?
?
?
?
?
?
?
?
?
?

s
s a
s
s b
s
s c ?
?
?
?
?
?
?
?
?
?
?
?
. . . . . .
>  n
2
?
s
s a
s
s b
s
s c ?
?
?
?
?
?
?
?
?
?
?
?
. . . . . .

n s s
s
? ?
?
?
?
?
?  >  n
2
or
s a
s
s b
s
s c
s
?
?
?
?
?
+ ......  >
n
n
2
1 ?
Ex. If  a, b, c, d  are all positive and the sum of any three is greater than twice the fourth, then show that,
a b c d

>

(b + c + d ? 2a)

(c + d + a ? 2b)

(d + a + b ? 2c)

(a + b + c ? 2d).
Sol. Use    A.M.  >  G.M.
a + b + c ? 2d  =  m
1
,  b + c + d ? 2a  =  m
2
,  c + d + a ? 2b  =  m
3
,  d + a + b ? 2c  =  m
4
Now m
1
+ m
2
+ m
3
= 3

c ?    c  =
m m m
1 2 3
3
? ?
>  (m
1
m
2
m
3
)
1/3
m
2
+ m
3
+ m
4

=  3

d ?     d  =
m m m
2 3 4
3
? ?
>  (m
2
m
3
m
4
)
1/3
m
3
+ m
4
+ m
1

=  3

a ?     a  =
m m m
3 4 1
3
? ?
>  (m
3
m
4
m
1
)
1/3
m
4
+ m
1
+ m
2

=  3

b ?     b  =
m m m
4 1 2
3
? ?
>  (m
4
m
1
m
2
)
1/3
Hence   a b c d  >  m
1
m
2
m
3
m
4
?   Result
Ex. If the polynomial  f (x) = 4x
4
– ax
3
+ bx
2
– cx + 5 where a,b,c ? R has four positive real roots say
r
1
, r
2
, r
3
and r
4
, such that
2
r
1
+
4
r
2
+
5
r
3
+
8
r
4
= 1. Find the value of 'a'.
Sol. Consider 4 positive terms
2
r
1
,
4
r
2
,
5
r
3
,
8
r
4
A.M. =
4
1
?
?
?
?
?
?
?
?
? ? ?
8
r
5
r
4
r
2
r
4 3 2 1
=
4
1
× 1 =
4
1
G.M. =
4 1
4 3 2 1
8
r
·
5
r
·
4
r
·
2
r
?
?
?
?
?
?
?
?
=
4 1
4 3 2 1
8 · 5 · 4 · 2
r · r · r · r
?
?
?
?
?
?
?
?
now, r
1
r
2
r
3
r
4
=
4
5
? ? ?G.M. =
4 1
) 8 · 5 · 4 · 2 ( 4
5
?
?
?
?
?
?
?
?
=
4 1
8
2
1
?
?
?
?
?
?
=
4
1
.  Hence A.M. = G.M. ? ? ? ?All numbers are equal
2
r
1
=
4
r
2
=
5
r
3
=
8
r
4
= k ? ? ?r
1
= 2k;   r
2
= 4k;    r
3
= 5k;   r
4
= 8k
? ? ?
?
1
r = (2 · 4 · 5 · 8)k
4
?
4
5
=  (2 · 4 · 5 · 8)k
4
? k = 1/4
hence r
1
=
2
1
;  r
2
= 1;  r
3
=
4
5
;  r
4
= 2 ?
?
1
r =
4
19
but r
1
+ r
2
+ r
3
+ r
4
=
4
a
?
4
19
=
4
a
? a = 19
Ex. If  x > 0  and  n ? N ,  show that
x
x x x
n
n
1
2 2
? ? ? ? . . . . . .
?
1
1 2 ? n
.
Sol. x
k
+ x
?k
?  2,   k = 1, 2, 3, ...... , n  ?  1 +
k
n
?
?
1
(x
k
+ x
?k
)  ?  2

n + 1  Expand  ? and interpret
Ex. If a
1
, a
2
,......,a
n
are n distinct odd natural numbers not divisible by any prime greater than 5, then
prove that
n 2 1
a
1
...
a
1
a
1
? ? ?
< 2.
Sol. Since each a
i
is an odd number not divisible by a prime greater than 5, a
i
can be written as
a
i
= 3
r
5
s
where r, s are non-negative integers. Thus, for all n ? N.
n 2 1
a
1
...
a
1
a
1
? ? ?
<
?
?
?
?
?
?
? ? ? ?
?
?
?
?
?
? ? ? ....
5
1
5
1
1 ....
3
1
3
1
1
2 2
=
?
?
?
?
?
?
? 3 / 1 1
1

?
?
?
?
?
?
? 5 / 1 1
1
=
?
?
?
?
?
?
2
3

?
?
?
?
?
?
4
5
=
8
15
< 2.
Ex. A
1
, A
2
,...., A
n
are n A.M’s, and H
1
, H
2
,....., H
n
are n H.M’s inserted between a and b. Prove that
2
2
n 1
n 1
G
A
H H
A A
?
?
?
, where A is the arithmetic mean and G is the geometric mean of a and b.
Sol. Since A
1
, A
2
,...., A
n
are n arithmetic means between a and b,
A
1
=
1 n
nb a
A ,
1 n
b na
n
?
?
?
?
?
?  A
1
+ A
n
= a + b    ...(1) Also,
1 n
1 a nb 1 an b
,
H (n 1)ab H (n 1)ab
? ?
? ?
? ?
....(2)
From (1) and (2)
1 n
1 n
A A a b
(n 1)ab (n 1)ab
H H
a nb b na
? ?
?
? ?
?
?
? ?
=
? ? ?
? ? ? ?
(b na)(a nb)(a b)
(n 1)ab(na b nb a)
? ?
?
?
2
(b na)(a nb)
(n 1) ab
<
2
2
na b a nb 1
2
(n 1) ab
? ? ? ? ?
? ?
? ? ?
2 2
2
(a b) A
4ab
G
?
? ?
?
2
2
n 1
n 1
G
A
H H
A A
?
?
?
.
Ex If  a

,

b

,

c  are real and positive

,

prove that the inequality ,
1 1 1
a b c
? ?

<

a b c
a b c
8 8 8
3 3 3
? ?
.
So This is equivalent to showing that , a
8
+

b
8
+

c
8

>

a
2

b
2

c
2
(a

b

+

b

c

+

c

a)
or
a
b c
6
2 2

+

b
a c
6
2 2

+

c
a b
6
2 2
>  a

b

+

b

c

+

c

a    ?  (1)
Now
a
b c
b
a c
3 3
2
?
?
?
?
?
?
? >

0 ?
a
b c
6
2 2
+
b
c a
6
2 2
>
2
2 2
2
a b
c
Thus
a
b c
6
2 2
+
b
a c
6
2 2
+
c
a b
6
2 2
>
a b
c
2 2
2
+
b c
a
2 2
2
+
c a
b
2 2
2
again
a b
c
b c
a
?
?
?
?
?
?
?
2

>

0  ?
a b
c
2 2
2
+
b c
a
2 2
2
>  2 b
2
Thus
a b
c
2 2
2
+
b c
a
2 2
2
+
c a
b
2 2
2
>  a
2

+

b
2

+

c
2
Finally  it is well known that ,  a
2

+

b
2

+

c
2

>

a

b + b

c + c

a
Ex. Establish the inequality

,  1 +
1
2
1
3
1
? ? ? . . . . . .
n
>  2

[
n ? 1
? n ].
Sol. r r r r ? ? ? ? 1 or 2

r r r ? ? ? 1
or
1
2
1
1 r r r
?
? ?
or
1
2
1
r
r r ? ? ?      or
? ?
1
2 1
r
r r ? ? ?
Ex. Prove that for any positive integer n > 1 the inequality 1 +
2
1
+
3
1
+ .... +
n
1
> 2 (
1 n ?
– n )
holds true.
Sol. To prove this, reduce each term of the sum in the left-hand member:
1 k k
2
k
1
? ?
?
= 2(
1 k ?
–
k
)
Therefore, the left side of the inequality we want to prove can be reduced:
1 +
2
1
+
3
1
+ .....+
n
1
> 2(
2
–
1
)  + 2(
3
–
2
) + ..... + 2(
n
–
1 n ?
+ 2 (
1 n ?
–
n
)
Since the right side of this latter inequality is exactly equal to 2 (
1 n ?
– n ), the original inequality is valid.
Ex. Prove that for every positive integer n the following inequality holds true:
4
1
) 1 n 2 (
1
.....
25
1
9
1
2
?
?
? ? ?
```

## Mathematics (Maths) Class 11

85 videos|243 docs|99 tests

### Up next

 Video | 12:37 min
 Doc | 3 pages
 Doc | 14 pages

## FAQs on A.M., G.M., H.M. Inequalities - Mathematics (Maths) Class 11 - Commerce

 1. What are A.M., G.M., and H.M. inequalities?
Ans. A.M., G.M., and H.M. inequalities are mathematical concepts that relate to the arithmetic mean (A.M.), geometric mean (G.M.), and harmonic mean (H.M.) of a set of numbers. These inequalities state that for any set of positive numbers, the arithmetic mean is always greater than or equal to the geometric mean, which is always greater than or equal to the harmonic mean.
 2. How can A.M., G.M., and H.M. inequalities be applied in real-life situations?
Ans. A.M., G.M., and H.M. inequalities have various applications in real-life situations. For example, in finance, these inequalities can be used to analyze investment returns or risk. In physics, they can be used to study the relationships between different physical quantities. Additionally, these inequalities are also used in statistics, economics, and engineering to analyze and interpret data.
 3. Can you provide an example to illustrate A.M., G.M., and H.M. inequalities?
Ans. Certainly! Let's consider the set of positive numbers: 2, 4, and 8. The arithmetic mean (A.M.) of these numbers is (2 + 4 + 8)/3 = 14/3. The geometric mean (G.M.) is the cube root of the product of these numbers, which is ∛(2 * 4 * 8) = ∛(64) = 4. Lastly, the harmonic mean (H.M.) is the reciprocal of the average of the reciprocals of these numbers, which is 3/(1/2 + 1/4 + 1/8) = 3/(4/8 + 2/8 + 1/8) = 3/(7/8) = 24/7. In this example, we can observe that A.M. ≥ G.M. ≥ H.M.
 4. Are there any specific formulas to calculate A.M., G.M., and H.M.?
Ans. Yes, there are specific formulas to calculate A.M., G.M., and H.M. For a set of n positive numbers, the arithmetic mean (A.M.) is calculated by summing all the numbers and dividing by n. The geometric mean (G.M.) is calculated by taking the nth root of the product of all the numbers. The harmonic mean (H.M.) is calculated by dividing n by the sum of the reciprocals of all the numbers.
 5. Can A.M., G.M., and H.M. inequalities be extended to any set of numbers, including negative numbers?
Ans. No, A.M., G.M., and H.M. inequalities are applicable only to sets of positive numbers. These inequalities do not hold true when negative numbers are included in the set. Therefore, it is important to ensure that the numbers used in these inequalities are positive to obtain meaningful results.

## Mathematics (Maths) Class 11

85 videos|243 docs|99 tests

### Up next

 Video | 12:37 min
 Doc | 3 pages
 Doc | 14 pages
 Explore Courses for Commerce exam

### Top Courses for Commerce

Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Track your progress, build streaks, highlight & save important lessons and more!
Related Searches

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;