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) 
?
?
?
?
?
?
? ? ? ?
ad
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) 
?
?
?
?
?
?
? ? ? ?
ad
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
cd ab bd ac ad
? ?
? ? ? ?
? ?
? ?
[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) 
?
?
?
?
?
?
? ? ? ?
ad
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
cd ab bd ac ad
? ?
? ? ? ?
? ?
? ?
[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) 
?
?
?
?
?
?
? ? ? ?
ad
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
cd ab bd ac ad
? ?
? ? ? ?
? ?
? ?
[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) 
?
?
?
?
?
?
? ? ? ?
ad
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
cd ab bd ac ad
? ?
? ? ? ?
? ?
? ?
[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
?
?
? ? ?