Inequalities
A.M. ≥
G. M. ≥ H. M. :
Let a1,
a2, . .. . . . , an be n positive real numbers,
then we define their arithmatic mean (A) , geometric mean (G) and
harmonic mean (H) as A = 
,
G = (a1 a2 . . . . an)1/n and H
= 
.
It
can be shown that A ≥ G ≥ H . Moreover equality holds at
either place if and only if a1 = a2 = ……. = an
.
Weighted Means:
Let a1, a2, .
.. . . . , an be n positive real numbers and m1, m2,
. .. . . . , mn be n positive rational numbers. Then we define
weighted Arithmetic mean (A*), weighted Geometric mean (G*) and weighted
harmonic mean (H*) as .
A* = 
, G* = 
and
H* = 
.
It can be shown that A* ≥ G* ≥ H* . Moreover equality holds at either place if and
only
if a1 = a2 = . . . . . = an . .
Cauchy's Inequality:
If ai's and bi's are
reals, then (a12+...+an2)(b12+...+bn2)
≥ (a1b1
+ a2b2 +...+ anbn)2.
Equality holds when 
.
Example 1
Let ABC and PQR be two given
triangles such that
sinA = 
, where a, b, c and p,
q, r are the sides of the triangle ABC and PQR respectively. If p, q, r(p
< q < r) are consecutive natural numbers and perimeter of the
triangle ABC is equal to 12l , then find the sides of the
triangle PQR and that of triangle ABC.
Solution:
Since (a2
+ b2 + c2).( r2 + p2 + q2)
≥ (ar + bp + cq)2
, sinA ≥ 1.
But sinA £ 1
&⇒
there is only one possibility , which is sinA = 1
&⇒ A = 90° and 
&⇒ 
( say
) . . . . (1)
&⇒ D
ABC and D PQR are similar
since ∠A = p/2, ∠R = p/2
(as r is the greatest side). .
In DPQR, q = p +1, r = p+2 and also r2
= p2 + q2
&⇒ (p+2)2 = p2 +
(p +1)2
&⇒ p =-1, 3
&⇒ q = 4, r = 5. .
Now a = 5t, b = 3t
and c = 4t from (1)
&⇒ 5t +3t + 4t = 12l
&⇒
t = l
&⇒ a = 5l , b = 3l,
c = 4l. .
Alternate:
Let 
and 
, where 
= ar + bp + cq
&⇒ 
cosq = ar + bp + cq , where q is the angle between 
.
&⇒ cos2q = 
&⇒ 
.
. . (1)
But sinA = 
. . . (2)
From (1) and (2)
only equality will hold
i.e.
&⇒ cos2q =1 and sinA =1
&⇒ q
= 0 or p and A = p/2
&⇒ 
are either collinear
or parallel
&⇒ 
&⇒ .
. . (3)
&⇒ D
ABC and D PQR are similar
since ∠A = p/2, ∠R = p/2
(as r is the greatest side). .
In DPQR, q = p +1, r = p+2 and also r2
= p2 + q2
&⇒ (p+2)2 = p2 +
(p +1)2
&⇒ p =-1, 3
&⇒ q = 4, r = 5. .
Now a = 5t, b = 3t
and c = 4t from (3)
&⇒ 5t +3t + 4t = 12l
&⇒
t = l
&⇒ a = 5l , b = 3l,
c = 4l. .
Proving Inequalities:
(i) Any inequality has to be solved using a clever
manipulation of the previous results.
(ii) Any inequality
involving the sides of a triangle can be reduced to an inequality involving
only positive reals, which is generally easier to prove
For the triangle we
have the constraints a + b > c, b + c > a, a + c > b
Do the following :
put x = s- a, y = s – b, z = s - c
then, x + y + z = 3s
- 2s = s and a = y +z, b = x + z, c = x + y.
Substitute a = y +
z, b = x + z, c = x + y in the inequality involving a,b,c to get an inequality
involving x,y, z.
Also note that the
condition a + b > c is equivalent to
a + b + c > 2c
i.e., 2s > 2c OR s - c > 0, i.e., z > 0.
Similarly b + c >
a º x > 0, a + c
> b º y > 0
\
The inequality obtained after the substitution is easier to prove. (involving
only positive reals without any other constraints).
Example 1
If a,b,c are the sides of a triangle and 
.
Prove that 8(s - a) (s - b) (s - c) £ abc
Solution:
Let x = s - a, y = s
- b, z = s - c, then a = y + z, b = x + z, c = x + y
then the inequality
reduces to 8xyz £
(x + y)(y + z)(x + z) x,y,z ≥
0
which follows easily from A.M. ≥
G.M. inequality.
\
(x + y) (x + z)(x + z) ≥ 8xyz. Hence proved.
Arithmetic Mean of mth
Power:
Let a1,
a2, . .. . . . , an be n positive real numbers ( not
all equal) and let m be a real number , then .
>
if m ∈ R –[0, 1].
However if m∈ (0, 1), then 
< .
Obviously if m∈ {0, 1} , then = .
Example 1
Show that x +
≥
2, if x > 0 and x +£
-2 , if x < 0 .
Solution:
Since x > 0 , 
≥ 
( A.M. ≥ G. M. )
&⇒ x +
≥ 2.
If x
< 0 , let y = -x , then y > 0 and y + 
≥ 2
&⇒ – x – ≥ 2
&⇒ x +£
-2.
Example 2
If ai ’s
are all positive real numbers then prove that
(1+ a1 +a12
) (1+ a2 +a22 ) . . . . (1+ an +an2
) ≥ 3n a1a2 ….. an
.
Solution:
≥ (1. ai . ai2 )1/3
= aI ( i = 1, 2, 3, . . . . , n)
&⇒ 1+ai +ai2
≥ 3ai
&⇒ (1+ a1
+a12 ) (1+ a2 +a22 ) . .
. . (1+ an +an2 ) ≥ 3n a1a2
….. an .
Example 3
If x, y, z are
positive real numbers , such that x+y+z = a , then prove that 
.
Solution:
Since
A.M. ≥ H.M. .
&⇒
.
Example 4
If a, b, c are positive real
numbers such that a+b+c = 18,
find the maximum
value of a2 b3c4 . .
Solution:
a+b+c
= 18
&⇒ 
&⇒ 
£
(theorem of weighted means)
&⇒ a2 b3 c4
£ 29 . 22
. 33 .44.
Equality will hold,
when 
&⇒ a= 4, b = 6, c = 8.
Thus maximum value
of a2 b3 c4 is 42. 63
.84 = 219 .33 .
Example 5
If a, b, c are
positive real numbers such that a+b+c = 1 , then prove that 
.
Solution:
We have to show
that
i.e 
Now A.M. of mth
power ≥ mth power of
arithmetic mean (m= -1 here).
&⇒ 
&⇒ 
.
