Other Inequalities Exercise Solution PDF

A.M. G.M.
, CBS inequality, etc

1.
Obviously, equality holds when

Without lost of generality,
Since
> 0,
(1),
.(1)
a > b > 0.
a b
<b
a+b
a+b
A.M. > G.M. ,
Combining
2.
assume
a b
ab
a=b
(2) and
<b
a +b

>a b
2
a +b
a+b
a+b
a+b
>a
a+b
> a b ba
.(2)
.(3)
a+b
Obviously the inequality holds when x = y.

Without lost of generality,
x
x
< 1
y
y
m n
assume
But
0 < x < y.
x
x
<
y
y
<1
Since
.(2)
x m
x n
1
1
m
m n
n
n m
1 + < 1 + mn [x + y ] < mn [x + y ]
y
y
y
y
(xm + ym)n < (xn + yn)m.
As ymn > 0,
Alternatively,
Let
m = k + n,
w.l.o.g.
x y,
assume
(x n )k
(xm + ym)n = (xnxk + ynyk)n (xnxk + ynxk)n = xkn (xn + yn)n =

< (xn + yn)k+n
W.l.o.g.,
(I)
(x n + yn )k
( x n + y n ) k+n
= (xn + yn)m.
a1 a2 an.
let
If p, q > 0 , then a1p a2p anp
m
n
x m
x n
x
x
1 < 1 + < 1 + 1 + < 1 + ....(1)
y
y
y
y
x
1 < 1 + and m > n
y
Combining (1) and (2),
m n > 0, we have
x m
x m
1 + < 1 +
y
y
3.
a+b
a+b
a b b a
(3),
(aip ajp) (aiq ajq) 0
a1q a2q anq
and
aipajq
aip+q + ajp+q
+ ajpaiq
n
Summing the above inequalities from i = 1, 2, , n
(fix
j),
ai
p+q
.(1)
+ na j
p+q
aj
i =1
Summing again from j = 1, 2, , n,
p+q
i
+n
i =1
Since
i, j
are dummy variables, replace
2n
a
i =1
If
p+ q
i
a a
p
i =1
q
i
i =1
still holds
p+q
j
j=1
p+q
i
i =1
and
a i +a j
p
i =1
a a
q
j=1
p
i
i =1
q
i
i =1
a a
p
j=1
a
q
i =1
by i, we have,
p, q < 0 , then a1p a2p anp

The inequality (1)
i =1
and
(2)
a a
q
i
.(2)
i =1
a1q a2q anq.
follows.
1
(II) If
p<0
and
q > 0 , then
a1p a2p anp
and
a1q a2q anq.
If
p>0
and
q < 0 , then
a1p a2p anp
and
a1q a2q anq.
(aip ajp) (aiq ajq) 0.
In both cases,
As in (I), we have
p+q
i =1
4.
If a, b, c, , k Q .
+
Let
a a
p
i =1
q
i
i =1
m be the L.C.M.
of the denominators of
a, b, c, , k.
Then ma, mb, mc, , mk N+. Apply A.M. G.M., we have,

1
1
1
1
1
ma + mb + mc + ... + mk
ma
mb
mc
mk
ma + mb+ mc+...+ mk
1
1
1
1
k ...
c
b
a

k
ma + mb + mc + ... + mk
a b c
a + b + c + .... + k
On simplification, we have
then infinite sequences
If a, b, c, , k R+,
such that ai a, bi b, ci c, , ki k
Now, since
(1 + x r ) = 1 +
r =1
.(1)
(i = 1, 2, 3, )
as i .
a i + bi + c i +....+ k i
a i i b i i c i i ....k i
a
ki
.(2)
a=b=c==k.
x + x x + x x x
i
i =1
i j
+ ... > 1 +
i j k
r =1
1 tan
tan
1
A + B
A + B =
2
2
=
= tan
= cot
A
B
A
B
2 tan +
2
2
2
tan + tan
2 2
2
2
C
A
A
B
B
C
tan tan + tan tan + tan tan = 1
2
2
2
2
2
2
C
A
B
xy + yz + zx = 1 x2 + y2 + z2 1.
x = tan ,
y = tan ,
z = tan , it remains to show:
2
2
2
A + B + C = tan
Put
But (x y)2 + (y z)2 + (z x)2

7.
ai, bi, ci, , ki Q+,
a + b i + c i + .... + k i
ai, bi, ci, , ki Q+, i
Equality holds
6.
a a b b cc ....k k
i on both sides of (2), we have (1) holds for a, b, c, , k R+.
Taking
5.
a + b +c+....+ k
x2 + y2 + z2 xy + yz + zx = 1.
Proof 1
sin
A
2
sin
B
2
sin
C
2
(s b)(s c) (s c)(s a ) (s a )(s b)

bc
( 2s 2a )( 2s 2 b)( 2s 2c)
8abc
Since
b + c a,
ca
c+ab
( b + c a )( c + a b)
ab
(s a )(s b)(s c)
abc
( b + c a )( c + a b)(a + b c)
8abc
and
a+bc
( b + c a ) + ( c + a b)
2
where s =
a+b+c
2
.(1)
are positive numbers,

=c
(A.M.G.M.)
.(2)
Similarly,
( c + a b)( a + b c) a
....( 3)
( a + b c)( b + c a ) b
....( 4)
2
( b + c a )( c + a b)( a + b c) abc
(2)(3)(4),
.(5)
Result follows by putting (5) to (1).

3
sin A + sin B + sin C
A
B
C
2
2
2 ,
sin sin sin
2
2
2
3
Proof 2
sin
A /2 + B/2 + C/2
,
= sin
1
A+ B+C
1
= sin = =

6
6 2
8
(A.M.G.M.)
sine function is concave on [0, ]
since
3
Proof 3
2
1 sin C + sin C
A
B
C 1
AB
A + B
C 1
C
C 1
2
2 =1
sin sin sin = cos
cos
sin
sin
sin
2
2
2 2
2
2
2 2
2
2 2
2
8
8.
(i)
cos A + cos B + cos C = 2 cos

= 2 cos
4
(ii)
1
8
A+B
2
cos
AB
2
cos( A + B) = 2 cos
A+B
2
cos
AB
2
2 cos 2
A+B
2
AB
A + B
A+B
A
B
A
B
C
A+B
cos
cos
+ 1 = 2 cos
2 sin sin + 1 = 4 sin sin sin + 1
2
2
2
2
2
2
2
2
2
+1 =
, by No. 7.
A+ B+C
(sin A + sin B + sin C ) sin
Proof 1
cos
Proof 2
cos A + cos B + cos C
A
B
C
2
2
2 ,
cos cos cos
2
2
2
3
cos
cos
3
4
sin
3 3
8
(A.M.G.M.)
3
3
3
1 A B C
3 3
=
cos + + = cos =
,
3 2 2 2
6 2
8
9.
(a)
(i)
Let f(x) = ln x,
f(x)
and
x ,
Now,
(ii)
ln a
then
cosine function is concave on [0, /2]
since
x > 0.
f(x) = 1/x > 0
is an increasing function.
0<x1
ln b
f(x) f(1)
ln a ln b
Let g(x) = x 1 ln x,
g(x) = 1 1/x
g(x) 0
g(1) = 0 with
This implies that
g(1) = 0
g(x) = x 1 ln x g(1) = 0
ln x x 1 x > 0.
ln x ln 1 = 0
ln (a/b) 0
0<x<1
(1)
a/b 1,
and g(x) 0
by (1)
a b.
x > 1.
is the minimum value of g(x).

x > 0
The equality holds if and only if
x = 1.
(b)
By the above result, when
t1, t2, , tn
1, 2, , n
and
are positive numbers, we have
1 ln t1 + 2 ln t2 + + n ln tn 1 (t1 1) + 2 (t2 1)+ + n (tn 1)

= (1 t1 + 2 t2 + + n tn) (1 + 2 + + n) = 1 1 = 0
ln (t 1 1 t 2 2 ....t n
m1a1 + m 2a 2 + ... + m n a n
i =1
(i)
x p+1 1
p +1
t 1 1 t 2 2 ....t n
m1
m2
a 1 a 2 ....a n
xp 1
1
mi
i =
i = 1, 2, , n.
m1 + m 2 + ... + m n
mn
1
mi
m1 + m2 +...+ m n
( m1 + m 2 + ... + m n )
1
m1a 1 + m 2 a 2 + ... + m n a n
m a + m 2 a 2 + .... + m n a n
1 1
m1 + m 2 + .... + m n
pxP+1 p pxp + xp p 1
(a )
n
mi
m1 + m 2 +...+ m n
i =1
m 1 + m 2 +...+ m n
pxP+1 pxp + xp + 1 0
(x 1) [ pxp xp-1 xp-2 x 1] 0 (x 1)2 [pxp-1 + (p 1)xp-2 + + 3x2 + 2x + 1] 0
True statement since the expression is the product of a complete square and a positive quantity
Equality holds
x1=0
xi 1
m
(ii)
( m1 + m 2 + ... + m n )a i
m1a 1 + m 2 a 2 + ... + m n a n
10.
By the above,
n
1 t1 + 2 t2 + . + n tn = 1 + 2 + . + n = 1
It can be seen that
t 1 1 t 2 2 ....t n
( m1 + m 2 + ... + m n )a i
ti =
Now, let
) 0
(a)
By (i),
m
n
xi
m 1
...
m 1
xi n 0
m
i =1
x=0
xi 1
xi 1
i =1
xi 1
i =1
xj
xi 1
m
we have
xi 1
i =1
xj 1
xj 1, then
s.t.
xi 1
1
>
x j 1
1
>
xi 1
i =1
, since equality holds
jth
xi =
i =1
i =1
n
n
i =1
yi
A
A
1
yi =
i =1
yi n 1

A
n
xi =
m
i
=nn
Am
m=1
or
m
i
= n.
Then
xi n ,
m
y1 + y 2 + .... + y n
m
i =1
or
by (ii)(a).
i =1
If the equality holds, by (ii)(b), then either

i.e. either
term.
i =1
i = 1, 2, , n.
nA
yi
jth
n
x i > n , which contradicts to the given
i =1
y1 + y 2 + ... + y n
x = 1.
term, we have
Strict inequality holds since there is a strict inequality in the

n
A=
Following the same method in (a), we have
(iii) Let
xi n = n n = 0
i =1
Summing up all inequalities from i = 1, 2, , n including the

n
i =1
Suppose on the contrary

i j,
xi n
m
(b)
m=1
or
y1 + y 2 + .... + y n
xi = 1 i = 1, 2, , n.
y1 = y2 = . = yn.
4

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A.M. G.M.

, CBS inequality, etc

Obviously, equality holds when

A.M. > G.M. ,

Obviously the inequality holds when x = y.

(xm + ym)n < (xn + yn)m.

(xm + ym)n = (xnxk + ynyk)n (xnxk + ynxk)n = xkn (xn + yn)n =

If p, q > 0 , then a1p a2p anp

Combining (1) and (2),

(aip ajp) (aiq ajq) 0

a1q a2q anq

Summing the above inequalities from i = 1, 2, , n

Summing again from j = 1, 2, , n,

are dummy variables, replace

p, q < 0 , then a1p a2p anp

a1q a2q anq.

a1p a2p anp

a1q a2q anq.

a1p a2p anp

a1q a2q anq.

(aip ajp) (aiq ajq) 0.

Then ma, mb, mc, , mk N+. Apply A.M. G.M., we have,

then infinite sequences

But (x y)2 + (y z)2 + (z x)2

ai, bi, ci, , ki Q+,

i on both sides of (2), we have (1) holds for a, b, c, , k R+.

(s b)(s c) (s c)(s a ) (s a )(s b)

are positive numbers,

Result follows by putting (5) to (1).

sin A + sin B + sin C

sine function is concave on [0, ]

cos A + cos B + cos C = 2 cos

(sin A + sin B + sin C ) sin

cos A + cos B + cos C

cosine function is concave on [0, /2]

f(x) = 1/x > 0

This implies that

is the minimum value of g(x).

The equality holds if and only if

By the above result, when

are positive numbers, we have

1 ln t1 + 2 ln t2 + + n ln tn 1 (t1 1) + 2 (t2 1)+ + n (tn 1)

(x 1) [ pxp xp-1 xp-2 x 1] 0 (x 1)2 [pxp-1 + (p 1)xp-2 + + 3x2 + 2x + 1] 0

It can be seen that

, since equality holds

If the equality holds, by (ii)(b), then either

x i > n , which contradicts to the given

Strict inequality holds since there is a strict inequality in the

Following the same method in (a), we have

Summing up all inequalities from i = 1, 2, , n including the

Suppose on the contrary

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