Some proofs of
A.M. ≥ G.M.
Yue Kwok Choy
1.
(a) Let k be any positive integer and x any positive number.
Put x0 = 1. Show that:
(i) kxk < xk-1 + …. + x + 1
(ii) kxk > xk-1 + …. + x + 1
if x < 1
if x > 1
Hence show that kxk+1 ≥ (k + 1)xk
and determine the value of x for which equality holds.
x1, …., xk, xk+1 be any k + 1 (k ≥ 1) positive numbers.
x x2
x
By putting x k ( k +1) = 1
and using (a) , show that:
.... k
x k +1 x k +1 x k +1
(b) Let
k ( x 1 x 2 ....x k )
1k
+ x k +1 ≥ ( k + 1)( x 1 x 2 ....x k x k +1 )
1 k +1
and the equality holds if and only if
x1
x2
x k +1 x k +1
....
xk
x k +1
=1
(c) Using (b) or otherwise, show that, for any n (n ≥ 1) positive numbers
x1, x2, …. , xn,
x 1 + x 2 + .... + x n
1n
≥ (x 1 x 2 ....x n )
n
and that the equality holds if and only if x1= x2= …. = xn.
2.
(a) Show that, if p > 0, q > 0 and n is a positive integer, then
pn+1 – (n + 1)pqn + nqn+1 = p(pn – qn) + nqn (q – p) > 0
unless p = q.
(b) Hence, or otherwise, show that, if ai > 0 for all i, and
An =
then
1
(a 1 + a 2 + .... + a n ), G n = (a 1a 2 ....a n )1 n
n
n(An – Gn) < (n + 1)( An+1 – Gn+1)
(c) Deduce that An > Gn
3.
unless
ai = a
for
i = 1, 2, …., n.
(a) Show that for two positive numbers a, b , an + bn ≥ an-1b + bn-1a
and that equality holds if and only if a = b.
(b) Let a1, a2, …. , an be positive numbers. Using (a), show that:
(n –1) (a1n + a2n + …. + ann) ≥ a1(a2n-1 + …. + ann-1) + a2(a1n-1 + a3n-1 + …. + ann-1)
+ …. + an(a1n-1 + a2n-1 + …. + an-1n-1)
and that equality holds if and only if a1 = a2 = …. = an.
(c) Using (b) and mathematical induction, show that:
a1n + a2n + …. + ann ≥ n a1 a2 …. an,
and hence show that A.M. ≥ G..M.
1
(a) Show that
x1 + x 2
≥ x 1x 2 , where
2
(b) Show that
x1 + x 2 ≥ x1x 2
2
4.
k
(c) Use (b)
(x
(k − 1)
2
k
k
k −1
+ x1 x 2 , where k ∈
)
(
+ x 2 + ... + x k ≥ x1 x 2
k
k
(
k −1
+ x k x1
+ x3
k −1
k −1
+ x2
+ ... + x k
k −1
k −1
) + x (x
).
+ ... + x k −1
2
k −1
1
x1 + x 2 + ... + x n
≥ x1 x 2 ...x n , where xi ≥ 0 .
2
(e) Prove that
a 1 + a 2 + .... + a n
1n
≥ (a 1a 2 ....a n ) ,
n
2
+ x3
k −1
+ ... + x k
k −1
) + ...
k −1
(d) Prove that
2
5.
.
x 1 , x 2 ,..., x k ≥ 0
to show that for
1
k −1
x1 , x 2 ≥ 0 .
2
where ai ≥ 0 .
(a) Show that if x < 1 < y , then x + y –xy – 1 > 0 .
Hence prove that for any n +1 (n ≥ 1) real numbers x 1 ,..., x n +1 ,
if x 1 < 1 < x n +1 and
x 1 x n +1 + x 2 + ... + x n ≥ n , then x 1 + x 2 + ... + x n + x n +1 ≥ n + 1 .
(b) Using (a) and mathematical induction prove that for any n (n ≥ 1) positive real
numbers x 1 ,..., x n , if x 1 x 2 ...x n = 1 , then x 1 + x 2 + ... + x n ≥ n .
Hence, or otherwise show that for any n positive real numbers, a 1 ,..., a n ,
a 1 + a 2 + .... + a n n
≥ a 1a 2 ....a n .
n
6.
(a) For any non-negative number
x + k − 1 ≥ kx
When does the equality holds ?
k
x and for any integer k > 1, prove that
…. (*)
(b) Let n be an integer greater than 1 , {a 1 ,..., a n } be a set of positive numbers.
For m = 1,2, …, n , let
⎛ m ⎞
G m = ⎜⎜ ∏ a i ⎟⎟
⎝ i =1 ⎠
1 m
Am = ∑ai ,
m i =1
1/ m
.
⎛ Gm ⎞
mA m − (m − 1)A m −1
⎜⎜
⎟⎟ =
G m −1
⎝ G m −1 ⎠
m
(i)
Show that, for m = 2, …, n ,
…. (**)
(ii) Making use of (*) and (**) , or otherwise, prove that
Am − Gm ≥
m −1
(A m−1 − G m −1 )
m
(iii) Deduce that A n ≥ G n
a 1 = a 2 = ... = a n .
for m = 2, 3, …, n .
and show that the equality holds if and only if
2
7.
If
f (x ) = f (θ) + (x − θ) f ' (θ) +
where 0 < m < 1 and if
(x − θ)2 f " (θ + m(x − θ))
, that is, the mean value theorem holds,
2!
f (x ) ≥ 0 , prove that
f (x1 ) + f (x 2 ) + .... + f (x n )
⎛ x + x 2 + .... + x n ⎞
≥ f⎜ 1
⎟
n
n
⎝
⎠
x 1 + x 2 + .... + x n
≥
and hence show that if f(x) = – ln x , then
n
8.
(b) If a 1 , a 2 ,..., a n are positive numbers, let
Prove that if n >1,
bn
≥ an
b n −1
a 1 + a 2 + .... + a n
≥
n
if and only if a 1 = a 2 = ... = a n .
∀n ∈
⎛ a + a 2 + .... + a n ⎞
bn = ⎜ 1
⎟
n
⎝
⎠
n
x 1 x 2 .... x n
.
n
and c n = a 1a 2 ....a n ,
b n b n −1
≥
≥ 1.
c n c n −1
and deduce that
(c) Hence show that
9.
(1 + x )n ≥ 1 + nx
(a) Prove by induction that if x > – 1 , then
n
a 1 a 2 .... a n
and show that the equality holds
For positive numbers , a 1 , a 2 ,..., a n let
G(a 1 ,..., a n ) = n a 1a 2 ....a n (G.M.) and
A (a 1 ,..., a n ) =
a 1 + a 2 + .... + a n
(A.M.)
n
G (a 1 ,..., a n ) = A(a 1 ,..., a n ) = a .
(a) Show that if
a 1 = a 2 = ... = a n = a , then
(b) Show that if
G (a 1 ,..., a n )
a 1 , a 2 ,..., a n are not all equal, then there is one ai
and there is one aj
which is greater than
which is less than G (a 1 ,..., a n ) .
a 1 > G (a 1 ,..., a n ) > a n and let a 1 ' = G (a 1 ,..., a n ), a 2 ' = a 2 ,..., a n −1 ' = a n −1 and
a 1a n
an ' =
, show that G (a 1 ' ,..., a n ') = G (a 1 ,..., a n ) and A(a 1 ' ,..., a n ') < A(a 1 ,..., a n ) .
G (a 1 ,..., a n )
(c) Let
(d) Using the results above, show that
and only if a 1 = a 2 = ... = a n .
10. (a) Show that for a, b > 0 ,
(a + b )n ≥ a n + na n −1b
(c) For
.
⎡ 1 n −1
1 n −1 ⎞⎤
1⎛
⎡1 n ⎤
a i ⎟⎥
a
a
a
=
+
−
⎜
∑ i n ⎝ n n −1 ∑
⎢
⎢n ∑ i ⎥
i =1
⎣ i =1 ⎦
⎠⎦
⎣ n − 1 i =1
n
(b) Show that
G (a 1 ,..., a n ) ≤ A(a 1 ,..., a n ) and that the equality holds if
a 1 ≥ a 2 ≥ ... ≥ a n , show that
n
a 1 + a 2 + .... + a n
≥
n
n
a 1 a 2 .... a n
.
3
11. Let Gn
and An denote the geometric mean and arithmetic mean of
(a) Let a 1 , a 2 ..., a r +1
a1 ≤ G r ≤ a r ,
Prove that
(i)
be positive numbers such that
a1 ≤ A r ≤ a r
a 1 = a 2 = ... = a r . Deduce that
a 1 ,..., a n respectively .
a 1 ≤ a 2 ≤ ... ≤ a r +1 .
and that the equality holds iff
A r = a r +1 iff
a 1 = a 2 = ... = a r = a r +1 .
(ii) Show that when k is a positive integer and x > 0, (1 + x ) > 1 + kx .
k
r +1
Also show that
(A r+1 )
Deduce that if
Gr ≤ Ar
r +1
a − Ar ⎞
⎛
= ⎜ A r + r +1
⎟ .
r +1 ⎠
⎝
A r ≠ a r +1 , then
and
G r +1 < A r +1 .
(b) Using (a), or otherwise, show that, for any n (n ≥ 1) positive numbers
G n ≤ A n and that the equality holds iff
12. If
a 1 , a 2 ,..., a n ,
a 1 = a 2 = ... = a r .
are positive numbers ,show that the function
a 1 ,..., a n
1
1
f ( x ) = a 1 + ... + a n + x − n (a 1 ....a n x ) n +1 has the minimum value
a 1 + ... + a n − n (a 1 ....a n ) n
1
x = (a 1a 2 ....a n ) n . Deduce
and that this value is taken when
13. Let f (x 1 , x 2 ,...., x n )
∑ f (x , x
1
of
2
be a function of n variables
,...., x n )
f (x 1 x 2 ....x n ) , e.g.
and that
∑
x 1 x 2 ....x n = n!x 1 x 2 ....x n .
x + x 2 + .... + x n
ϕ n = ϕ(x 1 , x 2 ,...., x n ) = 1
− x 1 x 2 ....x n .
n
n
Show that
(b) Let
x 1 , x 2 ,...., x n
be the sum of n! quantities obtained from all possible permutations
x 1 , x 2 ,...., x n in
(a) Let
A.M. ≥ G.M.
n
(
(
Show that
n
n!ϕ n = ∑ x 1 − ∑ x 1 x 2 ....x n
ϕ1 = ∑ x 1
ϕ3 = ∑ x1
n
n −3
n −1
− x2
− x2
ϕn =
n −3
n −1
)(x
1
)(x
1
− x 2 ),
(
ϕ 2 = ∑ x1
n −2
− x2
n −2
)(x
1
− x 2 )x 3 ,
− x 2 )x 3 x 4 , ..., ϕ n −1 = ∑ (x 1 − x 2 )(x 1 − x 2 )x 3 x 4 ...x n ,
1
(ϕ1 + ϕ 2 + ... + ϕ n −1 ) .
2(n!)
(c) Let x 1 , x 2 ,...., x n be non-negative numbers. Show that ϕ n ≥ 0 and show that the
equality holds if and only if x 1 = x 2 = ... = x n . Hence show that A.M. ≥ G.M.
4
14. (a) Prove that for any positive integer n (n > 1) and any positive number x ,
n
x ⎞
⎛ x⎞ ⎛
⎜1 + ⎟ > ⎜1 +
⎟
⎝ n ⎠ ⎝ n −1 ⎠
n −1
.
(b) For any n (n > 1) positive numbers
a m = min{a 1 , a 2 ,..., a n }
⎡
⎤
1 n
⎡1
⎤
⎢
⎥
⎢n ∑ ar ⎥ > am ⎢n −1∑ ar ⎥
r =1
⎣ r =1 ⎦
⎢⎣
⎥⎦
r≠m
n −1
n
n
Use (a) , show that
a 1 , a 2 ,..., a n , not all equal, let
.
(c) Hence, using induction or otherwise, prove that, for any n positive numbers
n
n
⎡1 n ⎤
p
>
pr .
∑
∏
r
⎢n
⎥
⎣ r =1 ⎦
r =1
p1 , p 2 ,..., p n , not all equal,
15. (a) Let f(x) = ln x – x + 1 , x > 0. Use the fact that
d
1
,
ln x =
dx
x
prove that f(x) ≤ 0, for all x > 0.
(b) Let A be the arithmetic mean and G be the geometric mean of n positive real
numbers
a1 , a2, … , an . By putting x =
ai
A
in (a), prove that A ≥ G .
16. (a) Given a function f(x) = e x −1 − x where x is a real number. Find the minimum
value of f(x).
(b) Let a 1 , a 2 , L , a n and b1 , b 2 , L , b n be positive numbers.
(i)
Show that e
⎧⎛ n a ⎞ ⎫
⎪⎜
i⎟ ⎪
⎨⎜ ∑ ⎟ − n ⎬
b
⎪⎩⎝ i =1 i ⎠ ⎪⎭
≥
n
ai
∏b
i =1
.
i
n
(ii) Hence , or otherwise, show that if
(c) By using the result in
a 1 , a 2 ,L, a n .
ai
≤ n , then
∑
i =1 b i
n
n
i =1
i =1
∏ a i ≤∏ b i
(b), prove that AM ≥ GM for any n positive numbers
5
Solution
1.
(
) (
) (
)
(
) (
)
(a) kx k − x k −1 + ... + x + 1 = x k − x k −1 + x k − x k −2 + ... + x k − x + x k − 1
⎧< 0 , x < 1
= (x − 1)(kx k −1 + (k − 1)x k −2 + ... + 2 x + 1) ⎨
⎩> 0 , x > 1
(since x > 0)
Obviously,
(x − 1)2 (kx k −1 + (k − 1)x k −2 + ... + 2x + 1) ≥ 0
(
(
(with equality holds iff x = 1)
))
(
)
⇔ (x − 1) kx k − x k −1 + ... + x + 1 ≥ 0 ⇔ (x − 1)kx k − (x − 1) x k −1 + ... + x + 1 ≥ 0
(
)
⇔ kx k +1 + 1 ≥ (k + 1)x k
⇔ kx k +1 − kx k − x k − 1 ≥ 0
(x x ...x )
x
x1 x 2
.... k , x k +1 = 1 2 k
x k +1
x k +1 x k +1 x k +1
1/ k
(b) Put
x k ( k +1) =
By (a) ,
k
k (x 1 x 2 ...x k )
(x1x 2 ...x k )1/ k
1/ k
x k +1
+ 1 ≥ (k + 1)
(x1x 2 ...x k )1/ (k+1)
x k +1
k / ( k +1)
(x1x 2 ...x k )1/ (k +1)
x k +1
+ x k +1 ≥ (k + 1)(x 1 x 2 ...x k )
1 / ( k +1)
Equality holds ⇔ x k ( k +1) = 1 ⇔
, xk =
k / ( k +1)
1− k / ( k +1)
x k +1
= (k + 1)(x 1 x 2 ...x k x k +1 )
1 / ( k +1)
x1 x 2
x
.... k = 1
x k +1 x k +1 x k +1
(c) Use induction. The assertion is obviously true for n = 1 .
Assume that the assertion is true for n = k. i.e. x 1 + x 2 + .... + x k ≥ k (x 1 x 2 ....x k ) .
1k
For n = k + 1,
x 1 + x 2 + .... + x k + x k +1 ≥ k (x 1 x 2 ....x k ) + x k +1
1k
≥ (k + 1)(x 1 x 2 ...x k x k +1 )
1 / ( k +1)
Equality holds ⇔
x
x1 x 2
.... k = 1,
x k +1 x k +1 x k +1
(inductive hypothesis)
and the proof completes.
∀k = 1,2,..., (n − 1)
⇔ x 1 = x 2 = ... = x n .
6.
(a) Let
f (x ) = x k + k − 1 − kx
(
)
(
)
f ' (x ) = kx k −1 − k = k x k −1 − 1 = k (x − 1) x k −2 + x k −3 + ... + x + 1
For turning point,
f "(x ) = k (k − 1)x k −2 ,
f ' (x ) = 0 . Since x > 0, there is one root, i.e. x = 1.
f "(1) = k (k − 1) > 0 , since k is an integer > 1.
6
∴
∴
f(x) is a minimum when x = 1 . It is an absolute min.
f(x) ≥ f(1) . ∀x > 0 .
∴
x k + k − 1 ≥ kx
and equality holds when x = 1 .
⎡ (a 1a 2 ....a m )1 / m ⎤
⎛ Gm ⎞
a 1a 2 ....a m
a a ....a m −1a m
⎜⎜
⎟⎟ = ⎢
=
= 1 2
1 / ( m −1) ⎥
1+1 / ( m −1)
a 1a 2 ....a m −1G m −1
(a 1a 2 ....a m−1 )
⎝ G m −1 ⎠
⎣ (a 1a 2 ....a m −1 )
⎦
m
m
(b) (i)
=
(a + a 2 + .... + a m ) − (a 1 + a 2 + .... + a m−1 ) = mA m − (m − 1)A m−1
am
= 1
G m −1
G m −1
G m −1
x m + m − 1 ≥ mx , put
(ii) From (*) ,
From (**) ,
xm =
x=
mA m − (m − 1)A m −1
G m −1
Gm
,
G m −1
⎛ G ⎞
mA m − (m − 1)A m −1
+ m − 1 ≥ m⎜⎜ m ⎟⎟
G m −1
⎝ G m −1 ⎠
We therefore have
m −1
(A m−1 − G m−1 ) .
m
(iii) We like to use mathematical induction to show that A n − G n ≥ 0 .
Am − Gm ≥
Simplify, we get
The proposition is obviously true for n = 1.
We assume that A m − G m ≥ 0 is true for some integer m ≥ 2 .
m −1
(A m−1 − G m−1 ) ≥ 0 , by inductive hypothesis .
m
An − Gn ≥ 0 ∀ n ∈ .
From (b) (ii),
∴
Am − Gm ≥
G
Equality holds ⇔ m = 1
G m −1
m
and
⎛ Gm ⎞
a
⎟⎟ = m = 1
⎜⎜
G m −1
⎝ G m −1 ⎠
, by (b) (ii).
a m = G m ∀ m = 2, 3, …, n .
⇔ G1 = G 2 = ... = G n and
⇔ a 1 = a 2 = .... = a n
12. f ' ( x ) = 1 − (a 1a 2 ...a n )(a 1a 2 ...a n x )
1
−1
n +1
1
n +1
= 1 − (a 1a 2 ...a n ) x
−
n
n +1
For critical values, f’(x) = 0
1
n +1
∴ 1 − (a 1a 2 ...a n ) x
−
n
n +1
1
Now,
(a 1a 2 ...a n )
= 0,
1
n +1
x
−
n
n +1
=1
,
∴ x = (a 1a 2 ...a n )
1
n
2 n +1
−
n +1
n
(a1a 2 ...a n ) x n +1
f ' ' (x) =
n +1
∴ f ’’(x) > 0 when
1
n
x = (a 1a 2 ...a n ) .
1
n
∴ f(x) attains its min value at x = (a 1a 2 ...a n ) .
7
1
n +1
⎧
n ⎫
a 1 + a 2 + ... + a n + (a 1...a n ) − (n + 1)⎨a 1 ...a n (a 1a 2 ...a n ) ⎬
⎩
⎭
1
1
n
And min(f(x)) =
= a 1 + a 2 + ... + a n − n (a 1a 2 ...a n )
1
n
To prove A.M. ≥ G..M., use Mathematical Induction,
1
n
Let P(n) be the statement : “ a 1 + a 2 + ... + a n − n (a 1a 2 ...a n ) ≥ 0
Obviously P(1) is true.
Assume P(k) is true for some ∀k ∈ N .
1
k
i.e. a 1 + a 2 + ... + a k − n (a 1a 2 ...a k ) ≥ 0
(*)
Put an = x, we get f ( x ) = a 1 + a 2 + ... + a k + x − (k + 1)(a 1a 2 ...a n x )
For P(k+1),
1
k +1
1
k
By (a) f(x) ≥ Min {f(x)} = a 1 + a 2 + ... + a k − k (a 1a 2 ...a n ) ≥ 0 , by (*)
∴
∴
P(k+1) is also true.
By the Principle of Mathematical Induction, P(n) is true ∀n ∈ N
n
a 1 + a 2 + .... + a n
≥ (a 1a 2 ....a n )
n
1
∴
13. (a) It is easily seen that whatever permutation is used, the function ϕn remains unchanged.
x 1 + x 2 + .... + x n
− x 1 x 2 ....x n , summing up, we have
n
1
n
n
n
n
x 1 + x 2 + ... + x n − ∑ x 1 x 2 ....x n = × ∑ x 1 − ∑ x 1 x 2 ....x n
n
n
Since
ϕn =
n!ϕ n =
1
∑
n
n
n
(
)
= ∑ x 1 − ∑ x 1 x 2 ....x n
n
(b) We get, from definition,
ϕ1 = 2∑ x 1 − 2∑ x 1 x 2 ,
ϕ 2 = 2∑ x 1 x 2 − 2 ∑ x 1
ϕ 3 = 2∑ x 1
n −1
n
n −1
n −2
n −2
x 2 x 3 − 2∑ x 1
x2x3 ,
n −3
x 2 x 3x 4 ,
…..
ϕ n −1 = 2∑ x 1 x 2 x 3 ...x n −1 − 2∑ x 1 x 2 x 3 ...x n
2
Summing up,
By (a) ,
ϕ1 + ϕ 2 + ... + ϕ n −1 = 2∑ x 1 − 2∑ x 1 x 2 ....x n
n
ϕn =
1
(ϕ1 + ϕ 2 + ... + ϕ n −1 ) .
2(n!)
(
(c) From the definition of ϕi , ϕ1 = ∑ (x 1 − x 2 ) x 1
2
n −2
+ x1
n −3
x 2 + ... + x 2
n −2
) ≥ 0,
8
(
ϕ 2 = ∑ (x 1 − x 2 ) x 1
2
n −3
+ x1
n −4
x 2 + ... + x 2
n −3
)x
3
≥ 0, ……..,
ϕ n −1 = ∑ (x 1 − x 2 ) x 3 x 4 ...x n ≥ 0
2
∴
ϕn =
1
(ϕ1 + ϕ 2 + ... + ϕ n −1 ) ≥ 0
2(n!)
It is evident that ϕ1 , ϕ 2 ,..., ϕ n −1 vanish iff
ϕ n vanish iff x 1 = x 2 = .... = x n .
a 1 = x 1 , a 2 = x 2 ,...., a n = x n
n
Putting
Equality holds if
14. (a) Let
n
ϕ n ≥ 0 , we get A.M. ≥ G.M.
in
a 1 = a 2 = .... = a n .
⎛ x⎞
f ( x ) = ⎜1 + ⎟
⎝ n⎠
f ' (x) =
n
x 1 = x 2 = .... = x n .
n
x ⎞
⎛
⎜1 +
⎟
⎝ n −1 ⎠
n −1
n −1
n
(
(
n − 1)
n + x)
=
×
nn
(n − 1 + x )n −1
(n − 1)n −1 × x (n − 1 + x )n −2 (n + x )n −1 > 0
nn
(n − 1 + x )2(n −1)
, or any positive integer n (n > 1)
∴
f(x) is increasing in (0, ∞) and f(x) > f(0) = 1.
∴
x ⎞
⎛ x⎞ ⎛
⎟
⎜1 + ⎟ > ⎜1 +
⎝ n ⎠ ⎝ n −1 ⎠
n
n −1
n
(b) Put
x=
∑a
r =1
r
am
− n , then
x=
a1 − a m a 2 − a m
a − am
+
+ ... + n
>0 .
am
am
am
n
Substitute in (a), we get
⎛ n
⎜ ar
1 ⎜∑
⇒ n r =1
am ⎜ n
⎜
⎝
n
n
⎛
⎞ ⎛
⎞
⎜ ∑ar am − n ⎟ ⎜ ∑ar am − n ⎟
⎜1 + r =1
⎟ > ⎜1 + r =1
⎟
⎜
⎟
⎜
⎟
n
n −1
⎜
⎟ ⎜
⎟
⎝
⎠ ⎝
⎠
n
⎞
⎛ n
⎟
⎜ ∑ar − am
⎟ > 1 ⎜ r =1
n −1
⎟
a m ⎜ n −1
⎟
⎜
⎠
⎝
⎞
⎟
⎟
⎟
⎟
⎠
n −1
n −1
.
⎡
⎤
n
1
⎡1
⎤
⎢
ar⎥
⇒⎢ ∑ ar⎥ > am ⎢
∑
⎥
n
n
1
−
r =1
⎣ r =1 ⎦
r≠m
⎣⎢
⎦⎥
n
n −1
n
n
(c) Let
s = ∑ pr .
r =1
Use induction. The assertion is easily proved for n = 2 , for (p1 − p 2 ) > 0 ,
2
where p1 , p 2 > 0 and unequal.
Assume that the assertion is true for the case with (n – 1) numbers.
9
⎛ s − pm ⎞
⎛s⎞
Now, ⎜ ⎟ > p m ⎜
⎟
⎝n⎠
⎝ n −1 ⎠
n
n −1
, by (b) where
p m = min{p1 , p 2 ,..., p n } .
n
> p m ∏ p r , by inductive hypothesis.
r =1
r ≠m
n
=
∏
pr
r =1
15. (a) f(x) = ln x – x + 1 , x > 0.
f ' (x ) =
1
−1 = 0 ,
x
when x = 1.
For 0 < x < 1, f’(x) > 0 and for x > 1 , f’(x) < 0.
Therefore f(x) attains its maximum value at x = 1 .
∴ f(x) ≤ f(1) = ln 1 + 1 – 1 = 0, ∀ x > 0 .
∴ f(x) ≤ 0, ∀ x > 0 .
(b) Let A =
a 1 + a 2 + ... + a n
.
n
Then for i = 1, 2, …, n,
∴
ln
a a
⎛a ⎞
f ' ⎜ i ⎟ = ln i − i + 1 ≤ 0
A A
⎝A⎠
ai ai
≤ −1
A A
n
ai
⎛ ai ⎞
≤
⎜ − 1⎟
∑
∑
A k =1 ⎝ A ⎠
k =1
n
n
a a ...a
a
1 n
1
ln 1 2 n n ≤ ∑ i − ∑ 1 = ∑ a i − n = nA − n = 0
A k =1
A
A
k =1 A
k =1
n
ln
∴
a 1a 2 ...a n
a + a 2 + ... + a n
≤ 1 ⇒ n a 1a 2 ...a n ≤ A = 1
n
A
n
16. (a) f (x ) = e x −1 − x ⇒ f ' (x ) = e x −1 − 1
f ' (x ) = 0
For x < 1 ,
For x > 1 ,
∴
(b) (i)
⇒ e x −1 = 1
⇒ x =1
f ' (x ) < 0 .
f ' (x ) > 0 .
At x = 1, f(x) attains an absolute minimum value of f(1) = 0 .
By (a) ,
a
Put x = i
bi
e x −1 ≥ x
∀x∈
.
where i = 1, 2, …., n .
10
e
ai
−1
bi
n
(ii) If
n
a
≥ i
bi
∑
i =1
Hence
⇒∏ e
ai
−1
bi
i =1
a
≥∏ i
i =1 b i
n
ai
≤n,
bi
then
∑
i =1
n
n
i =1
i =1
n
⇒e
n⎛a
⎞
∑ ⎜⎜ i −1 ⎟⎟
bi ⎠
i =1⎝
⎛
n
n
a
≥∏ i
i =1 b i
⇒e
⎛ n ai
⎜⎜ ∑
⎝ i =1 b i
⎞
⎟⎟ − n
⎠
n
≥∏
i =1
ai
bi
ai ⎞
n
⎜⎜ ∑ b ⎟⎟ − n
a
ai
− n ≤ 0 ⇒ e ⎝ i =1 i ⎠ ≤ 1 ⇒ ∏ i ≤ 1
bi
i =1 b i
∏ a i ≤∏ b i
n
bi =
(c) Put
n
∑
then
i =1
∑
i =1
ai
∀i = 1, 2,…., n .
n
n
ai
=∑
b i i =1
ai
n
∑
i =1
ai n
⎛ n
a
⎜ ∑ ai
i
n
n ∑
i =1
(b) , ∏ a i ≤∏
= ⎜ i =1
⎜ n
n
i =1
i =1
⎜
⎝
n
By
=n
⎞
⎟
⎟
⎟
⎟
⎠
n
n
1/ n
Hence
⎛ n ⎞
⎜⎜ ∏ a i ⎟⎟
⎝ i =1 ⎠
≤
∑
i =1
n
ai
i.e. G..M. ≤ A.M.
11