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Bernoulli’s inequality
(for integer cases)
(1 + x)n ≥ 1 + nx
(1)
with conditions:
(a)
(1) is true
∀ n∈N ∪ {0}
(b)
(1) is true
∀ n∈2N
and ∀x∈ R,
x ≥ -1.
and ∀x∈ R.
(1 + x)n > 1 + nx
The strict inequality:
is true for every integer n ≥ 2
(2)
and every real number
x ≥ -1
with x ≠ 0.
(The strict inequality is not discussed in the following.)
Proof
Use Mathematical Induction
1
Condition (a)
Let P(n) be the proposition: (1 + x)n ≥ 1 + nx
0
(1 + x) = 1 ≥ 1 + 0x
For P(0),
∀n∈N ∪ {0}
and ∀x∈ R,
x ≥ -1.
∴ P(0) is true.
k ≥ 0,
Assume P(k) is true for some k∈Z,
k
that is,
(1 + x) ≥ 1 + kx
∀ k∈N ∪ {0}
For P(k+1),
(1 + x)k+1 = (1 + x)k(1 + x)
≥ (1 + kx) (1 + x)
,
by
(3)
and ∀x∈ R,
and also note that since
x ≥ -1.
(3)
x ≥ -1, the factor
(x + 1) ≥ 0.
2
= 1 + (k + 1) x + kx
≥ 1 + (k + 1) x
∴
P(k+1) is also true.
By the Principle of Mathematical Induction, P(n) is true ∀n∈N ∪ {0}
and ∀x∈ R,
x ≥ -1.
Condition (b)
Let P(n) be the proposition: (1 + x)n ≥ 1 + nx
(1 + x)0 = 1 ≥ 1 + 0x
For P(0),
2
and ∀x∈ R.
∴ P(0) is true.
2
(1 + x) = 1 + 2x + x ≥ 1 + 2x,
For P(2),
∴ P(2)
∀ n∈2N
since x2 ≥ 0, ∀x∈ R.
is true.
Assume P(k) is true for some k∈Z,
that is,
k ≥ 0,
k
(1 + x) ≥ 1 + kx
∀ k∈N ∪ {0}
and ∀x∈ R,
x ≥ -1.
(4)
For P(k + 2),
(1 + x)k+2 = (1 + x)k (1 + x)2 ≥ (1 + kx)(1 + 2x),
by (4) and P(2).
2
= 1 + (k + 2) x + 2kx
≥ 1 + (k + 2) x,
since k > 0
and x2 ≥ 0, ∀x∈ R.
∴ P(k + 2) is also true.
By the Principle of Mathematical Induction, P(n) is true ∀ n∈2N
and ∀x∈ R.
Condition (a) is discussed only in the following.
1
Proof
Use A.M. ≥ G.M.
2
Consider the A.M. and G.M. of n positive numbers (1 + nx), 1, 1, ….,1 [with (n-1) “1”s]
(1 + nx ) + 1 + 1 + .... + 1
n
n + nx
n
≥ n (1 + nx ).1.1....1
≥ n (1 + nx )
(1 + x)n ≥ 1 + nx
∴
(1)
Numbers should be positive before applying A.M. – G.M. theorem.
Note :
In the numbers used in A.M.-G.M. above, 1 > 0 and
However, if
1 + nx ≥ 0, i.e. x ≥ −
since it is given that x ≥ -1, or
1 + nx < 0,
1
n
.
x + 1 ≥ 0,
L.H.S. of (1) = (1 + x)n ≥ 0
R.H.S. of (1) =
1 + nx < 0
∴ (1 + x)n ≥ 0 > 1 + nx, which is always true.
∴ x ≥ -1 and not x ≥ −
Proof
1
n
can ensure (1) is correct.
Use Binomial Theorem
3
⎛n⎞
⎛n⎞
⎛n⎞
⎛n⎞
(1 + x ) n = 1 + nx + ⎜ ⎟ x 2 + ⎜ ⎟ x 3 + ⎜ ⎟ x 4 + .... + ⎜ ⎟ x n ≥ 1 + nx
⎝n⎠
⎝2⎠
⎝ 3⎠
⎝4⎠
(a)
For x > 0,
(b)
For x = 0,
(c)
For
-1 < x < 0,
Put
y = -x,
obviously (1 + x)n ≥ 1 + nx
(5)
is true.
(The proof below is not very rigorous.)
then
0 < y < 1,
⎛n⎞
⎛n⎞
⎛n⎞
⎛n⎞
(1 − y ) n = 1 − ny + ⎜ ⎟ y 2 − ⎜ ⎟ y 3 + ⎜ ⎟ y 4 − .... + ( −1) n ⎜ ⎟ y n
⎝n⎠
⎝ 2⎠
⎝ 3⎠
⎝4⎠
(6)
Put y = 0, we have:
⎛n⎞ ⎛n⎞ ⎛n⎞
⎛n⎞
0 = (1 − 1) n = 1 − n + ⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ − .... + ( −1) n ⎜ ⎟
⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠
⎝n⎠
⎛n⎞ ⎛n⎞ ⎛n⎞
⎛n⎞
∴ ⎜ ⎟ − ⎜ ⎟ + ⎜ ⎟ − .... + ( −1) n ⎜ ⎟ = n − 1 > 0
⎝ 2⎠ ⎝ 3⎠ ⎝ 4⎠
⎝n⎠
Now
(7)
0 < y < 1 , ∴ y2 > y3 > …> yn
Therefore in (7), each term is multiplied by a factor that is smaller than the term before,
n
⎛n⎞ 2 ⎛n⎞ 3 ⎛n⎞ 4
n⎛ ⎞ n
⎜ ⎟ y − ⎜ ⎟ y + ⎜ ⎟ y − .... + ( −1) ⎜ ⎟ y ≥ 0
⎝ 2⎠
⎝ 3⎠
⎝ 4⎠
⎝n⎠
(8)
From (6),
(1 – y)n ≥ 1 – ny
or
n
(1 + x) ≥ 1 + nx
for 0 < y < 1.
for -1 < x < 0.
2
Extension of Bernoulli’s inequality
Given x > -1, then
(a)
(b)
(1 + x)r ≤ 1 + rx
r
(1 + x) ≥ 1 + rx
for
0<r<1
for
r < 0 or
(9)
r>1
(10)
Firstly we give the proof that r is a rational number first.
Use A.M. ≥ G.M.
Proof
4
Since
r ∈ Q,
(a)
p
r=
q
Let 0 < r < 1, ∴ p < q, q – p > 0. Also 1 + x > 0,
(1 + x)r = (1 + x )
=
(b)
p q
px + q
q
q
= (1 + x ) p 1q − p ≤
=1+
p
q
p(1 + x ) + 1 ⋅ ( q − p )
q
(G.M. ≤ A.M.)
x = 1 + rx
Let r > 1,
(i)
If
1 + rx ≤ 0, then
(ii)
If
1 + rx > 0,
(1 + x)r > 0 ≥ 1 + rx.
rx > -1.
1
Since r > 1, we have 0 <
r
< 1 . By (a) we get:
1r
1
(1 + rx ) ≤ 1 + rx = 1 + x
r
r
∴ (1 + x) ≥ 1 + rx.
Let r < 0, then - r > 0.
Choose a natural number n sufficiently large such that 0 < -r/n < 1 and 1 > rx/n > -1.
By (a),
Since
0 < (1 + x )
−r n
≤1+
r
n
x > 0.
−1
2
r
r
r
r
r
1 ≥ 1 − ⎛⎜ x ⎞⎟ = ⎛⎜ 1 − x ⎞⎟⎛⎜ 1 + x ⎞⎟ ⇒ ⎛⎜ 1 − x ⎞⎟ ≥ 1 + x
⎝n ⎠ ⎝
n ⎠⎝
n ⎠ ⎝
n ⎠
n
(11)
Hence by (11),
r
n
r
(1 + x )r ≥ ⎛⎜ 1 + x ⎞⎟ ≥ 1 + n ⎛⎜ x ⎞⎟ = 1 + rx .
⎝
n ⎠
⎝n ⎠
Again, equality holds if and only if x = 0.
Note :
If r is irrational, we choose an infinite sequence of rational numbers r1, r2, r3, ….,
such that rn tends to r as n tends to infinity. For part (a), we can extend to irrational r:
(1 + x ) r = lim (1 + x ) rn ≤ lim (1 + rn x ) = 1 + rx .
n→∞
n→∞
Similar argument for part (b) completes the proof for the case where r ∈ R.
3
Proof
Use analysis
5
f(x) = (1 + x)r – 1 – rx
Let
where x > –1 and r ∈ R\ {0, 1}
(12)
Then f(x) is differentiable and its derivative is:
f ’(x) = r(1 + x)r-1 – r = r [(1 + x)r-1 – 1]
from (13) we can get
(a)
If
0 < r < 1,
f ’(x) = 0 ⇔ x = 0.
then f ’(x) > 0 ∀ x∈ (–1, 0) and
∴ x = 0 is a global maximum point of
(b)
∴
f(x) < f(0) = 0.
∴
(1 + x)r ≤ 1 + rx
If
(13)
for
f.
0 < r < 1.
r < 0 or r > 1, then f ’(x) < 0
∀ x∈ (-1, 0) and
∴ x = 0 is a global minimum point of
∴
f(x) > f(0) = 0.
∴
(1 + x)r ≥ 1 + rx
for
f ’(x) < 0 ∀ x∈ (0, +∞).
r < 0 or
f ’(x) > 0
∀ x∈ (0, +∞).
f.
r > 1.
Finally, please check that the equality holds for x = 0 or for r ∈ {0, 1}. The proof is complete.
4
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