Page 39 - RMGO 5

Refinements of Bernoulli’s inequality 39

Proposition 3 (Double refinement of Bernoulli’s inequality). Let x ≥ 0
and r ∈ R, r ≥ 3.

a) We have
r
(1 + x) ≥ (1 + 2x)(1 + x) r−2 ≥ (1 + 2x) [1 + (r − 2)x] ≥ 1 + rx. (B3)

The first and the third inequality become equalities if x = 0, and the second
inequality becomes equality if r = 3 or x = 0.
b) We have

r
2
(1 + x) ≥ (1 + x) [1 + (r − 2)x] ≥ (1 + 2x) [1 + (r − 2)x] ≥ 1 + rx. (B4)
The first inequality becomes equality if r = 3 or x = 0, and the last two inequalities
become equalities if x = 0.
Proof. a) We apply inequality (B) for the powers 2 and r − 2, successively:

(B) (B)
r
2
(1 + x) = (1 + x) · (1 + x) r−2 ≥ (1 + 2x) · (1 + x) r−2 ≥ (1 + 2x) [1 + (r − 2)x] ,
that is the first two inequalities in the statement. For the third inequality, we note
2
that: (1 + 2x) [1 + (r − 2)x] = 1 + rx + 2(r − 2)x ≥ 1 + rx.
b) Similar proof, but with the application of inequality (B) in turn, for the
powers r − 2 and 2.

A generalization of the last two sentences is given by the following result.
Proposition 4 (Double refinement of Bernoulli’s inequality). Let x ≥ 0
and r, p ∈ R, r ≥ 1 + p, p ≥ 1.
a) We have

r
(1 + x) ≥ (1 + px)(1 + x) r−p ≥ (1 + px) [1 + (r − p)x] ≥ 1 + rx. (B5)

The first inequality becomes equality if p = 1 or x = 0, the second inequality becomes
equality if r = p + 1 or x = 0, and the third inequality becomes equality if x = 0.

b) We have

r
p
(1 + x) ≥ (1 + x) [1 + (r − p)x] ≥ (1 + px) [1 + (r − p)x] ≥ 1 + rx. (B6)
The first inequality becomes equality if r = p + 1 or x = 0, the second inequality
becomes equality if p = 1 or x = 0, and the third inequality becomes equality if
x = 0.

Proof. a) We apply inequality (B) for the powers p and r − p, successively:

(B) (B)
r
p
(1 + x) = (1 + x) · (1 + x) r−p ≥ (1 + px) · (1 + x) r−p ≥ (1 + px) [1 + (r − p)x] ,