From "Discrete Mathematics with Graph Theory"
by Goodaire and Parmenter
Page 157 # 93
For any x∈R, x > -1, (1 + x)ⁿ ≥ 1 + nx
for all n∈ N
by Goodaire and Parmenter
Page 157 # 93
For any x∈R, x > -1, (1 + x)ⁿ ≥ 1 + nx
for all n∈ N
-
This is called Bernoulli's inequality.
Note that when n = 1, the inequality holds (as an equality) since
(1 + x)^1 = 1 + 1x.
Suppose that for some k ≥ 1, that the conclusion holds when n = k. That is, suppose that
(1 + x)^k ≥ 1 + kx.
Next, let n = k + 1. Note that as x > -1, the quantity 1 + x > 0. So using the induction hypothesis
(1 + x)^(k+1) = (1 + x)(1 + x)^k ≥ (1 + x)(1 + kx) = 1 + (k + 1)x + kx² ≥ 1 + (k + 1)x.
The last inequality is true because kx² ≥ 0. Now we have (1 + x)^(k+1) ≥ 1 + (k + 1)x; the required conclusion for the n = k + 1 case.
It follows that for any real x > -1, (1 + x)ⁿ ≥ 1 + nx for all natural numbers n.
Note that when n = 1, the inequality holds (as an equality) since
(1 + x)^1 = 1 + 1x.
Suppose that for some k ≥ 1, that the conclusion holds when n = k. That is, suppose that
(1 + x)^k ≥ 1 + kx.
Next, let n = k + 1. Note that as x > -1, the quantity 1 + x > 0. So using the induction hypothesis
(1 + x)^(k+1) = (1 + x)(1 + x)^k ≥ (1 + x)(1 + kx) = 1 + (k + 1)x + kx² ≥ 1 + (k + 1)x.
The last inequality is true because kx² ≥ 0. Now we have (1 + x)^(k+1) ≥ 1 + (k + 1)x; the required conclusion for the n = k + 1 case.
It follows that for any real x > -1, (1 + x)ⁿ ≥ 1 + nx for all natural numbers n.