Bernoulli's inequality

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An illustration of Bernoulli's inequality, with the graphs of

y = (1 + x) r

and

y = 1 + r x

shown in red and blue respectively. Here,

r = 3.

In real analysis, Bernoulli's inequality (named after Jacob Bernoulli) is an inequality that approximates exponentiations of 1 + x.

The inequality states that

$(1 + x)^r \geq 1 + rx\!$

for every integer r ≥ 0 and every real number x ≥ −1. If the exponent r is even, then the inequality is valid for all real numbers x. The strict version of the inequality reads

$(1 + x)^r > 1 + rx\!$

for every integer r ≥ 2 and every real number x ≥ −1 with x ≠ 0.

Bernoulli's inequality is often used as the crucial step in the proof of other inequalities. It can itself be proved using mathematical induction, as shown below.

1 Proof of the inequality
2 Generalization
3 Related inequalities
4 References
5 External links

[ Proof of the inequality

For r = 0,

$(1+x)^0 \ge 1+0x \,$

is equivalent to 1 ≥ 1 which is true as required.

Now suppose the statement is true for r = k:

$(1+x)^k \ge 1+kx. \,$

Then it follows that

$\begin{align} & {} \qquad (1+x)(1+x)^k \ge (1+x)(1+kx)\quad\text{(by hypothesis, since }(1+x)\ge 0) \\ & \iff (1+x)^{k+1} \ge 1+kx+x+kx^2, \\ & \iff (1+x)^{k+1} \ge 1+(k+1)x+kx^2. \end{align}$

However, as 1 + (k + 1)x + kx² ≥ 1 + (k + 1)x (since kx² ≥ 0), it follows that (1 + x)^k + 1 ≥ 1 + (k + 1)x, which means the statement is true for r = k + 1 as required.