Bernoulli's inequality

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In mathematics, Bernoulli's inequality 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:

Proof: For r=0, $(1+x)^0 \ge 1+0x$ equals $1\ge 1$ which is true as required.

Now suppose the statement is true for r=k: $(1+x)^k \ge 1+kx$ Then it follows that $(1+x)^{k+1} = (1+x)(1+x)^k \ge (1+x)(1+kx)$ (by hypothesis, since $(1+x)\ge 0$ ) $= 1+kx+x+kx^2 = 1+(k+1)x + kx^2 \ge 1+(k+1)x$ (since $kx^2 \ge 0$ ) So it follows that $(1+x)^{k+1} \ge 1+(k+1)x$ , which means the statement is true for r=k+1 as required.

By induction we conclude the statement is true for all $r\ge 0 \ \ \Box \;$

The exponent r can be generalized to an arbitrary real number as follows: if x > −1, then

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

for r ≤ 0 or r ≥ 1, and

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

for 0 ≤ r ≤ 1. This generalization can be proved by comparing derivatives. Again, the strict versions of these inequalities require x ≠ 0 and r ≠ 0, 1.