Bernoulli inequality

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Bernoulli's inequality

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An illustration of Bernoulli's inequality, with the graphs of y = (1 + x)r and y = 1 + rx shown in red and blue respectively. Here, r = 3.

In real analysis, 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, as shown below.

Contents

[ Proof of the inequality

For r=0,\,

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

is equivalent to 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)(1+x)^k \ge (1+x)(1+kx) (by hypothesis, since (1+x)\ge 0)
\begin{matrix}
& \iff & (1+x)^{k+1} \ge 1+kx+x+kx^2 \\
& \iff & (1+x)^{k+1} \ge 1+(k+1)x+kx^2
\end{matrix}.

However, as 1+(k+1)x + kx^2 \ge 1+(k+1)x (since kx^2 \ge 0), 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.

[ Generalization

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.

[ Related inequalities

The following inequality estimates the r-th power of 1 + x from the other side. For any real numbers x, r > 0, one has

(1 + x)^r \le e^{rx},\!

where e = 2.718.... This may be proved using thee inequality (1 + 1/k)k < e.

[ References

[ External links

Source: this wikipedia article, under CC-BY-SA.

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