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In real analysis, Bernoulli's inequality is an inequality that approximates exponentiations of 1 + x.## Proof of the inequality

## Generalization

## 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
^{k} < e.
## References

## External links

The inequality states that

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

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

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.

For $r=0,,$

- $(1+x)^0\; ge\; 1+0x$

Now suppose the statement is true for $r=k$:

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

- $(1+x)(1+x)^k\; ge\; (1+x)(1+kx)$ (by hypothesis, since $(1+x)ge\; 0$)

- $begin\{matrix\}$

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 $rge\; 0.$

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

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

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

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

- Carothers, N. (2000).
*Real Analysis*. Cambridge: Cambridge University Press. - Bullen, P.S. (1987).
*Handbook of Means and Their Inequalities*. Berlin: Springer. - Zaidman, Samuel (1997).
*Advanced Calculus*. City: World Scientific Publishing Company.

- Bernoulli Inequality by Chris Boucher, The Wolfram Demonstrations Project.

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Last updated on Friday June 20, 2008 at 05:59:40 PDT (GMT -0700)

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