Bernoulli's inequality

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.

Proof of the inequality

For r=0,,

(1+x)^0 ge 1+0x
is equivalent to 1ge 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)

& 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 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!
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 the inequality (1 + 1/k)k < e.


  • 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.

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