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 (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.

Proof of the inequality[edit]

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

& {} \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.

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.

By induction we conclude the statement is true for all r ≥ 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[edit]

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

Alternative form[edit]

An alternative form of Bernoulli's inequality for  t\geq 1 and  0\le x\le 1 is:

 (1-x)^t \ge 1-xt.

This can be proved (for integer t) by using the formula for geometric series: (using y=1-x)

t=1+1+\dots+1 \ge 1+y+y^2+\ldots+y^{t-1}=\frac{1-y^t}{1-y}

or equivalently xt \ge 1-(1-x)^t.

Proof for rational case[edit]

An "elementary" proof can be given using the fact that geometric mean of positive numbers is less than arithmetic mean

First assume t=\frac{a}{b}\leq 1

By comparing Arithmetic and Geometric mean of b numbers ( (1+x) occurs  a times):

1,1, \ldots, (1+x),(1+x),\ldots (1+x)

we get

 (1+x)^{a/b} \leq \left(1+\frac{a}{b}x\right)

or equivalently

 (1+x)^{t} \leq \left(1+t x\right)

This proves inequality for  t \leq 1 case.

For  s \geq 1 case, let  z=\frac{a}{b}x As x\geq -1, z\geq -\frac{a}{b} we get with s=\frac{1}{t}=\frac{b}{a}\geq 1,

 \left(1+ z \right)^{s}\geq 1+sz

This proves inequality for  s \geq 1 case.

As these inequalities are true for all rational numbers  t \leq 1 and  s \geq 1, they are also true for all real numbers, which follows from a density argument of the rationals in the reals, and the fact that the functions involved are continuous.


  • Carothers, N. (2000). Real Analysis. Cambridge: Cambridge University Press. p. 9. ISBN 978-0-521-49756-5. 
  • Bullen, P.S. (1987). Handbook of Means and Their Inequalities. Berlin: Springer. p. 4. ISBN 978-1-4020-1522-9. 
  • Zaidman, Samuel (1997). Advanced Calculus. City: World Scientific Publishing Company. p. 32. ISBN 978-981-02-2704-3. 

External links[edit]