Important inequalities: Bernoulli's inequality

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For a real number $x$ at least $- 1$ , and a positive integer $n$ , can $(1 + x)^{n}$ fall below the straight line $1 + n x$ ? The line is the value at $0$ , which is $1$ , plus $n$ times the step $x$ . A power of a binomial is rarely that cheap. The question is whether it is ever cheaper, inside this hypothesis.

This is Bernoulli's inequality. It is not the Bernoulli numbers that name sums of powers. An integer power $a^{n}$ itself sits with powers, roots, and logarithms. The new fact is the comparison with the line.

The square that sits above the line

Two extra assumptions, both on the board: $n$ is a positive integer, and $x$ is a real number with $x \geq - 1$ . Drag $x$ and it stops at $- 1$ . Drag $n$ and it rounds to an integer.

Start at $n = 1$ . Then $(1 + x)^{1}$ and $1 + 1 \cdot x$ are the same expression, so the gap is zero everywhere the board allows. The inequality holds because there is nothing to compare.

Move to $n = 2$ . Expand the square:

$(1 + x)^{2} = 1 + 2 x + x^{2} .$

The line for this $n$ is $1 + 2 x$ . What is left over is $x^{2}$ , and a square is at least $0$ . So

$(1 + x)^{2} \geq 1 + 2 x$

whenever the square makes sense, which for a real $x$ is everywhere — the restriction $x \geq - 1$ is not what saves the square. It is the hypothesis the general statement is willing to spend.

Drag n or x. n rounds to a positive integer from 1 through 4, and x stops at −1 on the left. The curve is (1+x)^n. The straight graph is 1+nx. The segment at the current x is the gap. At n=1 the gap is gone. At n=2 the gap is the square x².

Drag $n$ or $x$ . The curve is $(1 + x)^{n}$ . The straight graph is $1 + n x$ . The segment at the current $x$ is the gap between them. At $n = 1$ the segment has no length. At $n = 2$ its length is the square you just expanded. For $n = 3$ and $n = 4$ , still with $x \geq - 1$ , the curve stays on or above the line.

The inequality those pictures are paying for is Bernoulli's inequality: for every real $x \geq - 1$ and every integer $n = 1, 2, \dots$ ,

$(1 + x)^{n} \geq 1 + n x .$

On this board the first two integers show when the two sides meet: every allowed $x$ when $n = 1$ , and only $x = 0$ when $n = 2$ . At the endpoint $x = - 1$ with $n = 2$ , the power is $0$ and the line is $- 1$ , so the gap is the square $1$ , not zero.

One integer step past the hypothesis

Drop $x \geq - 1$ and the statement no longer claims anything. That does not mean every point to the left is a counterexample. For $n = 2$ the square is still nonnegative, so $(1 + x)^{2} \geq 1 + 2 x$ keeps holding for real $x < - 1$ . The failure has to be checked at a point.

Freeze $n = 3$ and set $x = - 4$ . Then

$(1 + x)^{3} = (1 - 4)^{3} = - 27, 1 + 3 x = 1 - 12 = - 11.$

Since $- 27 < - 11$ , the inequality has reversed. This is one point outside the hypothesis. It is not a second theorem about the half-line $x < - 1$ .

Drag x. n is frozen at 3, and x runs from −4 to −1, outside the hypothesis except at the right end. The left bar is (1+x)^3. The right bar is 1+3x. At the start x=−4, so the left bar is −27 and the right bar is −11. The power sits lower, and the inequality has failed. At x=−2 the power is above the line again.

Drag $x$ , with $n$ frozen at $3$ . At the start, $x = - 4$ , the left bar is $- 27$ and the right bar is $- 11$ , so the power sits lower. Move to $x = - 2$ and the left bar is $- 1$ while the right bar is $- 5$ : the power is above the line again, even though $x < - 1$ . The hypothesis is sufficient for the inequality on this page. Leaving it does not, by itself, point at the first place the comparison fails.

The object is the comparison of $(1 + x)^{n}$ with the line through $(0, 1)$ of slope $n$ , for a positive integer $n$ and a real $x \geq - 1$ . It does not order the harmonic, geometric, arithmetic, and quadratic means. That chain is means of positive numbers. A sum of moduli is the triangle inequality.