Bernoulli's Inequality

A Lower Bound for Powers

The Binomial Theorem gives an exact expansion of $(1+x)^n$ when $n$ is a nonnegative integer. Sometimes an exact expansion is more information than we need: a simple lower bound can be enough to compare quantities or estimate a power. Bernoulli’s Inequality supplies such a bound using only the constant and linear terms.

The sign of $x$ matters. When $x\geq0$, the inequality follows naturally from the nonnegative terms in the binomial expansion. It also remains valid for some negative values, but the induction proof must be handled carefully: multiplying an inequality by a factor preserves its direction only when that factor is nonnegative. This is why the condition $x\geq-1$ will be part of the result.

Bernoulli’s Inequality

We use the convention that $x^0=1$, including when $x=0$. The exponent in the theorem is an integer; no assumption about powers with noninteger exponents is needed.

Theorem (Bernoulli’s Inequality). Let $x\in\mathbb{R}$ satisfy $x\geq-1$, and let $n\geq0$ be an integer. Then $$ (1+x)^n\geq1+nx. $$

The assertion includes $n=0$, when both sides equal $1$. For $n=1$, it is also an equality. For larger exponents it is a genuine lower bound, with equality conditions described below.

Proof. We use induction on $n$, keeping the assumption $x\geq-1$ fixed. When $n=0$,

$$ (1+x)^0=1=1+0x. $$

Now suppose that for some integer $n\geq0$,

(1+x)^n\geq1+nx.

Because $x\geq-1$, we have $1+x\geq0$. Multiplying the induction hypothesis by $1+x$ therefore preserves the direction of the inequality. It gives

(1+x)^{n+1}\geq(1+nx)(1+x) =1+(n+1)x+nx^2.

Since $n\geq0$ and $x^2\geq0$, the remaining term satisfies $nx^2\geq0$. Thus

(1+x)^{n+1}\geq1+(n+1)x+nx^2\geq1+(n+1)x.

This proves the result for $n+1$. By induction, it holds for every integer $n\geq0$. $\square$

The proof identifies the main mechanism: after one more multiplication by $1+x$, the desired linear expression appears, along with the extra term $nx^2$. That term is nonnegative. The restriction $x\geq-1$ ensures that the multiplication used in the induction does not reverse the inequality.

Equality and Strictness

The inequality is an equality for every $x$ when $n=0$ or $n=1$. For $n\geq2$, $x=0$ still gives equality, since both sides are $1$. For all other allowed $x$, the inequality is strict.

Corollary (Equality Cases in Bernoulli’s Inequality). If $n\geq2$ and $x\geq-1$, then $$ (1+x)^n=1+nx $$ if and only if $x=0$. For $n=0$ and $n=1$, equality holds for every $x\geq-1$.

Proof. Define the difference between the two sides by

$$ D_n=(1+x)^n-1-nx. $$

We have $D_1=0$. For every integer $n\geq1$, direct rearrangement gives

D_{n+1} =(1+x)^{n+1}-1-(n+1)x =(1+x)D_n+nx^2.

Indeed, substituting $D_n=(1+x)^n-1-nx$ into the middle expression on the right yields $(1+x)^{n+1}-(1+x)(1+nx)+nx^2$, which simplifies to $(1+x)^{n+1}-1-(n+1)x$. For $n=1$, this recurrence gives $D_2=x^2$. If $x\ne0$, then $D_2>0$. Whenever $n\geq2$ and $D_n>0$, the assumptions $1+x\geq0$, $n>0$, and $x^2>0$ imply

D_{n+1}=(1+x)D_n+nx^2>0.

Thus $D_n>0$ for every $n\geq2$ when $x\ne0$. If $x=0$, then $D_n=1-1-0=0$ for every $n$. The cases $n=0$ and $n=1$ follow directly from their formulas. $\square$

Worked Applications

Worked Example: A Positive Increment

Use Bernoulli’s Inequality to find a lower bound for $(1+\frac{2}{5})^4$. Here $x=\frac{2}{5}$, which satisfies $x\geq-1$, and $n=4$. Therefore

\left(1+\frac{2}{5}\right)^4 \geq1+4\left(\frac{2}{5}\right) =1+\frac{8}{5} =\frac{13}{5}.

The bound is strict because $n\geq2$ and $x\ne0$. As a direct check, the left side is $(7/5)^4=2401/625$, while $13/5=1625/625$. Since $2401>1625$, the calculated value does exceed the stated lower bound.

Worked Example: A Negative Increment

Apply the inequality to $(1-\frac{1}{4})^3$. Set $x=-\frac{1}{4}$ and $n=3$. The domain condition holds because $-\frac{1}{4}\geq-1$. Then

\left(1-\frac{1}{4}\right)^3 \geq1+3\left(-\frac{1}{4}\right) =1-\frac{3}{4} =\frac{1}{4}.

This is a strict inequality: the left side is $(3/4)^3=27/64$, and the right side is $1/4=16/64$. In this case the bound is positive and informative. If the same $x$ is used with $n=6$, the right side becomes $1-6/4=-1/2$; the inequality remains correct, but a negative lower bound for a positive power is not very useful.

Worked Example: A Bound Valid for Every Positive Integer

Let $m$ be any positive integer, and consider $(1+\frac{1}{m})^m$. The value $x=\frac{1}{m}$ is nonnegative, so Bernoulli’s Inequality applies with exponent $n=m$:

\left(1+\frac{1}{m}\right)^m \geq1+m\left(\frac{1}{m}\right) =2.

Thus every one of these powers is at least $2$. If $m=1$, the expression is exactly $2$. If $m\geq2$, the equality-case result shows that it is strictly greater than $2$, since $x=1/m\ne0$. No expansion with $m$ terms is needed to get this lower bound.

Worked Example: Scaling the Inequality

Suppose $a>0$, $b\geq-a$, and $n\geq1$ is an integer. To estimate $(a+b)^n$, write it as $a^n(1+b/a)^n$. Since $a>0$, the ratio $b/a$ is defined, and $b\geq-a$ implies $b/a\geq-1$. Bernoulli’s Inequality gives

(a+b)^n =a^n\left(1+\frac{b}{a}\right)^n \geq a^n\left(1+n\frac{b}{a}\right) =a^n+na^{n-1}b.

For example, take $a=4$, $b=-1$, and $n=3$. The hypotheses hold, and the scaled bound is

(4-1)^3\geq4^3+3\cdot4^2(-1) =64-48=16.

Indeed, $3^3=27\geq16$. The scaling requires $a>0$: it lets us divide by $a$ and multiply the resulting inequality by the positive number $a^n$ without changing its direction.

Why the Domain Condition Matters

The hypothesis $x\geq-1$ is more than a technical detail. In the induction proof, it makes $1+x$ nonnegative, so the induction hypothesis can be multiplied by that factor without reversing its direction. If $x<-1$, the theorem can fail. For instance, take $x=-4$ and $n=3$. Then

(1+x)^n=(-3)^3=-27, \qquad 1+nx=1+3(-4)=-11,

and $-27<-11$, contrary to the claimed lower bound. The theorem therefore cannot be used without checking its domain. A separate check of $n$ is needed too: the result concerns nonnegative integer exponents, not arbitrary real exponents.

For $x\geq0$, the Binomial Theorem also helps explain why the inequality is natural. The expansion begins with $1+nx$; when $n\geq2$, its remaining terms are nonnegative because they involve nonnegative powers of $x$ and positive binomial coefficients. For $-1\leq x<0$, however, the later terms can have alternating signs. The induction proof handles that case without incorrectly assuming that every term in the expansion is nonnegative.

A useful habit is to check three items before applying Bernoulli’s Inequality: the exponent is a nonnegative integer, the increment satisfies $x\geq-1$, and the resulting lower bound is strong enough to be useful. A true lower bound may still be weak, as the negative right side in the second example illustrates.

Key takeaway. For $x\geq-1$ and an integer $n\geq0$, Bernoulli’s Inequality gives $(1+x)^n\geq1+nx$. Its induction proof depends on multiplying by the nonnegative factor $1+x$. When $n\geq2$, equality occurs exactly at $x=0$.

Check Your Understanding

Use the domain condition, induction argument, and equality cases to answer these questions.

For which real values of $x$ does Bernoulli’s Inequality apply, and why does the induction proof need this condition?
Apply the inequality with $x=\frac{1}{3}$ and $n=5$ to obtain a lower bound for $(1+\frac{1}{3})^5$.
For $x=-\frac{1}{2}$ and $n=2$, is the inequality an equality or a strict inequality? Explain using the equality cases.
Use the scaled form with $a=3$, $b=1$, and $n=4$ to obtain a lower bound for $(3+1)^4$.
Why does the induction step not establish Bernoulli’s Inequality for all $x<-1$?

What You'll Learn

A Lower Bound for Powers