Standard Inequalities Derived from the Mean Value Theorem

From a derivative estimate to an inequality

The Mean Value Theorem links the change in a function to one of its derivative values. This makes it a useful source of inequalities: if the derivative cannot be too large on an interval, then the function cannot change too much across that interval.

The key is to estimate the derivative on the entire interval between the inputs, not only at one endpoint or at a convenient point. Once that estimate is valid, the Mean Value Theorem converts it into a bound on the difference of function values.

The same statement applies with the endpoints in either order: \[ |f(y)-f(x)|\le M|y-x| \] whenever the segment between \(x\) and \(y\) lies in an interval where the stated hypotheses and derivative bound hold. The number \(M\) must be nonnegative.

Why the derivative bound gives a function bound

The Mean Value Theorem says that, under its continuity and differentiability hypotheses, there is some \(c\in(a,b)\) such that \[ f(b)-f(a)=f'(c)(b-a). \] Taking absolute values and using the bound on the derivative gives \[ |f(b)-f(a)|=|f'(c)|(b-a)\le M(b-a). \] This proves the theorem. The unknown point \(c\) does not need to be found: it is enough that the derivative estimate holds at every possible point in the interval.

When a derivative is continuous on the closed interval, it attains a largest absolute value there. In that situation one may take \(M\) to be the maximum of \(|f'|\) on the interval. Often, a simpler bound is sufficient, even if it is not the smallest possible one.

Bounds for differences of integer powers

The same reasoning works for every positive integer power. The derivative of \(t^n\) is \(nt^{n-1}\), so a bound on \(|t|\) over the interval gives a bound on the derivative.

To prove the result when \(x\ne y\), apply the Mean Value Theorem to \(f(t)=t^n\) on the interval with endpoints \(x\) and \(y\). Every point \(c\) between these endpoints satisfies \[ |c|\le\max(|x|,|y|). \] Consequently, \[ |f'(c)|=n|c|^{n-1} \le n\max(|x|,|y|)^{n-1}. \] The Mean Value Theorem gives \(|x^n-y^n|=|f'(c)||x-y|\), which proves the inequality. If \(x=y\), both sides are zero, so the inequality holds in that case too.

For nonnegative endpoints, the derivative also gives a useful lower bound. If \(x\ge y\ge0\), then every point \(c\) between \(y\) and \(x\) satisfies \(c\ge y\). For \(n\ge1\), this implies \(nc^{n-1}\ge ny^{n-1}\). Therefore, when \(x>y\), the Mean Value Theorem yields the two-sided estimate below; when \(x=y\), both sides of the estimate are equal to zero.

For the upper bound, the mean value point satisfies \(c\le x\), so \(nc^{n-1}\le nx^{n-1}\). For the lower bound, \(c\ge y\), so \(nc^{n-1}\ge ny^{n-1}\). Substituting these estimates into \(x^n-y^n=nc^{n-1}(x-y)\), with \(x-y\ge0\), proves both inequalities.

Bernoulli’s inequality from the Mean Value Theorem

A familiar inequality for integer powers also follows directly from a derivative estimate. It compares a power of \(1+x\) with the first-degree expression \(1+nx\).

For \(x=0\), equality holds. Suppose first that \(x>0\). Apply the Mean Value Theorem to \(f(t)=t^n\) between \(1\) and \(1+x\). There is a \(c\) with \(1<c<1+x\) such that \[ (1+x)^n-1=nc^{n-1}x. \] Since \(c\ge1\), \(c^{n-1}\ge1\), and since \(x>0\), \[ (1+x)^n-1=nc^{n-1}x\ge nx. \] Rearranging proves the claim for positive \(x\).

Now suppose \(-1\le x<0\). The endpoints \(1+x\) and \(1\) are nonnegative, so the Mean Value Theorem applies to \(t^n\) on that interval. For some \(c\) between \(1+x\) and \(1\), \[ (1+x)^n-1=nc^{n-1}x. \] Here \(0\le c\le1\), so \(c^{n-1}\le1\). Multiplication by the negative number \(nx\) reverses the comparison, giving \(nc^{n-1}x\ge nx\). Hence \((1+x)^n-1\ge nx\), as required. This also covers \(x=-1\); if \(n=1\), the derivative factor is \(c^0=1\).

A reciprocal-function bound

The method is not limited to powers. For example, the derivative of \(f(t)=1/t\) is \(-1/t^2\). On an interval bounded away from zero, this derivative has a finite absolute bound, which controls changes in reciprocal values.

If \(a=b\), the inequality is immediate. Otherwise, apply the Mean Value Theorem to \(f(t)=1/t\) on the interval between \(a\) and \(b\). Every point \(t\) in that interval satisfies \(t\ge m\), so \(|f'(t)|=1/t^2\le1/m^2\). The derivative-bound theorem then gives the stated result.

A common mistake: checking the derivative at only one point

A derivative bound must hold throughout the interval. It is not valid to evaluate the derivative at one endpoint and assume that value controls the rest of the interval.

Choosing and using a derivative bound

In applications, first identify the function and the interval joining the inputs. Then find a bound for the absolute value of its derivative everywhere on that interval. A bound that is larger than necessary still proves a valid inequality; a bound that fails at even one point cannot be used.

These inequalities matter because they turn local information about a derivative into concrete control over function values. The power bounds, Bernoulli’s inequality, and reciprocal estimate are examples of the same reasoning pattern. The essential discipline is to keep the estimate valid across the entire interval before applying the Mean Value Theorem.