Mean Value Arguments in Inequalities

From Derivatives to Inequalities Between Values

Taylor’s Theorem gives inequalities by controlling a remainder. Another useful approach is to apply the Mean Value Theorem directly to the difference between two function values. It identifies that difference as a derivative evaluated somewhere between the inputs, multiplied by the distance between them. If the derivative is bounded on that interval, the difference is bounded as well.

The Mean Value Theorem itself was established earlier in the course, as were its consequences for derivative bounds. Here we focus on a practical inequality technique: choose the function whose difference is the expression of interest, locate the intermediate point supplied by the theorem, and use the location of that point to bound the derivative. Two results illustrate the method for powers and logarithms.

Theorem (Bounds for Differences of Integer Powers): Let $n\geq1$ be an integer, and let $0\leq u<v$. Then $$ n u^{n-1}(v-u)\leq v^n-u^n\leq n v^{n-1}(v-u). $$ For $n=1$, both bounds are equalities. For $n>1$, the inequalities are strict if $u<v$.

Proof. Apply the Mean Value Theorem to $f(t)=t^n$ on $[u,v]$. This function is continuous on that closed interval and differentiable on its interior, so there is a point $c\in(u,v)$ such that

v^n-u^n=f'(c)(v-u)=n c^{n-1}(v-u).

Since $0\leq u<c<v$ and $n-1\geq0$, we have $u^{n-1}\leq c^{n-1}\leq v^{n-1}$; when $n=1$, all three powers with exponent zero equal $1$. Multiplying these inequalities by the nonnegative quantity $n(v-u)$ gives the claimed bounds. If $n>1$, the function $t^{n-1}$ is strictly increasing on the positive part of the interval, and $c$ lies strictly between its endpoints. Thus both bounds are strict. $\square$

This result is useful when a difference of powers is difficult to factor or when the factors in an exact factorization are not convenient to estimate. The theorem reduces the problem to bounding a derivative over the interval. Its endpoint assumptions matter: nonnegativity ensures that the power $t^{n-1}$ is ordered as the input increases. For odd powers on intervals containing negative numbers, the same displayed bounds need not follow in this form.

Worked Example: Bounding a Difference of Fourth Powers

Take $n=4$, $u=2$, and $v=3$. The theorem gives

4(2^3)(3-2)\leq 3^4-2^4\leq4(3^3)(3-2).

The left side is $4\cdot8\cdot1=32$, and the right side is $4\cdot27\cdot1=108$. Direct calculation gives $3^4-2^4=81-16=65$, so in this case

$$ 32<65<108. $$

The estimate is not meant to compute the difference exactly. It brackets it using only the derivative $4t^3$ at points between $2$ and $3$. In applications where one needs a safe upper or lower bound rather than an exact value, that can be enough.

Bounds for Logarithmic Increments

The derivative of the logarithm is $1/t$, which decreases on the positive real numbers. As a result, the change in $\ln t$ between two positive inputs is bounded using the reciprocal of either endpoint. This gives a convenient way to estimate logarithms without evaluating them.

Theorem (Bounds for a Logarithmic Increment): If $0<a<b$, then $$ \frac{b-a}{b}<\ln b-\ln a<\frac{b-a}{a}. $$

Proof. The function $f(t)=\ln t$ is continuous on $[a,b]$ and differentiable on $(a,b)$. By the Mean Value Theorem, some $c\in(a,b)$ satisfies

\ln b-\ln a=f'(c)(b-a)=\frac{b-a}{c}.

Because $a<c<b$, and all three numbers are positive, taking reciprocals reverses their order:

\frac{1}{b}<\frac{1}{c}<\frac{1}{a}.

Multiplication by $b-a>0$ preserves that order. Substituting the expression for the logarithmic increment proves both strict inequalities. $\square$

The positivity and order of the inputs do essential work in this proof. The derivative $1/t$ is defined and positive throughout the interval, and its endpoint values bound the derivative at the intermediate point. Reversing the order of the inputs without adjusting the signs would not give the stated inequalities; the theorem is deliberately written for $a<b$.

Worked Example: A Simple Bound for the Natural Logarithm of Two

Use the logarithmic increment theorem with $a=1$ and $b=2$. Since $\ln 1=0$, it yields

\frac{2-1}{2}<\ln 2-\ln 1<\frac{2-1}{1}, \qquad\text{so}\qquad \frac12<\ln 2<1.

Both endpoint comparisons are strict because the intermediate point $c$ lies strictly between $1$ and $2$. No decimal approximation to $\ln 2$ is needed. The estimate is often useful when only a rational upper or lower bound is required.

When the Increment Is Negative

A common source of errors is applying an inequality derived for a positive increment when the increment is negative. The Mean Value Theorem still applies, but multiplying by a negative number reverses the direction of an inequality. One safe strategy is to order the endpoints first, apply a result stated for an increasing pair, and only then rewrite the expression of interest.

Worked Example: Logarithm Bounds on Both Sides of Zero

We show that for $-1<x$ with $x\ne0$,

\frac{x}{1+x}<\ln(1+x)<x.

First suppose $x>0$. Apply the logarithmic increment theorem with $a=1$ and $b=1+x$. Since $\ln 1=0$, the result is

\frac{x}{1+x}<\ln(1+x)<x.

Now suppose $-1<x<0$. In this case $0<1+x<1$, so apply the theorem in increasing order with $a=1+x$ and $b=1$. It gives

\frac{1-(1+x)}{1}<\ln 1-\ln(1+x)<\frac{1-(1+x)}{1+x}.

Since $1-(1+x)=-x$ and $\ln1=0$, this becomes

-x<-\ln(1+x)<\frac{-x}{1+x}.

Multiplying all three parts by $-1$ reverses both inequality signs, giving

x>\ln(1+x)>\frac{x}{1+x}.

Reordering the two strict inequalities proves the stated result for negative $x$ as well. At $x=0$, both bounds are equalities, which is why the strict version excludes that point. The restriction $x>-1$ ensures that $1+x$ is positive and the logarithm is defined.

Using an Intermediate Derivative for Other Functions

The same reasoning applies whenever the derivative is easy to compare at points between the inputs. The intermediate point need not be found explicitly. It is enough to know where it lies and how the derivative behaves there.

Worked Example: An Exponential Inequality

We prove that $e^x>1+x$ for every $x\ne0$, with equality at $x=0$. Apply the Mean Value Theorem to $f(t)=e^t$ on the interval with endpoints $0$ and $x$. The theorem applies in either order: the function is continuous on the closed interval between the endpoints and differentiable in its interior. There is an intermediate point $c$ such that

e^x-e^0=e^c(x-0), \qquad\text{or equivalently}\qquad e^x-1=xe^c.

If $x>0$, then $0<c<x$. Since $e^t$ is strictly increasing, $e^c>1$, so $xe^c>x$. Therefore $e^x-1>x$, which is the required inequality. If $x<0$, then $x<c<0$, so $e^c<1$. Multiplication by $x<0$ reverses the comparison and gives $xe^c>x$ again. Thus $e^x-1>x$ also in this case. At $x=0$, direct substitution gives $e^0=1=1+0$.

This proof highlights why the sign of the increment must be tracked. The derivative value $e^c$ is below $1$ for negative $x$, but multiplying by the negative increment makes the resulting difference greater than $x$, not less.

Worked Example: A Square-Root Increment Bound

For $x>0$, apply the Mean Value Theorem to $f(t)=\sqrt{t}$ on $[1,1+x]$. This function is continuous on the interval and differentiable at every interior point, all of which are positive. For some $c\in(1,1+x)$,

\sqrt{1+x}-1=f'(c)x=\frac{x}{2\sqrt{c}}.

The location of $c$ implies $1<\sqrt{c}<\sqrt{1+x}$. Taking reciprocals and multiplying by $x/2>0$ gives

\frac{x}{2\sqrt{1+x}}<\frac{x}{2\sqrt{c}}<\frac{x}{2}.

Substituting the mean value identity yields

\frac{x}{2\sqrt{1+x}}<\sqrt{1+x}-1<\frac{x}{2}.

The upper bound says the square-root increment is smaller than the increment predicted by its derivative at $1$. The lower bound uses the derivative at the far endpoint, where the derivative has decreased.

Choosing and Checking a Mean Value Argument

A reliable proof begins by selecting the function whose endpoint difference matches the expression to be estimated. The Mean Value Theorem then turns that difference into a derivative at an intermediate point. The last step is to use the interval location to bound the derivative, while preserving the correct inequality direction.

Identify the increment.
Write the target expression as $f(b)-f(a)$, choosing the endpoint order carefully.

Check the hypotheses.
Verify continuity on the closed interval and differentiability on its interior. Also check that the function is defined throughout the interval.

Locate the intermediate point.
Use the theorem to write the increment as $f'(c)(b-a)$ for a point strictly between the endpoints.

Bound the derivative and track signs.
Use the location of $c$ to estimate $f'(c)$. If the increment is negative, remember that multiplication reverses inequalities.

The theorem supplies existence, not a formula for the intermediate point. A proof must not assume that $c$ is the midpoint or choose a convenient value for it. Instead, derive bounds valid for every point in the interval. This is why endpoint derivative bounds are effective when the derivative is monotone, as in the logarithm and square-root examples.

Mean value arguments complement Taylor estimates. Taylor’s Theorem can identify higher-order behavior near a center; the Mean Value Theorem compares two actual values through a derivative somewhere between them. In either method, the conclusion depends on controlling the error or intermediate derivative, not merely writing an approximation. Careful attention to interval hypotheses and signs turns the identity into a rigorous inequality.

Check Your Understanding

Use the interval location of the intermediate point and track signs carefully.

In the bounds for differences of integer powers, where is the intermediate point, and why can its power be bounded by the endpoint powers?
Why does the logarithmic increment theorem require $0<a<b$?
When proving the logarithmic inequality for negative $x$, why is it useful to apply the theorem first to the ordered endpoints $1+x$ and $1$?
In the exponential example with $x<0$, why does $e^c<1$ imply $xe^c>x$ rather than $xe^c<x$?
What must be checked before applying the Mean Value Theorem to a proposed function and interval?