Proving Inequalities With Derivatives

Turn the Inequality into a Function

A useful way to prove an inequality is to move all its terms to one side and study the resulting function. If the claim is $A(x)\geq B(x)$, define the difference $F(x)=A(x)-B(x)$. The inequality is then equivalent to $F(x)\geq0$. Differentiation can reveal where this difference increases or decreases and whether a known equality point is its minimum.

Earlier in the course, the Mean Value Theorem and the consequences of derivative signs were established. We will use those results here rather than prove them again. The main task is to choose a difference function whose derivative is simpler to analyze than the original comparison. The sign of the derivative may change, so it is important to divide the domain into the intervals on which that sign is fixed.

A common pattern is that two sides agree at a particular point. Subtracting them produces a function that vanishes there. If its derivative is negative on one side of that point and positive on the other, the function decreases toward zero and then increases away from zero. This establishes a global lower bound without needing to solve for any critical point other than the known equality point.

A Logarithmic Comparison

Theorem (Logarithm Lies Below Its Tangent at One): For every $x>0$, $$ \ln x\leq x-1, $$ with equality if and only if $x=1$.

Proof. Define $F(x)=x-1-\ln x$ for $x>0$. The desired inequality is exactly $F(x)\geq0$. Since $F(1)=0$, we examine the derivative on each side of $1$:

F'(x)=1-\frac{1}{x}=\frac{x-1}{x}.

The denominator is positive throughout the domain. Thus $F'(x)<0$ for $0<x<1$, and $F'(x)>0$ for $x>1$. By the derivative-sign results established earlier in the course, $F$ is decreasing on $(0,1)$ and increasing on $(1,\infty)$. Consequently, $F(x)>F(1)=0$ when $x\ne1$, while $F(1)=0$. This proves the inequality and its equality condition. $\square$

The proof relies on both intervals adjoining the equality point. Merely observing that $F'(1)=0$ would not establish that $F$ is nonnegative elsewhere. What matters is the sign of $F'$ on the entire domain on either side of $1$. The domain restriction $x>0$ is also essential: it ensures that the logarithm is defined and that the denominator in $F'(x)$ is positive.

Worked Example: Comparing a Positive Number with an Exponential

For $y>0$, apply the logarithmic comparison to $y$:

\ln y\leq y-1.

The exponential function is strictly increasing, so applying it to both sides preserves the inequality:

y\leq e^{y-1}.

Equality occurs exactly when $y=1$, because that is the equality case in the logarithmic comparison. For example, at $y=3$ the result is $3\leq e^2$. This application illustrates how a derivative-proved inequality can be transferred through an increasing function, provided the direction of the comparison is preserved.

Bernoulli’s Inequality for Real Exponents

A second useful comparison involves a power function and a linear function. For real exponents at least one, the power curve lies above its tangent line at $1$ on the nonnegative half-line. The following derivative argument proves the inequality, including its endpoint case.

Theorem (Bernoulli’s Inequality for Real Exponents at Least One): Let $r\geq1$ be real. For every $t\geq0$, $$ t^r\geq 1+r(t-1). $$ If $r=1$, equality holds for every $t\geq0$. If $r>1$, equality holds only at $t=1$.

Proof. First suppose $r=1$. Both sides are $t$, so the statement holds with equality for every $t\geq0$. Now let $r>1$, and define

G(t)=t^r-1-r(t-1), \qquad t\geq0.

The function is continuous at $0$, with $G(0)=r-1>0$, and differentiable for $t>0$. For positive $t$,

G'(t)=r(t^{r-1}-1).

Because $r-1>0$, the quantity $t^{r-1}$ is less than $1$ when $0<t<1$, equals $1$ at $t=1$, and exceeds $1$ when $t>1$. Therefore $G'(t)<0$ on $(0,1)$ and $G'(t)>0$ on $(1,\infty)$. Also $G(1)=0$. It follows that $G(t)>G(1)=0$ for positive $t\ne1$. At the remaining endpoint, $G(0)=r-1>0$. Hence $G(t)\geq0$ for all $t\geq0$, with equality only at $1$. Rearranging the definition of $G$ proves the result. $\square$

To express the same result in the more familiar shifted form, put $t=1+x$. The condition $t\geq0$ becomes $x\geq-1$, and the inequality reads $(1+x)^r\geq1+rx$. The endpoint $x=-1$ is included: there the left side is $0$ and the right side is $1-r\leq0$. Checking this endpoint matters because the derivative of $t^r$ was used only for $t>0$.

Worked Example: A Fractional Power with a Negative Shift

Take $r=\frac32$ and $x=-\frac35$. Then $1+x=\frac25$, which is positive, so Bernoulli’s inequality applies. Its two sides are

(1+x)^r=\left(\frac25\right)^{3/2} \qquad\text{and}\qquad 1+rx=1+\frac32\left(-\frac35\right)=\frac1{10}.

The left side is $\frac25\sqrt{\frac25}$. To verify the comparison directly, both sides are positive, so we can square them:

\left(\frac25\sqrt{\frac25}\right)^2=\frac{8}{125} \qquad\text{and}\qquad \left(\frac1{10}\right)^2=\frac1{100}.

Since $\frac{8}{125}>\frac1{100}$, the positive quantities satisfy $\left(\frac25\right)^{3/2}>\frac1{10}$. This agrees with the strict inequality predicted by the theorem, since $x\ne0$ and $r>1$.

Young’s Inequality from a Minimum

Derivative arguments can also prove inequalities involving two variables. A useful strategy is to hold one variable fixed and view the difference between the two sides as a function of the other. Finding where its derivative changes sign identifies its minimum and, in this case, the precise equality condition.

Theorem (Young’s Inequality): Let $p>1$, and define $q=p/(p-1)$. For all $a,b\geq0$, $$ ab\leq\frac{a^p}{p}+\frac{b^q}{q}. $$ Equality holds if and only if $b=a^{p-1}$.

Proof. If $a=0$, the claimed inequality becomes $0\leq b^q/q$, which holds because $b\geq0$. Equality in this case holds exactly when $b=0$, which is also the condition $b=a^{p-1}$. Now suppose $a>0$. With $a$ fixed, define, for $b\geq0$,

H(b)=\frac{a^p}{p}+\frac{b^q}{q}-ab.

For $b>0$, differentiation gives

H'(b)=b^{q-1}-a.

Since $q-1=1/(p-1)$, the equation $H'(b)=0$ has the unique positive solution $b_*=a^{p-1}$. If $0<b<b_*$, then $b^{q-1}<a$, so $H'(b)<0$. If $b>b_*$, then $H'(b)>0$. Thus $H$ decreases up to $b_*$ and increases after $b_*$. Its minimum on the positive half-line occurs at $b_*$; continuity at $b=0$ also gives $H(0)\geq H(b_*)$.

To evaluate the minimum, note that $(p-1)q=p$. Therefore $b_*^q=a^p$ and $ab_*=a^p$, giving

H(b_*)=\frac{a^p}{p}+\frac{a^p}{q}-a^p =a^p\left(\frac1p+\frac1q-1\right)=0,

because $1/p+1/q=1$. Hence $H(b)\geq0$ for all $b\geq0$, proving Young’s inequality. The derivative is negative before $b_*$ and positive after it, so this minimum is attained only at $b=b_*$. Together with the $a=0$ case, this proves the stated equality condition. $\square$

Worked Example: The Quadratic Form of Young’s Inequality

Choose $p=q=2$, which satisfies $1/p+1/q=1$. Young’s inequality becomes

ab\leq\frac{a^2}{2}+\frac{b^2}{2},\qquad a,b\geq0.

For $a=3$ and $b=4$, the left side is $12$, while the right side is

\frac{3^2}{2}+\frac{4^2}{2}=\frac92+8=\frac{25}{2}.

Thus $12\leq25/2$. The equality condition is $b=a^{p-1}=a$; since $4\ne3$, the inequality is strict here. In contrast, taking $a=b=3$ gives $9=9/2+9/2$, confirming equality in the predicted case.

Choosing the Difference and Checking the Argument

The examples use the same broad strategy, but the useful difference function depends on the form of the claim. For a one-variable inequality, subtract the proposed lower bound from the expression being bounded. For a two-variable inequality, it can be effective to fix one variable and differentiate with respect to the other. In either case, the derivative is a tool for locating the minimum; it does not replace the need to check the domain or equality cases.

Form a difference.
Arrange the claim as $F(x)\geq0$, or as $H(b)\geq0$ with other variables fixed.

Record the domain and equality point.
Check where every term is defined and identify any input at which equality is expected.

Analyze the derivative on each relevant interval.
A sign change can show that the difference decreases toward a minimum and then increases away from it.

Check endpoints and equality.
If differentiation was carried out only in the interior, verify boundary values separately. State exactly when the inequality becomes an equality.

A frequent mistake is to infer a global inequality from a derivative calculation at just one point. Knowing $F'(a)=0$ says only that the derivative vanishes there; it does not show that $F(a)$ is a minimum. Another common error is to overlook a boundary point where the function is defined but the derivative formula is unavailable. The Bernoulli proof handles $t=0$ separately, and the Young proof treats $a=0$ before differentiating with respect to $b$.

The purpose of these methods is not to differentiate every inequality mechanically. A good choice of difference makes the derivative manageable, and a known equality point often suggests where to split the domain. Once the derivative sign is established on the necessary intervals, the monotonicity results developed earlier in the course convert that sign information into a rigorous comparison.

Check Your Understanding

For each question, identify the difference function or derivative-sign information that makes the comparison work.

Why must the logarithmic comparison be proved separately on the intervals to the left and right of $1$?
In the real-exponent Bernoulli inequality, what is the value of the difference function at the endpoint $t=0$, and why must it be checked?
For which values of $t$ does the derivative of the Bernoulli difference have negative sign when $r>1$?
In Young’s inequality, why is it useful to hold $a$ fixed and differentiate with respect to $b$?
What condition on $a$ and $b$ gives equality in Young’s inequality?