Introduction
Many inequalities compare the values of two functions across an interval. A useful way to prove such a comparison is to examine how the functions change: if one function’s derivative is at least the other’s, then its values cannot fall behind, provided the values are correctly ordered at a common starting point.
The endpoint condition is essential. A comparison of derivatives controls the difference in how the functions change, not their initial separation. The Mean Value Theorem makes this idea precise and supplies a reliable proof method for inequalities.
The Derivative-Comparison Principle
Let \(I\) be an open interval, let \(a\in I\), and let \(f,g:I\to\mathbb{R}\) be differentiable. Suppose that for every \(t\in I\) with \(t\ge a\),
and suppose their values at \(a\) satisfy \(f(a)\ge g(a)\). Then \(f(x)\ge g(x)\) for every \(x\in I\) with \(x\ge a\).
For a strict conclusion, it is enough that \(f(a)\ge g(a)\) and \(f'(t)>g'(t)\) throughout the interior of \([a,x]\) when \(x>a\). Alternatively, a strict starting inequality \(f(a)>g(a)\), together with the weak derivative inequality, also gives \(f(x)>g(x)\). These conditions describe ways to obtain strictness; the weak conclusion does not require them.
Why the Principle Works
Define the difference \(h=f-g\). Then \(h(a)=f(a)-g(a)\ge0\), and \(h'(t)=f'(t)-g'(t)\ge0\). For any \(x>a\), the Mean Value Theorem applies to \(h\) on \([a,x]\): differentiability on an interval ensures continuity on this closed subinterval and differentiability in its interior. Therefore some \(c\in(a,x)\) satisfies
Since \(h'(c)\ge0\) and \(x-a>0\), the right side is nonnegative. Thus \(h(x)\ge h(a)\ge0\), which means \(f(x)-g(x)\ge0\), or \(f(x)\ge g(x)\). If \(x=a\), the conclusion is exactly the assumed endpoint inequality. This proves the result for every \(x\ge a\) in the interval.
In words, the derivative comparison says that the difference \(f-g\) cannot decrease as the input moves to the right. The endpoint condition says that this difference starts at or above zero. Together, they keep it at or above zero.
Worked Examples
Worked Example: A Quadratic Inequality for \(x\ge1\)
Prove that \(x^2\ge 2x-1\) for every \(x\ge1\) using derivative comparison.
Set \(f(x)=x^2\) and \(g(x)=2x-1\), on \(I=(0,\infty)\), and choose the common endpoint \(a=1\). Their derivatives are
For \(x\ge1\), \(2x\ge2\), so \(f'(x)\ge g'(x)\). At the endpoint,
The endpoint values agree, so the theorem gives \(f(x)\ge g(x)\) for all \(x\ge1\). That is, \(x^2\ge2x-1\). As an independent algebraic check, the difference factors as
At \(x=1\), both sides of the claimed inequality equal \(1\), in agreement with the endpoint calculation.
Worked Example: A Cubic Inequality on a Closed Range
Prove that \(x^3\le3x\) for \(0\le x\le1\).
Choose \(f(x)=3x\) and \(g(x)=x^3\), so the desired comparison is \(f(x)\ge g(x)\). At the common endpoint \(a=0\),
The derivative comparison on the full interval is
For \(0\le x\le1\), we have \(x^2\le1\), hence \(3x^2\le3\), so \(f'(x)\ge g'(x)\). Both functions are differentiable on an open interval containing \([0,1]\); therefore the derivative-comparison theorem applies from \(0\) to every \(x\in[0,1]\). It gives \(3x\ge x^3\), as required.
Check the conclusion by subtracting the smaller expression from the larger one:
because \(x\ge0\) and \(3-x^2\ge2>0\) on this interval. At \(x=0\), both sides of the original inequality are \(0\); at \(x=1\), the inequality reads \(1\le3\).
Worked Example: A Rational Inequality for \(x\ge0\)
Prove that \(\frac{1}{1+x}\ge1-x\) for \(x\ge0\).
Let \(f(x)=\frac{1}{1+x}\) and \(g(x)=1-x\), with \(a=0\). These functions are differentiable on \(I=(-1,\infty)\), which contains every interval from \(0\) to a chosen \(x\ge0\). Their endpoint values are
For \(x>-1\), differentiation gives
When \(x\ge0\), \(1+x\ge1\), so \((1+x)^2\ge1\) and \(\frac{1}{(1+x)^2}\le1\). Negating both sides reverses the inequality, giving
Thus \(f'(x)\ge g'(x)\) for all \(x\ge0\). The endpoint values are equal, so the theorem proves the desired inequality. Check it directly by taking the difference:
The denominator is positive and the numerator is nonnegative for \(x\ge0\), confirming the comparison. At \(x=0\), both sides equal \(1\).
Choosing the Right Comparison
A proof by derivative comparison has two distinct tasks: establish an order between derivatives on the entire interval of interest, and check the order between function values at the starting endpoint. The functions need not be positive; it is their difference that matters. Nor does the method require the derivative of either function to have a particular sign by itself. What matters is the sign of \(f'-g'\).
When choosing the order of the functions, begin with the target inequality. If the target is \(f(x)\ge g(x)\), test whether \(f'(x)\ge g'(x)\) on the interval and whether \(f(a)\ge g(a)\) at a convenient endpoint. If the target is \(f(x)\le g(x)\), the corresponding useful conditions are \(f'(x)\le g'(x)\) and \(f(a)\le g(a)\). Sometimes the same comparison is easier to establish after reversing the names of the functions.
The proof can also be read in terms of a single function. Once \(h=f-g\) is formed, the requirements become \(h'(x)\ge0\) and \(h(a)\ge0\). The Mean Value Theorem shows that \(h\) is nondecreasing from \(a\), so it remains nonnegative. This is the essential reasoning behind the comparison.
Common Pitfalls
Omitting the endpoint check. The claim that \(f'\ge g'\) alone implies \(f\ge g\) is false. For example, on \([0,1]\), let \(f(x)=x-10\) and \(g(x)=0\). Then \(f'(x)=g'(x)=1\) is false as written: in fact, \(f'(x)=1\) and \(g'(x)=0\), so \(f'(x)\ge g'(x)\) throughout. But at \(x=0\), \(f(0)=-10<0=g(0)\), and at \(x=1\), \(f(1)=-9<0=g(1)\). The derivative order holds, yet the claimed function order does not. The correction is to check the endpoint and confirm it has the required order before applying the theorem.
Checking derivatives at only one point. In the quadratic example, the comparison \(2x\ge2\) is valid for every \(x\ge1\), not merely at \(x=1\). A derivative comparison must hold throughout the interval between the common endpoint and the point where the function inequality is claimed. A favorable derivative value at the endpoint alone is not enough.
Reversing an inequality while negating. In the rational example, \(\frac{1}{(1+x)^2}\le1\) implies \(-\frac{1}{(1+x)^2}\ge-1\), not \(\le-1\). Negating both sides reverses the order. A sign error at this step would invalidate the derivative comparison and therefore the proof.
Why This Method Is Useful
Derivative comparison converts a problem about values across an interval into a local calculation plus one reference value. It is especially useful when differentiating the two sides produces a simple inequality, while direct comparison of the original expressions is less transparent. The method is not automatic: the interval, derivative order, endpoint, and endpoint order all have to be verified. When they do align, the Mean Value Theorem supplies the logical link from change to value.
Check Your Understanding
For each question, identify the derivative comparison and endpoint information that would be needed.
- To prove \(f(x)\ge g(x)\) for \(x\ge a\), which inequalities should hold for \(f'\), \(g'\), \(f(a)\), and \(g(a)\)?
- Why does the derivative comparison theorem require information across the interval from \(a\) to \(x\), rather than only at \(a\)?
- Suppose \(f'(t)\ge g'(t)\) on an interval to the right of \(a\), but \(f(a)<g(a)\). What part of the theorem’s argument is unavailable?
- For \(f(x)=4x\) and \(g(x)=x^2\) on \([0,2]\), check their derivatives and values at \(0\). What comparison follows from the theorem?
- In the rational example, why does \(\frac{1}{(1+x)^2}\le1\) give the required direction after multiplying by \(-1\)?
