Proving inequalities with derivatives

An inequality can be proved by taking the difference and examining where it rises and falls. We confirm $e^x \geq 1 + x$ .

Examining the difference

Put $g(x) = e^x - 1 - x$ . What is to be shown is that $g(x) \geq 0$ for every $x$ .

g'(x) = e^x - 1

The function $e^x$ increases monotonically and equals $1$ at $x = 0$ .

Range	$e^x$	$g'$	Behavior of $g$
$x < 0$	$< 1$	negative	decreasing
$x > 0$	$> 1$	positive	increasing

So $g$ is smallest at $x = 0$ , where its value is $g(0) = 0$ . Since the minimum is $0$ we have $g(x) \geq 0$ , that is $e^x \geq 1 + x$ for every $x$ , with equality only at $x = 0$ .

Seeing it in the picture

The line $y = 1 + x$ is the tangent to $y = e^x$ at $x = 0$ . Since $g''(x) = e^x > 0$ , the function $e^x$ is convex, and a convex function lies above every one of its tangents¹. This inequality is one instance of that.

Substitutions give other forms

Substitution	Inequality obtained
$x \to -x$	$e^{-x} \geq 1 - x$
$x \to \ln t$	$\ln t \leq t - 1$
$x \to \dfrac{1}{n}$	$e^{1/n} \geq 1 + \dfrac{1}{n}$

The second is much used as a way of bounding a logarithm from above.

A link to the definition of $e$

Raising both sides of the third to the $n$ th power gives the following.

e \geq \left( 1 + \frac{1}{n} \right)^n

That the sequence appearing in the definition of $e$ never exceeds $e$ follows from this single inequality.

The procedure

Take the difference
Differentiate
Find the minimum or maximum from the change of sign
See whether that value is at least $0$

To prove something only for $x > 0$ , it is enough to examine the value at the end of that range together with the monotonicity. Proving an inequality turns into a question about rising and falling.

The convex curve on the graph is $y = e^x$ , its tangent is $y = 1 + x$ , the difference of the two is $y = e^x - 1 - x$ , and the large dots are the point of tangency $(0, 1)$ and the origin, where the difference is $0$ .

Convex function, Wikipedia