Alternative proofs of this Inequality

Question

So I was reading a paper which made the claim "It is easy to see that $\frac{1-e^{-\alpha}}{\alpha} > 1-\frac{\alpha}{2} > \frac{1}{1+\alpha}$ when $0 < \alpha < 1$."

Verifying that $1-\frac{\alpha}{2} > \frac{1}{1+\alpha}$ only involves some simple algebra, but in order to prove that $\frac{1-e^{-\alpha}}{\alpha} > 1-\frac{\alpha}{2}$ I had to use what I have only seen referred to as the "racetrack theorem\principle" from elementary calculus twice.

Namely, if $f(0) = g(0)$ and $f'(x) \geq g'(x)$ for all $x \geq 0$, then $f(x) \geq g(x)$ for all $x \geq 0$.

Taking derivatives involving the quotient rule is somewhat of a pain in the ass, so I was wondering if there were any slick ways of proving this inequality? Or maybe I missed a far simpler way of proving this inequality?

Answer 1

First, let's rearrange $\frac{1-e^{-\alpha}}{\alpha} > 1-\frac{\alpha}{2}$ to $e^{-\alpha} < 1 - \alpha + \frac{\alpha^2}{2}$ and try proving this second inequality.

We have an infinite series for $e^{-\alpha}$: it is $1 - \alpha + \frac{\alpha^2}{2!} - \frac{\alpha^3}{3!} + \frac{\alpha^4}{4!} - \cdots$ where the $k^{\text{th}}$ term is $(-1)^k \frac{\alpha^k}{k!}$. (This is one of the definitions of $e^x$.) When $0 < \alpha < 1$, the terms of this sum are decreasing in absolute value: we raise $\alpha$ to a higher power, and on top of that we increase the denominator. Therefore:

• Stopping the sum after a positive term is an overestimate: we can group all the tail terms we dropped into consecutive pairs that are less than $0$.
• Stopping the sum after a negative term is an underestimate: we can group all the tail terms we dropped into consecutive pairs that are greater than $0$.

In particular, stopping the sum after $\frac{\alpha^2}{2}$ is an overestimate: $e^{-\alpha} = 1 - \alpha + \frac{\alpha^2}{2}$.

Answer 2

Let $f(\alpha) = 1 - e^{-\alpha}, g(\alpha) = \alpha - \dfrac{\alpha^2}{2}, 0 \le \alpha < 1$. Check that $f(0) = g(0) = 0$, and also $f'(\alpha) = e^{-\alpha}, g'(\alpha) = 1-\alpha$, and $f''(\alpha) = -e^{-\alpha}, g''(\alpha) = -1$. Obverse that $e^{\alpha} \ge 1 \implies e^{-\alpha} \le 1\implies -e^{-\alpha} \ge -1\implies f''(\alpha) \ge g''(\alpha)$, and $f''(0) = g''(0)=1$. Thus apply the racetrack theorem twice to conclude: $f'(\alpha) \ge g'(\alpha)\implies f(\alpha) \ge g(\alpha)\implies \dfrac{1-e^{-\alpha}}{\alpha} > 1 - \dfrac{\alpha}{2}, \alpha > 0$. Done.