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?