TITLE: Proof of Aristarchus' Inequality
QUESTION [6 upvotes]: Does anyone know how to prove that if $0<\alpha<\beta<\frac{\pi}{2}$ then $\frac{\sin\alpha}{\alpha}>\frac{\sin\beta}{\beta}$. Any methods/techniques may be used.

REPLY [3 votes]: First we prove the classic inequalities
$$
\sin x < x <\tan x \quad (0<x<\pi/2).
$$
The the inequality $\sin x < x$ follows from comparing the lengths of the segment $XX'$and the arc $XX'$; $x<\tan x$ comes from the area of triangle $OIT$ and the aera of sector $OIX$.

Then we prove that
$$
\frac{\sin\beta-\sin\alpha}{\beta-\alpha} < \cos\alpha < \frac{\sin\alpha}{\alpha}.
$$
The second inequality is simply $\alpha<\tan\alpha$. The second one is true because
$$
\frac{\sin\beta-\sin\alpha}{\beta-\alpha}
= \frac{2\sin\frac{\beta-\alpha}2\cos\frac{\beta+\alpha}2}{\beta-\alpha}
< \frac{2\cdot \frac{\beta-\alpha}2 \cdot \cos\alpha}{\beta-\alpha}
= \cos\alpha.
$$
Finally,
$\frac{\sin\beta-\sin\alpha}{\beta-\alpha} < \frac{\sin\alpha}{\alpha}$
is equivalent with
$\frac{\sin\beta}{\beta} < \frac{\sin\alpha}{\alpha}$.