Proof of Aristarchus' Inequality 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.

The function $f(x)=\frac{\sin(x)}{x}$ is decreasing on $[0,\pi/2]$. Since its derivative is
$$
f^{'}(x)=\frac{\cos(x)x-\sin(x)}{x^2},
$$
we've reduce the problem to seeing that $\cos(x)x-\sin(x)\leq 0$. For small values of $x$, we have $\sin(x)\approx x$, so $\cos(x)x-\sin(x)\approx x(\cos(x)-1)$, which is negative. To see that it remains negative for all $x$ in $[0,\pi/2]$, you can check, for example via bissection method, that the first positive solution to $\cos(x)x-\sin(x)=0$ is around $4.5>\pi/2$.
