Prove $\sin x < x < \tan x$ for $0 < x < \pi/2$ algebraically I know there are some proofs showing that, for  $0 < x < \pi/2$:
$\sin x < x < \tan x $
with the use of the unit circle, however, can I prove this algebraically in some way as well?

One simple way could be to demonstrate that , over $[0, pi/2)$ $\sin$ is concave while $\tan$ is convex, and that $\sin(0)=0=\tan(0)$ and $\sin '(0) = 1 =\tan'(0)$.
