An arctangent inequality As in the title, how can I prove
$$
\frac{\arctan(x)}{x}\geq\frac{1}{2}
$$
for $x\in(0,1]$?
I think I can say:
$$\frac{\arctan(x)}{x}$$
is monotonically decreasing in the interval, so its value is greater than the value in $1$, which is $\frac{\pi}{4}$, greater than $\frac{1}{2}$.
There exists some elegant proof of this simple inequality, maybe using series expansions?

Use that $\sin(x) <x$ for $x > 0$ (which is clear from a picture of the unit circle) and so $$\frac{\arctan(x)}{x} >\frac{\sin(\arctan(x))}{x} = \frac{1}{\sqrt{1+x^2}}$$ which seems good enough an approximation on $(0,1]$.
