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 $(0,1]$.
When $-1\le x\le 1$, we have $$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots,$$ and therefore if $-1\le x\le 1$ and $x\ne 0$, $$\frac{\arctan x}{x}=1-\frac{x^2}{3}+\frac{x^4}{5}-\frac{x^6}{7}+\cdots. \tag{$1$}$$ The series $(1)$ is an alternating series, and therefore if we truncate just after the term $-\frac{x^2}{3}$ the error term is positive (and less than the first "neglected" term $\frac{x^4}{5}$). It follows that $$\frac{\arctan x}{x}\gt 1-\frac{x^2}{3}\ge \frac{2}{3}.$$
