<s>[INST] 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?
 [/INST] 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 
$$\frac{\arctan x}{x}\gt 1-\frac{x^2}{3}\ge \frac{2}{3}.$$
</s>