$\left|\frac{\arctan(x)}{2+e^x}\right| \leq \frac{\pi/2}{2+e^x}$

Since $\arctan(x) \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ for all $x \in \mathbb{R}$, we have $\left|\frac{\arctan(x)}{2 e^x}\right| \leq \frac{\pi/2}{2 e^x}$. To see why, note that $2e^x > 0$ for all $x$, so we can drop the absolute value signs. Then, since $\arctan(x)$ is bounded between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$, dividing by $2e^x$ can only make the quantity smaller. Therefore, we have the inequality $\frac{\arctan(x)}{2 e^x} \leq \frac{\pi/2}{2 e^x}$, and taking the absolute value of both sides gives the desired result.