Prove that $\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2}$ Prove that $$\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2}$$ for any number $x$.
My attempt:
$$\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2} \\ \iff \frac{x}{1+x^2} \geq -\frac{1}{2} \land \frac{x}{1+x^2} \leq \frac{1}{2}$$ $$\iff (x+1)^2 \geq 0 \land (x-1)^2 \geq 0$$ the last two inequalities are obviously true, which concludes my proof attempt.
Not sure if this is a correct way to prove the inequality, also it's clearly not very elegant.
Could someone please verify my solution, and maybe suggest a more elegant or efficient approach?

Your solution looks fine, as an alternative, by a single inequality, we have
$$\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2} \iff \left|1+x^2\right|\ge 2|x| 
 \iff x^4-2x^2+1\ge 0\iff (x^2-1)^2 \ge 0$$
or also
$$\left|\frac{x}{1+x^2}\right| \leq \frac{1}{2} \iff 1+x^2\ge 2|x| 
 \iff x^2-2|x|^2+1\ge 0\iff (|x|-1)^2 \ge 0$$

Another way by AM-GM
$$\frac{1+x^2}{2}\ge \sqrt{x^2}=|x|$$

Another way, by $x=\tan \theta$ we have
$$\left|\frac{x}{1+x^2}\right|= \left|\frac{\tan \theta}{1+\tan^2 \theta}\right|=\frac12|\sin 2\theta|\le \frac12$$

Another way, by rearrangement
$$\left|\frac{1+x^2}{x}\right|=\frac{1+x^2}{|x|}=\frac1{|x|}\cdot 1+|x|\cdot 1\ge \frac1{|x|}\cdot |x|+1\cdot 1= 2$$

Another one
$$\left|\frac{x}{1+x^2}\right|\le \frac12 \iff \frac{2|x|}{(|x|-1)^2+2|x|}\le 1$$
