# 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?

• The proof is fine as long as you replace the last implication by an equivalence. Otherwise, you're affirming the consequent (If $P \Rightarrow Q$, and $Q$ is true, then it does not follow that $P$ is true. But if $P \iff Q$, and $Q$ is true, then so is $P$). Commented Sep 13, 2022 at 20:29
• It looks good to me. Commented Sep 13, 2022 at 20:47
• There are too many ways! :)
– user
Commented Sep 13, 2022 at 20:48
• You can reduce to just one inequality $(|x|-1)^2\ge 0$ which is equivalent to ${|x|\over x^2+1}\le {1\over 2}.$ Commented Sep 13, 2022 at 20:57

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

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

This is my solution:

Put: $$\frac{x}{1+x^2}=a$$

So this can be rewritten as: $$ax^2-x+a=0$$

Now we can have: $$\Delta_x = 1-4a^2$$ which give: $$a \in [- \frac{1}{2} ; \frac{1}{2}]$$

Now the inequality is Q.E.D