5
$\begingroup$

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?

$\endgroup$
4
  • 3
    $\begingroup$ 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 $P$). $\endgroup$ Commented Sep 13, 2022 at 20:29
  • 1
    $\begingroup$ It looks good to me. $\endgroup$ Commented Sep 13, 2022 at 20:47
  • $\begingroup$ There are too many ways! :) $\endgroup$
    – user
    Commented Sep 13, 2022 at 20:48
  • 1
    $\begingroup$ You can reduce to just one inequality $(|x|-1)^2\ge 0$ which is equivalent to ${|x|\over x^2+1}\le {1\over 2}.$ $\endgroup$ Commented Sep 13, 2022 at 20:57

2 Answers 2

12
$\begingroup$

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

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

$\endgroup$
0
1
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .