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I am really stuck with this assignment and have very little idea on how to proceed. I think using derivative is not allowed. How should I proceed, could anyone give any tips?

I've thought something like.

If $x>0$ $$ x^2+1 > x \geq 1 $$ But this is really all I got, I've tried changing the form of the expression but nothing really. I would be grateful for tips!

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3 Answers 3

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We will prove that $$ \left| \frac{x}{x^2+1}\right|\le\frac{1}{2} $$ for every real number $x$. Square both sides of the inequality then we get $$ \frac{x^2}{(x^2+1)^2}\le\frac{1}{4}. $$ In fact, these two inequalities are equivalent, and last inequality is easy to prove: it is equivalent to $0\le (x^2-1)^2$.

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  • $\begingroup$ I don't really get how that shows that the largest number the expression can give is 1/2 and the smallest -1/2. Could you open it up some more? $\endgroup$ Oct 6, 2013 at 12:53
  • $\begingroup$ @SamuliLehtonen Let $a$ be a positive real number then $|x|\le a$ if and only if $-a\le x\le a$. $\endgroup$
    – Hanul Jeon
    Oct 6, 2013 at 12:55
  • $\begingroup$ In fact, the range of $f(x)=x/(1+x^2)$ is $[-1/2,1/2]$. But you want to show that the range of $f$ is a subset of $[-1/2,1/2]$. So we only find the upper bound and lower bound of the range. $\endgroup$
    – Hanul Jeon
    Oct 6, 2013 at 12:57
  • $\begingroup$ Yeah I got it now, thank you so much! $\endgroup$ Oct 6, 2013 at 13:02
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Hint: $x^2 +1 - 2x = (x-1)^2 \geq 0$, and $x^2+1 + 2x = (x+1)^2 \geq 0$

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$$\left|\dfrac{x^2+1}{x}\right| = \dfrac{|x|^2+1}{|x|} = |x|+\frac1{|x|}\ge 2\sqrt{|x|\frac1{|x|}}.$$

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