0

Suppose $x \neq 0 $, then $| x + \frac{1}{x} | \geq 2 $.

I have shown this using the am gm inequality $(a+b)/2 \geq \sqrt{ab} $. In fact, with $a = x^2 $ and $b=1$ works. So, for $x > 0 $ we have proved the ineqality above. How can we handle the case $x < 0$ ?

Martin Sleziak
  • 53,687
  • Have you checked on Math. SE ? Very similar questions have been answered before – Shailesh Sep 02 '15 at 13:26
  • 2
    $|x+\frac1x|=|-x+\frac 1{-x}|$ – Piquito Sep 02 '15 at 13:30
  • ataulfo, what u wrote doesnt make sense. –  Sep 02 '15 at 13:35
  • Why not? That a hint to handle the case $x<0$ (then $-x>0$. – Quang Hoang Sep 02 '15 at 13:45
  • Some related questions: http://math.stackexchange.com/questions/439671, http://math.stackexchange.com/questions/705064 and many other questions which you can find here. Although it seems that you are mainly asking what to do for $x>0$, which makes your question a bit different from the ones I linked. – Martin Sleziak Sep 02 '15 at 14:21
  • @Ataulfo Since the main problem of the OP seems to be how to get from $x>0$ to the case $x<0$, perhaps you could expand your comment a bit and post it as an answer. – Martin Sleziak Sep 02 '15 at 14:32
  • @user135395 You wrote that you used AM-GM for the case $x>0$. I just want to point out that you would get the same result a bit faster using it with $a=x$ and $b=\frac1x$. – Martin Sleziak Sep 02 '15 at 14:33

1 Answers1

5

You can prove the result at once by writing

$$\left(x + \frac{1}{x}\right)^2 = x^2 + \frac{1}{x^2} + 2 \ge 2\sqrt{x^2\cdot \frac{1}{x^2}} + 2 = 4,$$

then taking square roots.

kobe
  • 41,901