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This question already has an answer here:

prove:

$\left|x+\frac{1}{x}\right|\geq 2$

Can I just use

$\left|\left(\sqrt{x}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^2\right|\geq 2$

and

$\left|\left(\sqrt{-x}\right)^2+\left(\frac{1}{\sqrt{-x}}\right)^2\right|\geq 2$?

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marked as duplicate by Martin Sleziak, Michael Hoppe, user85798, mookid, user63181 Mar 22 '14 at 12:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ You can't use those inequalities until you prove them. $\endgroup$ – Gerry Myerson Jul 9 '13 at 13:16
  • $\begingroup$ @GerryMyerson Do you mean $\left|x+\frac{1}{x}\right|\geq 2$ and $\left|\left(\sqrt{x}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^2\right|\geq 2$ are the same? $\endgroup$ – HyperGroups Jul 9 '13 at 13:19
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    $\begingroup$ No, I mean you can't use those inequalities until you prove them. You can't use anything in a proof that hasn't itself already been proved (OK, you can use axioms, and rules of inference, but that's it). $\endgroup$ – Gerry Myerson Jul 9 '13 at 13:24
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    $\begingroup$ If $x=0$ this could get rather interesting. $\endgroup$ – JB King Jul 9 '13 at 19:01
  • $\begingroup$ I don't see any answer by anyone named Abbras. $\endgroup$ – Gerry Myerson Jul 9 '13 at 22:38
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Hint: Prove that $$\left(x+\frac{1}{x}\right)^2 \geq 4$$

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  • $\begingroup$ why you can square them, and could be valid, and mine are just Sqrt them and then square them are wrong? $\endgroup$ – HyperGroups Jul 9 '13 at 13:22
  • $\begingroup$ You're technique isn't wrong, it's just incomplete - you say "Can I just use..." and then state two inequalities that are no more easy or difficult to prove. How do you know these inequalities are true? $\endgroup$ – Thomas Andrews Jul 9 '13 at 13:28
  • $\begingroup$ Abbra's answer essentially completes your proof... $\endgroup$ – Thomas Andrews Jul 9 '13 at 13:35
  • $\begingroup$ Abbra's second part is right? but why GerryMyerson say I cannot use those inequalities? $\endgroup$ – HyperGroups Jul 9 '13 at 13:43
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    $\begingroup$ Because you didn't prove them. You have to prove them to use them. Abbra proved them. $\endgroup$ – Thomas Andrews Jul 9 '13 at 13:45
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Find an absolute minimum:

$$\frac{d}{dx}\left|x+\frac{1}{x}\right|=\left(1-\frac{1}{x^2}\right)\frac{|x+\frac{1}{x}|}{x+\frac{1}{x}}$$

$$0=1-\frac{1}{x^2}$$

$$x=\pm{1}$$

$$|x+\frac{1}{x}|_{x=\pm{1}}=2$$

Therefore $2$ is a local minimum.

It can be demonstrated that $(\pm{1},2)$ is the absolute minimum by observing the boundaries of the intervals $(0,\infty)$ and $(-\infty,0)$.

$$\lim_{x\to{-\infty}}|x+\frac{1}{x}|=\infty$$ $$\lim_{x\to{0^-}}|x+\frac{1}{x}|=\infty$$ $$\lim_{x\to{0^+}}|x+\frac{1}{x}|=\infty$$ $$\lim_{x\to{\infty}}|x+\frac{1}{x}|=\infty$$

Therefore $(\pm{1},2)$ is the absolute minimum.

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    $\begingroup$ @AbhraAbirKundu It is an completely trivial to show that it is an absolute minimum, however, on $\mathbb R^+$. $\endgroup$ – Thomas Andrews Jul 9 '13 at 13:36
  • $\begingroup$ Yes @ThomasAndrews I do agree with you but its always nice to have a complete proof. $\endgroup$ – Abhra Abir Kundu Jul 9 '13 at 13:38
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    $\begingroup$ @AbhraAbirKundu See my edit. $\endgroup$ – Ataraxia Jul 9 '13 at 13:38
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Without loss of genarality we can consider $x\ge 0$(Reason: $x$ and $\frac{1}{x}$ has same sign so if $x<0$ the in place of $x$ we will take $-x(>0)$ and it will not make any difference since $\left| x+\frac{1}{x}\right |=\left| -x+\frac{1}{-x}\right |$)

Then we have,

$\left| x+\frac{1}{x}\right |=x+\frac{1}{x}=(\frac{1}{\sqrt{x}}-\sqrt{x})^2+2\ge 2$

You can use(with its proof which is done by my above arguement) either one(not both ) of the following(if you want to handle only real stuff) according to $x$ is positive or negative,

$\left|\left(\sqrt{x}\right)^2+\left(\frac{1}{\sqrt{x}}\right)^2\right|\geq 2$ or $\left|\left(\sqrt{-x}\right)^2+\left(\frac{1}{\sqrt{-x}}\right)^2\right|\geq 2$

You cant use both since ,if $x$ is positive then $\sqrt{-x}$ is complex.And if $x<0$ then $\sqrt{x}$ is complex.

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For $x\neq 0$, we have $(x-1)^{2}\geq 0$. Therefore, $x^{2}-2x+1\geq 0$ and hence $x^{2}+1\geq 2x$. If $x>0$, then we are done. If $x< 0$, then $-x> 0$. Hence,

$\displaystyle x^{2}+1\geq 2(-x)\implies -x-\frac{1}{x}\geq 2\implies x+\frac{1}{x}\leq -2$.

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$$ a^2 - 2ab + b^2 = (a-b)^2 $$

$$ x + \frac1x \text{ is the same as }\sqrt{x}^2 + {\sqrt{\frac1x}}^2 $$ but there's no "$-2ab$" term in the middle. Since $a=\sqrt{x}$ and $b=\sqrt{\frac1x}$, we conclude that $-2ab=-2\sqrt{x}\cdot\sqrt{\frac1x}$. This simplifies to just $-2$.

So $$ \sqrt{x}^2 -2 + {\sqrt{\frac1x}}^2 $$ is a perfect square: $a^2-2ab+b^2=(a-b)^2=\left(\sqrt{x}-\sqrt{\frac1x}\right)^2=x-2+\frac1x$.

So $$ x+\frac1x = \left(x-2+\frac1x\right) + 2 = \left(\sqrt{x}-\sqrt{\frac1x}\right)^2 + 2 = \underbrace{\text{a square} + 2}. $$ Since it's a square plus $2$, it is $\ge 2$.

(All this assumes $x$ is positive; it $x$ is negative, then it's not true that $x+\frac1x\ge 2$.)

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  • $\begingroup$ In the last equation it will be $-$ in place of $+$. And the question is about the abs. value of the expression which is ofcourse bounded below. $\endgroup$ – Abhra Abir Kundu Jul 9 '13 at 17:51
  • $\begingroup$ fixed; thanks. ${{}}$ $\endgroup$ – Michael Hardy Jul 9 '13 at 17:53
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Denote $x+\frac{1}{x}=v$.

Obviously $|v|>2$ if either $v>2$ or $-v<-2$. The first case is therefore $x +\frac{1}{x}>2$ and it becomes $(x-1)^2>0$, which is always true. The second case is $-x-\frac{1}{x}<-2$ or $\frac{x^2-2x+1}{x}>0$ and also breaks down into 2 cases: either $(x-1)^2>0 \cap x>0$ or $(x-1)^2<0 \cap x<0$. Obviously the second case is never true and the first case is always true.

Hence $|x+\frac{1}{x}|$ is always larger than $2$.

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If $x>0$, then by AM-GM inequality, $x+1/x \geq 2\cdot \sqrt{x\cdot(1/x)}=2$, hence $|x+1/x| \geq 2$. If $x<0$, then $-x>0$ and therefore $|x+1/x|=|-x-1/x|\geq 2$ by the former result.

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