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Prove $\forall x\in \mathbb R$, if $x>0$ then $(x+\frac 1 x \ge 2)$

I think a proof by contradiction is the easiest in this case, so we have: $\forall x\in \mathbb R :x>0\wedge \neg(x+\frac 1 x \ge 2)\iff \forall x\in \mathbb R :x>0\wedge (x+\frac 1 x < 2)$

Take $x=100$ and we get an immediate contradiction $100+\frac 1 {100}< 2$

Did I use proof by contradiction correctly?

Is this statement $\forall x (p\to q)$ or $(\forall x (p)) \to q$?

shinzou
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  • How is this a contradiction? Didn't you want to prove that its greater than $2$ – AvZ Feb 01 '15 at 08:51
  • @AvZ this is the contradiction: $100<2$ – shinzou Feb 01 '15 at 08:52
  • Oh okay, you're saying it's a contradiction to the second condition. – AvZ Feb 01 '15 at 08:53
  • This does not prove that it's true for all real numbers. It only proves that its true for $100$ and the second condition is not true for $100$ – AvZ Feb 01 '15 at 08:58
  • @AvZ I assumed the conclusion is false for all $x$ and got a contradiction, thus it has to be true for all $x$. – shinzou Feb 01 '15 at 08:59
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    The contrapositive of $\forall x: P(x)$ is $\exists x: \lnot P(x)$, so: No, you didn't you use proof by contradiction correctly. – Henrik supports the community Feb 01 '15 at 09:02
  • @Henrik but this isn't proof by contraposition... – shinzou Feb 01 '15 at 09:03
  • This only proves that inverse/not of the first condition is not true for $100$ and therefore not true for ALL the real numbers. However, we can bot be sure if it is true for any number like $99$ or not. If it is, then the first condition is false. – AvZ Feb 01 '15 at 09:04
  • Basically what I am saying is that proving something is not true for ALL the real numbers doesn't mean that it is false for all real numbers – AvZ Feb 01 '15 at 09:05
  • In this case proof by contradiction works if, assuming $x+\frac{1}{x}<2$ you can conclude that $x\leq0$. This is a contradiction because you assume $x>0$. – Marco Cantarini Feb 01 '15 at 09:05
  • You could use AM-GM; but that's a bit much. – David Mitra Feb 01 '15 at 09:05
  • You do claim to do proof by contradiction, but what you actually are doing is closer to proof by contrapositive (and it's almost the same thing). @Marco's comment tells you how to do it by contradiction. – Henrik supports the community Feb 01 '15 at 09:10
  • For example, proving that $x^2>0\forall\mathbb R$ is not true by contradiction doesn't mean that $x^2\leq 0\forall\mathbb R$ – AvZ Feb 01 '15 at 09:11
  • @MarcoCantarini what quantifiers do you use on the statements? – shinzou Feb 01 '15 at 09:20
  • @kuhaku $A\Longrightarrow B\Leftrightarrow\lnot B\Longrightarrow\lnot A.$ This is the proof by contradiction. – Marco Cantarini Feb 01 '15 at 09:30
  • @MarcoCantarini what you wrote is definitely proof by contra position https://en.wikipedia.org/wiki/Contraposition – shinzou Feb 01 '15 at 09:32
  • @kuhaku a proof by contraposition can be interpretated as a proof by contradiction https://en.wikipedia.org/wiki/Proof_by_contrapositive – Marco Cantarini Feb 01 '15 at 09:38

4 Answers4

4

Multiply each side of the inequality by $x$.

Since $x>0$, the direction of the inequality will not invert, so we get:

  • It's sufficient to prove that $x>0 \implies x^2+1\geq2x$
  • It's sufficient to prove that $x>0 \implies x^2-2x+1\geq0$
  • It's sufficient to prove that $x>0 \implies (x-1)^2\geq0$

The last one is obviously true for every $x\in\mathbb{R}$, hence true for $x>0$.

barak manos
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  • Isn't this assuming what you want to prove? You can't assume that the inequality $(x+\frac 1 x \ge 2)$ is true... – shinzou Feb 01 '15 at 08:55
  • @kuhaku: I haven't assumed anything. I merely stated "it's sufficient to prove". – barak manos Feb 01 '15 at 08:56
  • Ah I see... Is my approach with a proof by contradiction wrong though? – shinzou Feb 01 '15 at 08:58
  • Typically, for such type of inequalities try to reach square of something. Since square is always bigger or equal to zero, you have proven it. – grapher Feb 01 '15 at 09:06
  • @Adam Hughes: Why did you change "it's sufficient" to "it sufficient"? This is grammatically incorrect as far as I know. – barak manos Feb 01 '15 at 09:23
  • @barakmanos check the edit history, I did not change that, only "inequity" to "inequality." – Adam Hughes Feb 01 '15 at 09:25
  • @AdamHughes: Got it, thanks. Unfortunately, I don't know how to check edit-history, so I'll just have to take your word for it. BTW, I thought "inequity" was the correct term... Second thought, it's a $\geq$, so perhaps "equation" is the correct term... who knows... – barak manos Feb 01 '15 at 09:30
  • @barakmanos no, one still says "inequality" for $\le$ as well. To check edit histories, you click on the phrased "edited [n] minutes ago by" above an edit picture, and it shows you exactly what was changed by whom. – Adam Hughes Feb 01 '15 at 09:32
  • @AdamHughes: Oh... right, thank you very much (damn, that was me who wrote "it sufficient", I will fix it now)... – barak manos Feb 01 '15 at 09:33
2

The easiest proof is by noting that $\left(x+\dfrac{1}{x}-2\cdot\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}\right)=\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2\ge0$.

1

If $x > 0$, then there is a number $y > 0$ such that $y^2 = x$, hence $$x + \frac{1}{x} = y^2 + \frac{1}{y^2} = y^2 - 2 + \frac{1}{y^2} + 2 = \left(y - \frac{1}{y}\right)^2 + 2 \ge 0 + 2 = 2.$$ The inequality step is true because no real square is negative. Equality is therefore attained when $y = \frac{1}{y}$, or $y = 1$; i.e., $x = 1$.

heropup
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1

By AM–GM inequality we have

$$ \frac{x+\dfrac{1}{x}}{2} \geq \sqrt{x \cdot \frac{1}{x}}=1. $$

Generalisation: for $a_i >0$, such that $a_1 a_2 \ldots a_n =1$ we have $$ a_1+a_2+\cdots+a_n \geq n. $$

Leox
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