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This question already has an answer here:

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 $(\forall x (p)) \to q$?

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marked as duplicate by Adam Hughes, David Mitra, Claude Leibovici, Najib Idrissi, Davide Giraudo Feb 1 '15 at 12:29

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.

  • $\begingroup$ How is this a contradiction? Didn't you want to prove that its greater than $2$ $\endgroup$ – AvZ Feb 1 '15 at 8:51
  • $\begingroup$ @AvZ this is the contradiction: $100<2$ $\endgroup$ – shinzou Feb 1 '15 at 8:52
  • $\begingroup$ Oh okay, you're saying it's a contradiction to the second condition. $\endgroup$ – AvZ Feb 1 '15 at 8:53
  • $\begingroup$ 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 $100$ $\endgroup$ – AvZ Feb 1 '15 at 8:58
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    $\begingroup$ The contrapositive of $\forall x: P(x)$ is $\exists x: \lnot P(x)$, so: No, you didn't you use proof by contradiction correctly. $\endgroup$ – Henrik Feb 1 '15 at 9:02
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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 $x>0$.

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  • $\begingroup$ Isn't this assuming what you want to prove? You can't assume that the inequality $(x+\frac 1 x \ge 2)$ is true... $\endgroup$ – shinzou Feb 1 '15 at 8:55
  • $\begingroup$ @kuhaku: I haven't assumed anything. I merely stated "it's sufficient to prove". $\endgroup$ – barak manos Feb 1 '15 at 8:56
  • $\begingroup$ Ah I see... Is my approach with a proof by contradiction wrong though? $\endgroup$ – shinzou Feb 1 '15 at 8:58
  • $\begingroup$ 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. $\endgroup$ – grapher Feb 1 '15 at 9:06
  • $\begingroup$ @Adam Hughes: Why did you change "it's sufficient" to "it sufficient"? This is grammatically incorrect as far as I know. $\endgroup$ – barak manos Feb 1 '15 at 9:23
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The easiest proof is by noting that $\left(x+\dfrac{1}{x}-2\cdot\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}\right)=\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2\ge0$.

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    $\begingroup$ I really doubt that the easiest. $\endgroup$ – Henrik Feb 1 '15 at 9:12
  • $\begingroup$ @Henrik: Can you tell me the reason? $\endgroup$ – user 170039 Feb 1 '15 at 9:17
  • $\begingroup$ Just look at some of the other answers. No radicals for instance. $\endgroup$ – Henrik Feb 1 '15 at 9:18
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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., $x = 1$.

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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 $$ a_1+a_2+\cdots+a_n \geq n. $$

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