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Proof that $x+\frac{1}{x}\geq2$ for $x>0$

Would this be correct?

$x*(x+\frac{1}{x}\geq2)$

$x^2+1\geq2x$

$x^2-2x+1\geq2x-2x$

$x^2-2x+1\geq0$

Plug in 1 for x:

$(1)^2-2(1)+1\geq0$

$1-2+1\geq0$

$0\geq0$ Therefore, $x+\frac{1}{2}\geq2$ is true for $x>0$

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4 Answers 4

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We know for positive $a,b$, $\frac{ a + b}{2} \geq \sqrt{ab} $. Put $ a = x^2 $ and $b = 1$ and we obtain

$$ x + \frac{1}{x} \geq 2 $$

Added: You can also use calculus: Put $f(x) = x + \frac{1}{x} $. we have $f'(x) = 1 - \frac{1}{x^2} $ with critical values (points where $f'$ vanishes ) : $x=\pm1$. It is easy to see that $x = 1 $ will furnish a global minimum. Hence $f(x) \geq f(1) $ for all $x > 0 $. It follows that $$ x + \frac{1}{x} \geq 2 $$

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  • $\begingroup$ The first thing your write obscures the proof - you're going to need to prove the AM-GM inequality in basically the same way that you'd prove the original statement! (Not to mention that, if you use it, applying it to $\frac{1}x$ and $x$ directly suffices with no additional algebraic manipulation except multiplying by two) $\endgroup$ Dec 10, 2014 at 2:35
  • $\begingroup$ So, what you want me to do? Delete my answer? $\endgroup$
    – user139708
    Dec 10, 2014 at 4:20
  • $\begingroup$ If it were me, I'd provide a proof of AM-GM. It's natural enough to do that for this question, since the proof of it is basically the same as the proof in the accepted answer that $x+\frac{1}x\geq 2$ - meaning such a proof could serve as a useful generalization without excessive complication and help elucidate the link between the two statements. $\endgroup$ Dec 10, 2014 at 4:35
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$$(x-1)^2\geq 0\\ \Longrightarrow x^2-2x+1\geq 0\\ \Longrightarrow x^2+1\geq2x$$ Then we assume $x$ is positive and divide by $x$. $$\Longrightarrow x+\frac1{x}\geq 2$$ See if you can prove it for negative $x$ and the case where $x=0$ then see if you can prove $$(x-1)^2\geq 0$$

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Hint: You have to prove that 1 is the minimum of the function.

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Proof by contradiction:

Assume that the hypothesis is wrong. Therefore, $ \exists\ x \gt 0 $ such that; $$ x + \dfrac{1}{x} \lt 2 $$

Now, follow along:

$$\begin{align} x + \dfrac{1}{x} &\lt 2 \\ x^2 + 1 &\lt 2x \tag{$ \because x \gt 0 $} \\ x^2 - 2x + 1 &\lt 0 \\ (x - 1)^2 &\lt 0 \end{align}$$ which is a contradiction as a perfect square can never be negative for real numbers.

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    $\begingroup$ This is a direct proof. It is silly to phrase it as an argument by contradiction. Just delete the bit where you assume its false, reverse all the equals signs, write the equations in the opposite order, and boom! $\endgroup$ Dec 10, 2014 at 2:13
  • $\begingroup$ this proof is sort of goofy $\endgroup$
    – user139708
    Dec 10, 2014 at 2:28

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