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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 $x>0$

Martin Sleziak
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Bob
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4 Answers4

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

  • 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) – Milo Brandt Dec 10 '14 at 02:35
  • So, what you want me to do? Delete my answer? –  Dec 10 '14 at 04:20
  • 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. – Milo Brandt Dec 10 '14 at 04:35
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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 $$(x-1)^2\geq 0$$

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

Teoc
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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.

hjpotter92
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    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! – Milo Brandt Dec 10 '14 at 02:13
  • this proof is sort of goofy –  Dec 10 '14 at 02:28