# Prove $x + \frac{1}{x} \geq 2$ for $x>0$.

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 $x>0$

• It seems to me that your started with your assumption. Dec 10, 2014 at 1:56
• xkcd has a post on this with a lot of proofs. Dec 10, 2014 at 2:14
• How to prove this inequality $x + \frac{1}{x} \geq 2$ Jun 20, 2015 at 5:57

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) Dec 10, 2014 at 2:35
• So, what you want me to do? Delete my answer?
– user139708
Dec 10, 2014 at 4:20
• 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. Dec 10, 2014 at 4:35

$$(x-1)^2\geq 0\\ \Longrightarrow x^2-2x+1\geq 0\\ \Longrightarrow x^2+1\geq2x$$ 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 $$(x-1)^2\geq 0$$

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

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