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This question already has an answer here:

To prove $x+\frac{1}{x}\geq2$ where $x$ is a positive real number. This is what i try: $$\text{We need to prove } \hspace{1cm} x+\frac{1}{x}-2\geq 0$$ now,$$\frac{x^2-2x+1}{x}=(x-2)+\frac{1}{x}$$ its enough to show that $$\frac{1}{x}\geq(x-2)\hspace{0.5cm} \text{ when } \hspace{0.2cm}0<x\leq 2$$ we can easily show it using the graph.

but my question is can we do it algebraically or using calculus to prove it without any reference to the graphs.

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marked as duplicate by Martin R, E. Joseph, Rohan, Community Jan 26 '17 at 9:14

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    $\begingroup$ It looks like you went from $A$ to $B$ and then came back to $A$! $\endgroup$ – NeedForHelp Jan 26 '17 at 9:11
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    $\begingroup$ Just solve this inequality. $$\left(\sqrt{x}+\frac{1}{\sqrt x}\right)^2\ge0$$ $\endgroup$ – Harsh Kumar Jan 26 '17 at 12:36
  • $\begingroup$ @HarshKumar I just wanted to let you know that I have made a post on meta about the tag (a.m.-g.m.-inequality) which you have recently created. $\endgroup$ – Martin Sleziak Jan 30 '17 at 13:41
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$\dfrac{x+\dfrac{1}{x}}{2} \ge \sqrt{x \times \dfrac{1}{x} }$

$\dfrac{x+\dfrac{1}{x}}{2} \ge 1$

$x+\dfrac{1}{x} \ge 2$

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