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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.


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

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

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

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


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