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 $$\frac{1}{x}\geq(x-2)\hspace{0.5cm} \text{ when } \hspace{0.2cm}0 but my question is can we do it algebraically or using calculus to prove it without any reference to the graphs.
Prove that $x+\frac{1}{x}\geq2$
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calculus
algebra-precalculus
elementary-number-theory
inequality
a.m.-g.m.-inequality
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3It looks like you went from $A$ to $B$ and then came back to $A$! – 2017-01-26
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2Just solve this inequality. $$\left(\sqrt{x}+\frac{1}{\sqrt x}\right)^2\ge0$$ – 2017-01-26
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0@HarshKumar I just wanted to let you know that I have made [a post on meta](http://meta.math.stackexchange.com/questions/25694/tag-management-2017/25797#25797) about the tag (a.m.-g.m.-inequality) which you have recently created. – 2017-01-30
1 Answers
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$\dfrac{x+\dfrac{1}{x}}{2} \ge \sqrt{x \times \dfrac{1}{x} }$
$\dfrac{x+\dfrac{1}{x}}{2} \ge 1$
$x+\dfrac{1}{x} \ge 2$