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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.

Harsh Kumar
  • 2,846

1 Answers1

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

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

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

Kiran
  • 4,198