Question:
Write the solution set of the inequation $x+\frac{1}{x} \geq 2$.
Solution:
We have,
$x+\frac{1}{x} \geq 2$
$\Rightarrow \frac{x^{2}+1}{x} \geq 2$
$\Rightarrow \frac{x^{2}+1}{x}-2 \geq 0$
$\Rightarrow \frac{x^{2}-2 x+1}{x} \geq 0$
$\Rightarrow \frac{(x-1)^{2}}{x} \geq 0$
$\Rightarrow$ Either $(x-1)^{2} \geq 0$ and $x>0$ or $(x-1)^{2}<0$ and $x<0$
But, $(x-1)^{2}$ is always greater than zero.
$\therefore(x-1)^{2} \geq 0$ and $x>0$
$\Rightarrow x>0$
$\Rightarrow x \in(0, \infty)$
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