Write the solution set of the inequation

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 $x>0$

$\Rightarrow x>0$

$\Rightarrow x \in(0, \infty)$