Example 5. Solve the inequality $2 \sqrt{x} \geq 3-\frac{1}{x}$.
The set of permissible values for the unknown $x$ is $x>0$.
Let $f(x)=2 \sqrt{x}-\left(3-\frac{1}{x}\right)$, then the inequality $f(x) \geqslant 0$ is equivalent to the original inequality.
$$
f^{\prime}(x)=x^{-\frac{1}{2}}-x^{-2}=x^{-2}\left(x^{-\frac{3}{2}}-1\right) \text {. }
$$

Solve the equation $f^{\prime}(x)=0$, i.e., $x^{-2}\left(x^{-\frac{3}{2}}-1\right)$ $=0$, to get $x=1$.

Solve the inequality $f^{\prime}(x)>0$, i.e., $x^{-2}\left(x^{-\frac{3}{2}}-1\right)$ $>0$, to get $x>1$;
Solve the inequality $f^{\prime}(x)<0$, i.e., $x^{-2}\left(x^{-\frac{3}{2}}-1\right)$ $<0$, to get $0<x<1$;
Thus, $f(x)$ has a minimum value $f(1)=0$ within the interval $(0, +\infty)$. When $x>0$, $f(x) \geqslant f(1)=0$, so the solution to the inequality $f(x) \geqslant 0$ is $x>0$, and thus the solution to the original inequality is $x>0$.