Example 7. Given the function $f(x)=a x^{2}+\frac{1}{x}-2 \ln x(x>0)$. If $f(x)$ is monotonically increasing on $[1,+\infty)$, find the range of the real number a;
Solution: Given $f(x)=a x^{2}+\frac{1}{x}-2 \ln x$, we get $f^{\prime}(x)=2 a x-\frac{1}{x^{2}}-\frac{2}{x}$.
Since the function $f(x)$ is a monotonically increasing function on $[1,+\infty)$, we have $f^{\prime}(x) \geq 0$ on $[1,+\infty)$,

which means the inequality $2 a x-\frac{1}{x^{2}}-\frac{2}{x} \geq 0$ holds for all $x$ in $[1,+\infty)$.
This is equivalent to $a \geq \frac{1}{2 x^{3}}+\frac{1}{x^{2}}$ holding for all $x$ in $[1,+\infty)$.
Let $g(x)=\frac{1}{2 x^{3}}+\frac{1}{x^{2}}$, the above problem is equivalent to $a \geq g(x)_{\text {max }}$.
Since $g(x)=\frac{1}{2 x^{3}}+\frac{1}{x^{2}}$ is a decreasing function on $[1,+\infty)$, then $g(x)_{\text {max }}=g(1)$ $=\frac{3}{2}$

Thus, $a \geq \frac{3}{2}$ is the required solution.