9. (16 points) Given a moving point $P$ on the parabola $y^{2}=4 x$, and the focus $F(1,0)$. Find the maximum value of the inradius $r$ of $\triangle O P F$.

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By symmetry, without loss of generality, let
$$
P\left(\frac{y^{2}}{4}, y\right)(y>0) \text {. }
$$

It is easy to see that $S_{\triangle O P F}=\frac{y}{2}$, and
$$
O F=1, P F=\frac{y^{2}}{4}+1, O P=\frac{y \sqrt{y^{2}+16}}{4} \text {. }
$$

Thus, $\frac{1}{r}=\frac{O F+P F+O P}{2 S_{\triangle O P F}}$
$$
=\frac{2}{y}+\frac{y}{4}+\frac{\sqrt{y^{2}+16}}{4}=f(y) \text {. }
$$

Then $f^{\prime}(y)=\frac{1}{4}\left(1-\frac{8}{y^{2}}+\frac{1}{\sqrt{1+2 \times \frac{8}{y^{2}}}}\right)$
$$
=\frac{1}{4}\left(1-t+\frac{1}{\sqrt{1+2 t}}\right)\left(t=\frac{8}{y^{2}}>0\right) \text {. }
$$

By solving $1-t+\frac{1}{\sqrt{1+2 t}}=0$, we get $t=\frac{3}{2}$.

At this point, $y=\frac{4 \sqrt{3}}{3}$.
Thus, $f^{\prime}\left(\frac{4 \sqrt{3}}{3}\right)=0$, and $f^{\prime}(y)$ is monotonically increasing on $(0,+\infty)$.
Therefore, $f(y)$ reaches its minimum value $\frac{3 \sqrt{3}}{2}$ at $y=\frac{4 \sqrt{3}}{3}$.
Hence, $r_{\max }=\frac{2 \sqrt{3}}{9}$.