Three, (20 points) Find the maximum value of the function $f(x)=\frac{\sqrt{x}}{x^{2}+1}$.


Three, take $k>0$ ( $k$ is a parameter to be determined), then
$$
\begin{array}{l} 
f(x)=\frac{\sqrt{x}}{x^{2}+1}=\frac{2 \sqrt{k x}}{2 \sqrt{k}\left(x^{2}+1\right)} \\
\leqslant \frac{x+k}{2 \sqrt{k}\left(x^{2}+1\right)} . \\
\text { Let } y=\frac{x+k}{x^{2}+1} . \text { Then } \\
y x^{2}-x+y-k=0 .
\end{array}
$$

Since the domain of the function $f(x)$ is $[0,+\infty)$, equation (2) has a solution in $[0,+\infty)$.
Thus, $\Delta=1-4 y(y-k) \geqslant 0$.
Hence $y \leqslant \frac{k+\sqrt{k^{2}+1}}{2}$.
When $y=\frac{k+\sqrt{k^{2}+1}}{2}$, $\Delta=0$, at this time, from equation (2) we get
$$
x=\frac{1 \pm \sqrt{0}}{2 y}=\frac{1}{2 y}=\frac{1}{\sqrt{k^{2}+1}+k}=\sqrt{k^{2}+1}-k .
$$

Inequality (1) holds with equality when $x=k$, and inequality (3) holds with equality when $x=\sqrt{k^{2}+1}-k$.

To make the equalities in inequalities (1) and (3) hold simultaneously, we should have $k=\sqrt{k^{2}+1}-k$, solving this gives $k=\frac{\sqrt{3}}{3}$. Therefore,
$$
\begin{array}{l}
f(x) \leqslant \frac{x+k}{2 \sqrt{k}\left(x^{2}+1\right)}=\frac{y}{2 \sqrt{k}} \\
\leqslant \frac{1}{2 \sqrt{k}} \cdot \frac{k+\sqrt{k^{2}+1}}{2} \\
=\frac{\sqrt{3}}{2} \cdot \frac{\frac{\sqrt{3}}{3}+\sqrt{\frac{1}{3}+1}}{2} \\
=\frac{\sqrt{3}}{2} \cdot \frac{\frac{\sqrt{3}}{3}+\frac{2 \sqrt{3}}{3}}{2}=\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt[4]{27}}{4} \text {, } \\
\end{array}
$$

equality holds when $x=\frac{\sqrt{3}}{3}$.
Therefore, the maximum value of $f(x)=\frac{\sqrt{x}}{x^{2}+1}$ is $\frac{\sqrt[4]{27}}{4}$.