$\square$ Example 3 Given that $\alpha, \beta$ are two distinct real roots of the equation $4 x^{2}-4 t x-1=0(t \in \mathbf{R})$, and the domain of the function $f(x)-\frac{2 x-t}{x^{2}+1}$ is $[\alpha, \beta]$.
(1) Find $g(t)=\max f(x)-\min f(x)$;
(2) Prove that for $u_{i} \in\left(0, \frac{\pi}{2}\right)(i=1,2,3)$, if $\sin u_{1}+\sin u_{2}+$ $\sin u_{3}=1$, then $\frac{1}{g\left(\tan u_{1}\right)}+\frac{1}{g\left(\tan u_{2}\right)}+\frac{1}{g\left(\tan u_{3}\right)}<\frac{3}{4} \sqrt{6}$. (2004 National High School Mathematics Competition Problem)
(1) Let $\alpha \leqslant x_{1}t\left(x_{1}+x_{2}\right)-2 x_{1} x_{2}+\frac{1}{2}>0$, so $f\left(x_{2}\right)-f\left(x_{1}\right)>0$, hence $f(x)$ is an increasing function on the interval $[\alpha, \beta]$.

Since $\alpha+\beta=t, \alpha \beta=-\frac{1}{4}, \beta-\alpha=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}=\sqrt{t^{2}+1}$, we have
$$\begin{aligned}
g(t) & =\max f(x)-\min f(x)=f(\beta)-f(\alpha)=\frac{2 \beta-t}{\beta^{2}+1}-\frac{2 \alpha-t}{a^{2}+1} \\
& =\frac{(2 \beta-t)\left(\alpha^{2}+1\right)-(2 \alpha-t)\left(\beta^{2}+1\right)}{\left(a^{2}+1\right)\left(\beta^{2}+1\right)} \\
& =\frac{(\beta-\alpha)[t(\alpha+\beta)-2 \alpha \beta+2]}{\alpha^{2} \beta^{2}+\alpha^{2}+\beta^{2}+1} \\
& =\frac{\sqrt{t^{2}+1}\left(t^{2}+\frac{5}{2}\right)}{t^{2}+\frac{25}{16}}=\frac{8 \sqrt{t^{2}+1}\left(2 t^{2}+5\right)}{16 t^{2}+25}
\end{aligned}$$
(2) $\begin{aligned} \text { Since } g\left(\tan u_{i}\right) & =\frac{\frac{8}{\cos u_{i}}\left(\frac{2}{\cos ^{2} u_{i}}+3\right)}{\frac{16}{\cos ^{2} u_{i}}+9}= \\ & \geqslant \frac{16 \sqrt{6}}{16+9 \cos ^{2} u_{i}}(i=1,2 \\ \text { Therefore } \frac{1}{g\left(\tan u_{1}\right)}+ & \frac{1}{g\left(\tan u_{2}\right)}+\frac{1}{g\left(\tan u_{3}\right)}\end{aligned}$
$$\begin{array}{l}
\leqslant \frac{16+9 \cos ^{2} u_{1}}{16 \sqrt{6}}+\frac{16+9 \cos ^{2} u_{2}}{16 \sqrt{6}}+\frac{16+9 \cos ^{2} u_{3}}{16 \sqrt{6}} \\
\leqslant \frac{48+9\left(\cos ^{2} u_{1}+\cos ^{2} u_{2}+\cos ^{2} u_{3}\right)}{16 \sqrt{6}} \\
=\frac{48+9 \times 3-9\left(\sin ^{2} u_{1}+\sin ^{2} u_{2}+\sin ^{2} u_{3}\right)}{16 \sqrt{6}}
\end{array}$$

Since $\sin u_{1}+\sin u_{2}+\sin u_{3}=1$, and $u_{i} \in\left(0, \frac{\pi}{2}\right)(i=1,2,3)$, we have
$$3\left(\sin ^{2} u_{1}+\sin ^{2} u_{2}+\sin ^{2} u_{3}\right) \geqslant\left(\sin u_{1}+\sin u_{2}+\sin u_{3}\right)^{2}=1$$

In the inequalities of the mean and Cauchy, the equality cannot hold simultaneously, so
$$\frac{1}{g\left(\tan u_{1}\right)}+\frac{1}{g\left(\tan u_{2}\right)}+\frac{1}{g\left(\tan u_{3}\right)}<\frac{75-9 \times \frac{1}{3}}{16 \sqrt{6}}=\frac{3}{4} \sqrt{6}$$