$\square$ Example 16 Given that $\alpha, \beta$ are the two distinct real roots of the equation $4 x^{2}-4 t x-1=0(t \in \mathbf{R})$, and the function $f(x)=\frac{2 x-t}{x^{2}+1}$ has a domain of $[\alpha, \beta]$.
(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 League)
(1) $f^{\prime}(x)=\frac{2\left(x^{2}+1\right)-2 x(2 x-t)}{\left(x^{2}+1\right)^{2}}=\frac{2 x^{2}+2 t x+2}{\left(x^{2}+1\right)^{2}}$.
Given that $\alpha, \beta$ are the two distinct real roots of the equation $4 x^{2}-4 t x-1=0(t \in \mathbf{R})$, we know $4 x^{2}-4 t x-1=4(x-\alpha)(x-\beta)$. Since $x \in[\alpha, \beta]$, we have $4(x-\alpha)(x-\beta) \leqslant 0$, i.e., $-4 x^{2}+4 t x+1=-4(x-\alpha)(x-\beta) \geqslant 0$, so
$$f^{\prime}(x)=\frac{-4(x-\alpha)(x-\beta)+3}{2\left(x^{2}+1\right)^{2}}>0,$$

which means $f(x)$ is an increasing function on the interval $[\alpha, \beta]$.
$$\begin{aligned}
& \text { Given } \alpha+\beta=t, \alpha \beta=-\frac{1}{4}, \beta-\alpha=\sqrt{(\alpha+\beta)^{2}-4 \alpha \beta}=\sqrt{t^{2}+1} \text {, we have } \\
g(t) & =\max f(x)-\min f(x)=f(\beta)-f(\alpha)=\frac{2 \beta-t}{\beta^{2}+1}-\frac{2 \alpha-t}{\alpha^{2}+1} \\
& =\frac{(2 \beta-t)\left(\alpha^{2}+1\right)-(2 \alpha-t)\left(\beta^{2}+1\right)}{\left(\alpha^{2}+1\right)\left(\beta^{2}+1\right)}=\frac{(\beta-\alpha)[t(\alpha+\beta)-2 \alpha \dot{\beta}+2]}{\alpha^{2} \beta^{2}+\alpha^{2}+\beta^{2}+1} \\
& =\frac{\sqrt{t^{2}+1}\left(t^{2}+\frac{5}{2}\right)}{t^{2}+\frac{25}{16}}=\frac{8 \sqrt{t^{2}+1}\left(2 t^{2}+5\right)}{16 t^{2}+25} .
\end{aligned}$$
(2) Given $g(t)=\frac{8 \sqrt{t^{2}+1}\left(2 t^{2}+5\right)}{16 t^{2}+25}$, then $\frac{1}{g(t)}=\frac{16 t^{2}+25}{8 \sqrt{t^{2}+1}\left(2 t^{2}+5\right)}$,

Thus,
$$\begin{aligned}
& \frac{1}{g\left(\tan _{i}\right)}=\frac{16 \tan ^{2} u_{i}+25}{8 \sqrt{\tan ^{2} u_{i}+1}\left(2 \tan ^{2} u_{i}+5\right)} \\
= & \frac{\frac{8}{\cos ^{2} u_{i}}\left(\frac{2}{\cos ^{2} u_{i}}+3\right)}{\frac{16}{\cos ^{2} u_{i}}+9}=\frac{\left(25-9 x_{i}^{2}\right) \sqrt{1-x_{i}^{2}}}{40-24 x_{i}^{2}}\left(x_{i}=\sin u_{i}\right) .
\end{aligned}$$

We will prove the strengthened proposition: 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 the inequality
$$\frac{5}{4}\sum_{i=1}^{3} \frac{5}{8}\left(1-x_{i}\right) \\
= & \frac{5}{8}\left[3-\left(x_{1}+x_{2}+x_{3}\right)\right]=\frac{5}{4},
\end{aligned}$$

holds, which means the left inequality of (2) holds, and the estimate is optimal. Also,
$$\begin{aligned}
& \sum_{i=1}^{3} \frac{\left(25-9 x_{i}^{2}\right) \sqrt{1-x_{i}^{2}}}{40-24 x_{i}^{2}} \leqslant \sum_{i=1}^{3} \frac{\sqrt{2}}{392}\left(179-33 x_{i}\right) \\
= & \frac{\sqrt{2}}{392}\left[179 \times 3-33\left(x_{1}+x_{2}+x_{3}\right)\right]=\frac{9}{7} \sqrt{2},
\end{aligned}$$

which means the right inequality of (2) holds, with equality if and only if $x=\frac{1}{3}$.
In summary, inequality (2) holds, thus inequality (1) holds. Noting that $\frac{9}{7} \sqrt{2}<\frac{3}{4} \sqrt{6}$, the original inequality holds.