15. 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 the domain $[\alpha, \beta]$.
( I ) Find $g(t)=\max f(x)-\min f(x)$;
( II ) Prove: 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 $\frac{1}{g\left(\tan u_{1}\right)}+$ $\frac{1}{g\left(\tan u_{2}\right)}+\frac{1}{g\left(\tan u_{3}\right)}<\frac{3}{4} \sqrt{6}$.
$$
\frac{8 \sqrt{t^{2}+1}\left(2 t^{2}+5\right)}{16 t^{2}+25}
$$
15. 【Analysis and Solution】( I ) Let $\alpha \leqslant x_{1}t\left(x_{1}+x_{2}\right)-2 x_{1} x_{2}+\frac{1}{2}>0 \therefore f\left(x_{2}\right)-f\left(x_{1}\right)>0 \text {, }
$$

Therefore, $f(x)$ is an increasing function on the interval $[\alpha, \beta]$.
$$
\begin{aligned}
\because \alpha+\beta & =t, \alpha \beta=-\frac{1}{4}, \\
\therefore g(t) & =\max f(x)-\min f(x)=f(\beta)-f(\alpha) \\
& =\frac{(\beta-\alpha)[t(\alpha+\beta)-2 \alpha \beta+2]}{\alpha^{2} \beta^{2}+\alpha^{2}+\beta^{2}+1} \\
& =\frac{\sqrt{t^{2}+1}\left(t^{2}+\frac{5}{2}\right)}{t^{2}+\frac{25}{16}}=\frac{8 \sqrt{t^{2}+1}\left(2 t^{2}+5\right)}{16 t^{2}+25} .
\end{aligned}
$$
( II ) Proof:
$$
\begin{array}{l}
g\left(\tan u_{i}\right)=\frac{\frac{8}{\cos u_{i}}\left(\frac{2}{\cos ^{2} u_{i}}+3\right)}{\frac{16}{\cos ^{2} u_{i}}+9}=\frac{\frac{16}{\cos u_{i}}+24 \cos u_{i}}{16+9 \cos ^{2} u_{i}} \geqslant \frac{2 \sqrt{16 \times 24}}{16+9 \cos ^{2} u_{i}} \\
=\frac{16 \sqrt{6}}{16+9 \cos ^{2} u_{i}}(i=1,2,3) . \\
\therefore \sum_{i=1}^{3} \frac{1}{g\left(\tan u_{i}\right)} \leqslant \frac{1}{16 \sqrt{6}} \sum_{i=1}^{3}\left(16+9 \cos ^{2} u_{i}\right)=\frac{1}{16 \sqrt{6}}(16 \times 3+9 \times 3-9) \sum_{i=1}^{3} \sin ^{2} u_{i} \\
\because \sum_{i=1}^{3} \sin u_{i}=1, \text { and } u_{i} \in\left(0, \frac{\pi}{2}\right), i=1,2,3 \therefore 3 \sum_{i=1}^{3} \sin ^{2} u_{i} \geqslant\left(\sum_{i=1}^{3} \sin u_{i}\right) 2=1 \text {, and the equality in the mean value inequality and Cauchy-Schwarz inequality cannot hold simultaneously, }
\end{array}
$$

$$
\therefore \frac{1}{g\left(\tan u_{i}\right)}+\frac{1}{g\left(\tan u_{2}\right)}+\frac{1}{g\left(\tan u_{3}\right)}<\frac{1}{16 \sqrt{6}}\left(75-9 \times \frac{1}{3}\right)=\frac{3}{4} \sqrt{6} \ldots . .20 \text { points }
$$