China Mathematical Competition (Hainan)

(1) Let $\alpha \le x_1 < x_2 \le \beta$, then
$$4x_1^2 - 4tx_1 - 1 \le 0, \quad 4x_2^2 - 4tx_2 - 1 \le 0.$$
Therefore,
$$4(x_1^2 + x_2^2) - 4(t(x_1 + x_2) - 2) \le 0, \\ 2x_1x_2 - t(x_1 + x_2) - \frac{1}{2} < 0.$$
But
$$f(x_2) - f(x_1) = \frac{2x_2 - t}{x_2^2 + 1} - \frac{2x_1 - t}{x_1^2 + 1}$$
$$= \frac{(x_2 - x_1)[t(x_1 + x_2) - 2x_1x_2 + 2]}{(x_2^2 + 1)(x_1^2 + 1)},$$
and $t(x_1 + x_2) - 2x_1x_2 + 2 > t(x_1 + x_2) - 2x_1x_2 + \frac{1}{2} > 0$, thus
$$f(x_2) - f(x_1) > 0.$$
Consequently, $f(x)$ is an increasing function on the interval $[\alpha, \beta]$.
$$\text{Since } \alpha + \beta = t \text{ and } \alpha\beta = -\frac{1}{4},$$
$$g(t) = \max\{f(x)\} - \min\{f(x)\} = f(\beta) - f(\alpha)$$
$$= \frac{\sqrt{t^2 + 1}\left(t^2 + \frac{5}{2}\right)}{t^2 + \frac{25}{16}} = \frac{8\sqrt{t^2 + 1}(2t^2 + 5)}{16t^2 + 25}.$$

(2)
$$g(\tan u_i) = \frac{\frac{8}{\cos u_i}(\frac{2}{\cos^2 u_i} + 3)}{\frac{16}{\cos^2 u_i} + 9} = \frac{\frac{16}{\cos u_i} + 24\cos u_i}{16 + 9\cos^2 u_i} \\ \ge \frac{2\sqrt{16 \times 24}}{16 + 9\cos^2 u_i} = \frac{16\sqrt{6}}{16 + 9\cos^2 u_i} \quad (i = 1, 2, 3),$$
so
$$\begin{align*} \sum_{i=1}^{3} \frac{1}{g(\tan u_i)} &\le \frac{1}{16\sqrt{6}} \sum_{i=1}^{3} (16 + 9\cos^2 u_i) \\ &= \frac{1}{16\sqrt{6}} (16 \times 3 + 9 \times 3 - 9 \sum_{i=1}^{3} \sin^2 u_i). \end{align*}$$
Since $\sum_{i=1}^{3} \sin u_i = 1$, and $u_i \in (0, \frac{\pi}{2})$, $i = 1, 2, 3$, we obtain
$$3 \sum_{i=1}^{3} \sin^2 u_i > \left( \sum_{i=1}^{3} \sin u_i \right)^2 = 1.$$
$$\begin{gathered} \frac{1}{g(\tan u_1)} + \frac{1}{g(\tan u_2)} + \frac{1}{g(\tan u_3)} \\ < \frac{1}{16\sqrt{6}} (75 - 9 \times \frac{1}{3}) = \frac{3}{4}\sqrt{6}. \end{gathered}$$