I3 (4-2, Hungary) Find all real numbers $x$ that satisfy the inequality
$$
\sqrt{3-x}-\sqrt{x+1}>\frac{1}{2}
$$
Solve: First, by
$$
\left\{\begin{array}{l}
3-x \geqslant 0 \\
x+1 \geqslant 0
\end{array}\right.
$$

we get $-1 \leqslant x \leqslant 3$. Since in the interval $-1 \leqslant x \leqslant 3$,
$$
\sqrt{3-x}-\sqrt{x+1}
$$

decreases as $x$ increases, solving the equation
$$
\begin{array}{c}
\sqrt{3-x}-\sqrt{x+1}=\frac{1}{2}, \\
\frac{7}{4}-2 x=\sqrt{x+1}, \\
64 x^{2}-128 x+33=0, \\
x=\frac{8 \pm \sqrt{31}}{8} .
\end{array}
$$

Upon verification, $x=\frac{8-\sqrt{31}}{8}$ is a solution to (1), meaning that when $x=\frac{8-\sqrt{31}}{8}$, the value of $\sqrt{3-x}-\sqrt{x+1}$ is $\frac{1}{2}$.
Therefore, the solution to the inequality is $-1 \leqslant x<\frac{8-\sqrt{31}}{8}$.