4. 1.1 * Find all real numbers $x$ that satisfy the inequality $\sqrt{3-x}-\sqrt{x+1}>\frac{1}{2}$.


First, from $\left\{\begin{array}{l}3-x \geqslant 0, \\ x+1 \geqslant 0,\end{array}\right.$ we get $-1 \leqslant x \leqslant 3$.
When $-1 \leqslant x \leqslant 3$, the value of $\sqrt{3-x}-\sqrt{x+1}$ decreases as $x$ increases. Solve 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 the 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 $-1 \leqslant x<\frac{8-\sqrt{31}}{8}$.