9. (4th IMO problem) Solve the inequality $\sqrt{3-x}-\sqrt{x+1}>\frac{1}{2}$.


9. Given $(\sqrt{3-x})^{2}+(\sqrt{x+1})^{2}=4$, we can set $\sqrt{3-x}=2 \sin \theta, \sqrt{x+1}=2 \cos \theta, \theta \in\left[0, \frac{\pi}{2}\right]$, the original inequality can be transformed into: $2 \sin \theta-2 \cos \theta>\frac{1}{2}$, which is $2 \sin \theta>2 \cos \theta+\frac{1}{2}$. (*) Since $\theta \in\left[0, \frac{\pi}{2}\right]$, then $2 \cos \theta+\frac{1}{2}>0$. Squaring both sides of (*) and rearranging, we get $32 \cos ^{2} \theta+8 \cos \theta-15<0$, solving this yields $0 \leqslant \cos \theta<\frac{1}{8}(\sqrt{31}-1)$, thus $x=4 \cos ^{2} \theta-1<1-\frac{\sqrt{31}}{8}$ and $x \geqslant-1$, hence the solution set of the original inequality is $\left\{x \left\lvert\,-1 \leqslant x<1-\frac{\sqrt{31}}{8}\right.\right\}$.