991

已知 ${\displaystyle \mathrm{\Delta }ABC}$ 的内角满足 ${\displaystyle \sin A+2\sin B=2\sin C}$. 求 ${\displaystyle {\displaystyle \frac{1}{\sin A}+\frac{1}{\sin B}-\frac{1}{\sin C}}}$ 的最小值.

利用和差化积公式，条件等价于$$\begin{array}{r}\sin A=2(\sin C-\sin B)=4\cos \frac{C+B}{2}\sin \frac{C-B}{2}\Rightarrow \cos \frac{A}{2}=2\sin \frac{C-B}{2}.\end{array}$$
记 ${\displaystyle t={\displaystyle \tan \frac{A}{2}}}$, 则$$\begin{array}{rl}\text{原式}=& \frac{1}{\sin A}+\frac{\sin A}{2\sin B\sin C}=\frac{1}{\sin A}+\frac{\sin A}{\cos (B-C)-\cos (B+C)}\\ =& \frac{1}{\sin A}+\frac{\sin A}{(1-2{\sin }^2{\displaystyle \frac{B-C}{2}})+(2{\cos }^2{\displaystyle \frac{A}{2}}-1)}=\frac{1}{\sin A}+\frac{\sin A}{{\displaystyle \frac{3}{2}}{\cos }^2{\displaystyle \frac{A}{2}}}\\ =& \frac{1+t^2}{2t}+\frac{4t}{3}=\frac{1}{2t}+\frac{11t}{6}\ge \frac{\sqrt{33}}{3}.\end{array}$$