1054

在 ${\displaystyle \mathrm{\Delta }ABC}$ 中，求 ${\displaystyle 2\sin A+\sin B+\sin C}$ 的最大值的取等条件.

注意到$$\begin{array}{rl}2\sin A+\sin B+\sin C=& 2\sin A+2\sin \frac{B+C}{2}\cos \frac{B-C}{2}\le 2\sin A+2\cos \frac{A}{2}\\ =& \frac{2}{\lambda }\lambda \cos \frac{A}{2}(1+2\sin \frac{A}{2})\le \frac{2}{\lambda }\cdot \frac{1}{4}{(1+2\sin \frac{A}{2}+\lambda \cos \frac{A}{2})}^2\\ \le & \frac{1}{2\lambda }(1+\sqrt{{\lambda }^2+4}{)}^2.\end{array}$$
这里用了基本不等式和辅助角公式放缩，取等条件为$$\begin{array}{r}\{\begin{array}{rl}& \lambda \cos \frac{A}{2}=1+2\sin \frac{A}{2},\\ & \sin \frac{A}{2}=\frac{2}{\sqrt{{\lambda }^2+4}},\\ & \cos \frac{A}{2}=\frac{\lambda }{\sqrt{{\lambda }^2+4}}\end{array}⟺\lambda \cdot \frac{\lambda }{\sqrt{{\lambda }^2+4}}=2\cdot \frac{2}{\sqrt{{\lambda }^2+4}}+1⟺{\lambda }^2=4+\sqrt{{\lambda }^2+4},\end{array}$$
因此解得 ${\displaystyle {\lambda }^2={\displaystyle \frac{9+\sqrt{33}}{2}}}$ (依据 ${\displaystyle \lambda >6}$ 舍去增根.) 因此取等条件是 ${\displaystyle {\displaystyle A=2\arcsin \frac{\sqrt{33}-1}{8}}}$, ${\displaystyle B=C}$.