1734

在 ${\displaystyle \mathrm{\Delta }ABC}$ 中, 内角 ${\displaystyle A,B,C}$ 的对边分别为 ${\displaystyle a,b,c}$, 若 ${\displaystyle a^2+2b^2=4c^2}$, 则 ${\displaystyle \frac{2}{{\sin }^{2}A}+\frac{1}{{\sin }^{2}B}}$ 的最小值为_____.

由正弦定理条件等价于 $${\sin }^{2}A+2{\sin }^{2}B=4{\sin }^{2}C=4(\sin A\cos B+\sin B\cos A{)}^2,$$ 同除 ${\displaystyle {\sin }^{2}A{\sin }^{2}B}$ 得 $$(1+{\cot }^{2}B)+2(1+{\cot }^{2}A)=4(\cot A+\cot B{)}^2,$$ 整理得 $$\begin{array}{rl}3& =2{\cot }^{2}A+3{\cot }^{2}B+8\cot A\cot B\\ & \le 2{\cot }^{2}A+3{\cot }^{2}B+(8{\cot }^{2}A+2{\cot }^{2}B)\\ & =5(2{\cot }^{2}A+{\cot }^{2}B),\end{array}$$ 所以 $$\text{原式}=3+2{\cot }^{2}A+{\cot }^{2}B\ge 3+\frac{3}{5}=\frac{18}{5}.$$ 取等条件为 $$\{\begin{array}{rl}& 8{\cot }^{2}A=2{\cot }^{2}B,\\ & 2{\cot }^{2}A+{\cot }^{2}B=\frac{3}{5}\end{array}\Rightarrow \{\begin{array}{rl}& \cot A=\frac{\sqrt{10}}{10},\\ & \cot B=\frac{\sqrt{10}}{5}.\end{array}$$