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在锐角 ${\displaystyle \mathrm{\Delta }ABC}$ 中, 有 ${\displaystyle \overrightarrow{BC}\cdot (2\overrightarrow{BA}-\overrightarrow{BC})=\overrightarrow{AC}\cdot (\overrightarrow{BA}+\overrightarrow{AC})}$, 则 ${\displaystyle \tan A\cdot \tan B\cdot \tan C}$ 的最小值为_____.

条件式等价于 $$\overrightarrow{BC}\cdot (\overrightarrow{BA}+\overrightarrow{CA})=\overrightarrow{AC}\cdot \overrightarrow{BC}⟺\overrightarrow{BC}\cdot (\overrightarrow{BA}+2\overrightarrow{CA})=0.$$ 取 ${\displaystyle BC}$ 边上靠近 ${\displaystyle C}$ 的三等分点 ${\displaystyle D}$, 则这意味着 ${\displaystyle \overrightarrow{BC}\cdot (3\overrightarrow{DA})=0}$, 所以 ${\displaystyle AD\perp BC}$. 因为这是锐角三角形, 所以 ${\displaystyle \tan B=\frac{AD}{BD}=\frac{AD}{2DC}=\frac{1}{2}\tan C}$. 记 ${\displaystyle \tan B=x}$. 则 ${\displaystyle \tan C=2x}$, ${\displaystyle \tan A=-\frac{\tan B+\tan C}{1-\tan B\tan C}=\frac{3x}{2x^2-1}}$, 于是 $$\tan A\tan B\tan C=\frac{6x^3}{2x^2-1}\equiv f(x).$$ 由于 ${\displaystyle f^{\prime }(x)=\frac{6x^2(2x^2-3)}{(2x^2-1{)}^2}}$, 所以 ${\displaystyle f(x)}$ 有极小值 ${\displaystyle f\left(\frac{\sqrt{6}}{2}\right)=\frac{9\sqrt{6}}{4}}$.