【山东青岛高三第一次适应性检测】记 \(\triangle ABC\) 内角 \(A,B,C\) 的对边分别为 \(a,b,c\),\(\dfrac{c \cos A}{a \cos C}+\dfrac{2 c \cos B}{b \cos C}=3\),则 \(\dfrac{1}{\tan A}+\dfrac{2}{\tan B}\) 的最小值为______.
解.
\[\dfrac{\sin C \cos A}{\sin A \cos C}+\dfrac{2 \sin C \cos B}{\sin B \cos C}=3
\]
即 \(\dfrac{\tan C}{\tan A}+\dfrac{2 \tan C}{\tan B}=3\),
\[\dfrac{1}{\tan A}+\dfrac{2}{\tan B}=\dfrac{3}{\tan C},
\]
而 \(bc \cos A+2ac \cos B=3ab \cos C\),由余弦定理,
\[c^{2}=\dfrac{1}{3}a^{2}+\dfrac{2}{3}b^{2},
\]
从而有,
\[\cos C=\dfrac{a^{2}+b^{2}-c^{2}}{2ab} \geq \dfrac{\sqrt{2}}{3},
\]
进而得到,
\[\sin C \leq \dfrac{\sqrt{7}}{3}, \tan C \leq \dfrac{\sqrt{7}}{\sqrt{2}}=\dfrac{\sqrt{14}}{2},
\]
所以得,
\[\dfrac{3}{\tan C} \geq \dfrac{3}{\dfrac{\sqrt{14}}{2}}=\dfrac{6\sqrt{14}}{14}=\dfrac{3\sqrt{14}}{7},
\]
即 \(\left(\dfrac{1}{\tan A}+\dfrac{2}{\tan B}\right)_{\min }=\dfrac{3\sqrt{14}}{7}\).□