在三棱锥 $D-ABC$ 中，已知 $AB=2$，$\overrightarrow{AC}\cdot\overrightarrow{BD}=-3$．设 $AD=a$，$BC=b$，$CD=c$，则 $\dfrac{c^2}{ab+1}$ 的最小值为______．
正确答案是$2$．
分析与解 思路一 根据题意，有\[\begin{split} \overrightarrow{AB}&=\overrightarrow{AD}+\overrightarrow{DB}=\overrightarrow{AD}-\overrightarrow{BC}-\overrightarrow{CD}\\\overrightarrow{AC}&=\overrightarrow{AD}-\overrightarrow{CD}\\\overrightarrow{BD}&=\overrightarrow {BC}+\overrightarrow{CD},\end{split}\]记 $\overrightarrow{AD},\overrightarrow{BC},\overrightarrow{CD}$ 分别为 ${\bf x},{\bf y},{\bf z}$，则\[\begin{split} \left({\bf x}-{\bf y}-{\bf z}\right)^2&=4,\\\left({\bf x}-{\bf z}\right)\cdot \left({\bf y}+{\bf z}\right)&=-3,\end{split}\]即\[\begin{split} {\bf x}^2+{\bf y}^2+{\bf z}^2-2{\bf x}\cdot {\bf y}-2{\bf x}\cdot {\bf z}+2{\bf y}\cdot {\bf z}&=4,\\{\bf x}\cdot {\bf y}-{\bf y}\cdot {\bf z}+{\bf x}\cdot {\bf z}-{\bf z}^2&=-3,\end{split}\]因此\[{\bf x}^2+{\bf y}^2-{\bf z}^2=-2,\]从而\[\dfrac{c^2}{ab+1}=\dfrac{{\bf z}^2}{|{\bf x}|\cdot |{\bf y}|+1}=\dfrac{|{\bf x}|^2+|{\bf y}|^2+2}{|{\bf x}|\cdot |{\bf y}|+1}\geqslant 2,\]等号当且仅当 $|{\bf x}|=|{\bf y}|$，即 $a=b$ 时取得．因此所求的最小值为 $2$．
思路二 分别记 $\overrightarrow {DA},\overrightarrow{DB},\overrightarrow{DC}$ 为 ${\bf x},{\bf y},{\bf z}$，则根据题意，有\[\begin{split} {\bf x}^2-2{\bf x}\cdot {\bf y}+{\bf y}^2&=4,\\ {\bf y}\cdot {\bf z}-{\bf x}\cdot {\bf y}&=3,\end{split}\]所求代数式\[\dfrac{c^2}{ab+1}=\dfrac{{\bf z}^2}{|{\bf x}|\cdot |{\bf y}-{\bf z}|+1}.\]注意到\[\left({\bf x}^2-2{\bf x}\cdot {\bf y}+{\bf y}^2\right)-2\left({\bf y}\cdot {\bf z}-{\bf x}\cdot {\bf y}\right)={\bf x}^2-2{\bf y}\cdot {\bf z}+{\bf y}^2=-2,\]于是\[\begin{split} \dfrac{c^2}{ab+1}&=\dfrac{{\bf z}^2+{\bf x}^2-2{\bf y}\cdot{\bf z}+{\bf y}^2+2}{|{\bf x}|\cdot |{\bf y}-{\bf z}|+1}\\&=\dfrac{|{\bf x}|^2+|{\bf y}-{\bf z}|^2+2}{|{\bf x}|\cdot |{\bf y}-{\bf z}|+1}\\&\geqslant 2,\end{split}\]等号当 $|{\bf x}|=|{\bf y}-{\bf z}|$ 时，也即 $AD=BC$ 时取得．因此所求的最小值为 $2$．
思路三 分别记 $\overrightarrow {AB},\overrightarrow{AC},\overrightarrow{BD}$ 为 ${\bf x},{\bf y},{\bf z}$，则\[{\bf x}^2=4,{\bf y}\cdot {\bf z}=-3,\]而\[\begin{split}\dfrac{c^2}{ab+1}&=\dfrac{\left({\bf x}-{\bf y}+{\bf z}\right)^2}{\left|{\bf x}+{\bf z}\right|\cdot \left|{\bf y}-{\bf x}\right|+1}\\&=\dfrac{\left({\bf x}+{\bf z}\right)^2-2{\bf y}\cdot \left({\bf x}+{\bf z}\right)+{\bf y}^2}{\left|{\bf x}+{\bf z}\right|\cdot \left|{\bf y}-{\bf x}\right|+1}\\&=\dfrac{\left({\bf x}+{\bf z}\right)^2+\left({\bf x}-{\bf y}\right)^2-{\bf x}^2-2{\bf y}\cdot {\bf z}}{\left|{\bf x}+{\bf z}\right|\cdot \left|{\bf y}-{\bf x}\right|+1}\\&=\dfrac{\left({\bf x}+{\bf z}\right)^2+\left({\bf x}-{\bf y}\right)^2+2}{\left|{\bf x}+{\bf z}\right|\cdot \left|{\bf y}-{\bf x}\right|+1}\\&\geqslant 2,\end{split} \]等号当 $\left|{\bf y}-{\bf x}\right|=\left|{\bf x}+{\bf z}\right|$ 时，即 $a=b$ 时取得．因此所求的最小值为 $2$．