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Last edited by hbghlyj 2025-5-3 21:45解答(成都-zhcosin, 2017-06-09)
取$BC$边中点为$D$，则$\vv{PB}+\vv{PC}=2\vv{PD}$，于是原式即为$2\vv{PA}\cdot \vv{PD}=PA^2+PD^2-AD^2 \geqslant \frac{1}{2}(PA+PD)^2 - AD^2 \geqslant \frac{1}{2}AD^2 - AD^2 = - \frac{1}{2} AD^2 = -\frac{3}{2}$.
解答(成都-zhcosin, 2017-06-09)
把所有向量都用$\vv{AB}$和$\vv{AC}$来表示，设$\vv{PA}=x\vv{AB}+y\vv{AC}$，则$\vv{PB}=\vv{PA}+\vv{AB}=(1+x)\vv{AB}+y\vv{AC}$，$\vv{PC}=\vv{PA}+\vv{AC}=x\vv{AB}+(1+y)\vv{AC}$，于是
\begin{eqnarray*}
\vv{PA} \cdot (\vv{PB}+\vv{PC}) & = & (x\vv{AB}+y\vv{AC}) \cdot ((1+2x)\vv{AB}+(1+2y)\vv{AC}) \\
& = & x(1+2x)\vv{AB}^2+y(1+2y)\vv{AC}^2+(x+y+4xy)(\vv{AB} \cdot \vv{AC}) \\
& = & 4x(1+2x)+4y(1+2y)+2(x+y+4xy) \\
& = & 8x^2+8y^2+8xy+6x+6y
\end{eqnarray*}
为求此式最小值，作代换
\[ x=u+v, \ y=u-v \]
于是上式成为
\begin{eqnarray*}
& & 8(u+v)^2+8(u-v)^2+8(u+v)(u-v)+12u \\
& = & 24u^2+12u+8v^2 \\
& \geqslant & 24u^2+12u \\
& = & 24 \left( u+\frac{1}{4} \right)^2-\frac{3}{2} \geqslant -\frac{3}{2}
\end{eqnarray*} |
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zhcosin
Posted 2017-6-10 11:33
色k
Posted 2017-6-10 23:38
把几何解答的第二行代码丢上来呀，文字版的才能长期留存。。。
