MT【292】任意存在求最值

已知向量$\textbf{a},\textbf{b}$满足:$|\textbf{a}|=|\textbf{b}|=1,\textbf{a}\cdot\textbf{b}=\dfrac{1}{2},\textbf{c}=(m,1-m),\textbf{d}=(n,1-n),(m,n\in R)$,
存在$\textbf{a},\textbf{b}$,对于任意的实数$m,n$,不等式$|\textbf{a}-\textbf{c}|+|\textbf{b}-\textbf{d}|\ge T$ 恒成立,则$T$的取值范围_____

解析:设$\textbf{a}=\overrightarrow{OA},\textbf{b}=\overrightarrow{OB},\textbf{c}=\overrightarrow{OC},\textbf{d}=\overrightarrow{OD}$,
由题意$A,B$在单位圆上运动,且$\angle AOB=\dfrac{\pi}{3}$.$C,D$在$x+y-1=0$上运动.
由题意只需$T\le \max\limits_{A,B}\{\min\limits_{C,D}\{|AC|+|BD|\}\}$.
方法一:几何意义,先固定$A,B$,则$AC\perp CD,BD\perp CD$时取得$\min\limits_{C,D}(|AC|+|BD|)$,
再取$A,B$中点$M$;$C,D$ 中点$N$,故$\min\limits_{C,D}(|AC|+|BD|)=2|MN|$,作$OE\perp CD$,连接$OM$,则当$A,B$动起来时$MN\le OM+OE=\dfrac{\sqrt{3}+\sqrt{2}}{2}$,故$T\le \sqrt{3}+\sqrt{2} $
方法二:设$A(cos \theta,sin\theta),B(\cos(\theta+\dfrac{\pi}{3}),sin(\theta+\dfrac{\pi}{3})),$
设$d=\dfrac{|cos\theta+sin\theta-1|}{\sqrt{2}}+\dfrac{|cos(\theta+\dfrac{\pi}{3}+\sin(\theta+\dfrac{\pi}{3})-1|}{\sqrt{2}}$
题目转换为存在$\theta$对$T\le d $成立,即$T\le d_{max}$
$$\dfrac{|cos\theta+sin\theta-1|}{\sqrt{2}}+\dfrac{|cos(\theta+\dfrac{\pi}{3}）+\sin(\theta+\dfrac{\pi}{3})-1|}{\sqrt{2}}=\dfrac{1}{\sqrt{2}}*\max$$

\begin{equation*}
\{|cos\theta+sin\theta+cos(\theta+\frac{\pi}{3})+\sin(\theta+\frac{\pi}{3})-2|,|cos\theta+sin\theta-cos(\theta+\frac{\pi}{3})-\sin(\theta+\frac{\pi}{3})| \}
\end{equation*}

$$=\dfrac{\max\{|\sqrt{6}sin(\theta+\phi_1)-2|,|\sqrt{2}sin(\theta+\phi_2)|\}\le \sqrt{3}+\sqrt{2}}{\sqrt{2}}$$
即$T\le\sqrt{3}+\sqrt{2}$