或者换个写法,其实也没什么本质区别:
不妨设 $\bm e_1=(\cos30\du,\sin30\du)$, $\bm e_2=(\cos30\du,-\sin30\du)$, $\bm c=(x,y)$,则依题意有
\begin{align*}
1\geqslant\abs{\bm c\cdot\bm e_1}+\abs{\bm c\cdot\bm e_2}\geqslant\abs{\bm c\cdot\bm e_1+\bm c\cdot\bm e_2}=2\abs x\cos30\du&\riff\abs x\leqslant\frac1{\sqrt3},\\
1\geqslant\abs{\bm c\cdot\bm e_1}+\abs{\bm c\cdot\bm e_2}\geqslant\abs{\bm c\cdot\bm e_1-\bm c\cdot\bm e_2}=2\abs y\sin30\du&\riff\abs y\leqslant1,
\end{align*}
所以
\[
\abs{\bm c}=\sqrt{x^2+y^2}\leqslant\sqrt{\frac13+1}=\frac2{\sqrt3}.
\] |