1608

若 ${\displaystyle \mathrm{\exists }\phi \in \mathbb{R}}$, 对 ${\displaystyle \mathrm{\forall }n\in \mathbb{Z}}$, 都有 ${\displaystyle \cos (\frac{2n\pi }{3}+\phi )\le m}$ 成立, 求实数 ${\displaystyle m}$ 的最小值.

记 $$\begin{array}{rl}F(\phi )& =max\{\cos \phi ,\cos (\phi +\frac{2\pi }{3}),\cos (\phi +\frac{4\pi }{3})\}\\ & =max\{\cos \phi ,-\frac{1}{2}\cos \phi +\frac{\sqrt{3}}{2}|\sin \phi |\}.\end{array}$$

• 当 ${\displaystyle \cos \phi \ge \frac{1}{2}}$,则 ${\displaystyle F(\phi )\ge \cos \phi \ge \frac{1}{2}}$.
• 当 ${\displaystyle \cos \phi \le \frac{1}{2}}$, 我们证明 ${\displaystyle -\frac{1}{2}\cos \phi +\frac{\sqrt{3}}{2}|\sin \phi |>\frac{1}{2}}$. 记 ${\displaystyle t=\cos \phi \in [-1,1]}$, 则 ${\displaystyle |\sin \phi |=\sqrt{1-t^2}}$, 此时变为证明 $$\begin{array}{rl}& -\frac{1}{2}t+\frac{\sqrt{3}}{2}\sqrt{1-t^2}\ge \frac{1}{2}\\ \Leftarrow & \sqrt{3(1-t^2)}\ge 1+t⟺3(1-t)\ge 1+t⟺t\le \frac{1}{2},\end{array}$$ 所以 ${\displaystyle F(\phi )>\frac{1}{2}}$.

从上面的讨论看出, ${\displaystyle F(\phi {)}_{min}=\frac{1}{2}}$,所以 ${\displaystyle m}$ 的最小值为 ${\displaystyle \frac{1}{2}}$.