📝 题目
例 4 求证:
$$ \pi < \frac{\sin {\pi t}}{t\left( {1 - t}\right) } \leq 4\;\left( {\forall t \in \left( {0,1}\right) }\right) . \tag{3.5} $$
分析 作变量代换 $t = x + \frac{1}{2}\left( {\left| x\right| < \frac{1}{2}}\right)$ ,则
$$ \frac{\sin {\pi t}}{t\left( {1 - t}\right) } = \frac{\cos {\pi x}}{\frac{1}{4} - {x}^{2}}\;\left( {\left| x\right| < \frac{1}{2}}\right) . $$
再注意到 $\frac{\cos {\pi x}}{\frac{1}{4} - {x}^{2}}\left( {\left| x\right| < \frac{1}{2}}\right)$ 是偶函数,要证不等式 (3.5),只要证
$$ \pi < \frac{\cos {\pi x}}{\frac{1}{4} - {x}^{2}} \leq 4\;\left( {\forall x \in \left( {0,\frac{1}{2}}\right) }\right) . \tag{3.6} $$
💡 答案与解析
证 记 $f\left( x\right) = \cos {\pi x},g\left( x\right) = \frac{1}{4} - {x}^{2}$ ,
$$ h\left( x\right) = \left\{ \begin{array}{ll} \frac{f\left( x\right) }{g\left( x\right) }, & x \in \left( {0,\frac{1}{2}}\right) , \\ \pi , & x = \frac{1}{2}. \end{array}\right. $$
因为
$$ {\left\lbrack \frac{{f}^{\prime }\left( x\right) }{{g}^{\prime }\left( x\right) }\right\rbrack }^{\prime } = \frac{\pi \cos {\pi x}}{2{x}^{2}}\left( {{\pi x} - \tan {\pi x}}\right) < 0\;\left( {x \in \left( {0,\frac{1}{2}}\right) }\right) , $$
所以 $\frac{{f}^{\prime }\left( x\right) }{{g}^{\prime }\left( x\right) }$ 在 $\left( {0,\frac{1}{2}}\right)$ 上严格单调下降. 于是,根据