📝 题目
例 4 求证: 当 $x > 0$ 时 $\frac{1 + {x}^{2} + {x}^{4} + \cdots + {x}^{2n}}{x + {x}^{3} + {x}^{5} + \cdots + {x}^{{2n} - 1}} \geq \frac{n + 1}{n}$ ,且等号当且仅当 $x = 1$ 时成立.
💡 答案与解析
证 设
$$ f\left( x\right) = \frac{1 + {x}^{2} + {x}^{4} + \cdots + {x}^{2n}}{x + {x}^{3} + {x}^{5} + \cdots + {x}^{{2n} - 1}} - \frac{n + 1}{n} $$
$$ = \frac{1 - {x}^{{2n} + 2}}{x\left( {1 - {x}^{2n}}\right) } - \frac{n + 1}{n} $$
$$ = \frac{n\left( {1 - {x}^{{2n} + 2}}\right) - \left( {n + 1}\right) \left( {x - {x}^{{2n} + 1}}\right) }{{nx}\left( {1 - {x}^{2n}}\right) }. $$
令其分子为 $g\left( x\right) = n\left( {1 - {x}^{{2n} + 2}}\right) - \left( {n + 1}\right) \left( {x - {x}^{{2n} + 1}}\right)$ ,则
$$ {g}^{\prime }\left( x\right) = - \left( {n + 1}\right) \left\lbrack {{2n}{x}^{2n}\left( {x - 1}\right) - {x}^{2n} + 1}\right\rbrack , $$
$$ {g}^{\prime \prime }\left( x\right) = - \left( {n + 1}\right) {x}^{{2n} - 1}\left( {{2n} + 4{n}^{2}}\right) \left( {x - 1}\right) \left\{ \begin{array}{ll} > 0 & (0 < x < \\ = 0 & \left( {x = 1}\right) , \\ < 0 & \left( {x > 1}\right) . \end{array}\right. $$
点 $x = 1$ 是函数 ${g}^{\prime }\left( x\right)$ 在 $\left( {0, + \infty }\right)$ 内的惟一极值点,并且是极大值点,从而在点 $x = 1$ 处达到函数 ${g}^{\prime }\left( x\right)$ 在 $\left( {0, + \infty }\right)$ 内的最大值. 故有
$$ {g}^{\prime }\left( x\right) < 0\left( {x \neq 1}\right) ,{g}^{\prime }\left( 1\right) = 0 \Rightarrow g\left( x\right) \left\{ \begin{array}{ll} > 0 & (0 < x < \\ = 0 & \left( {x = 1}\right) , \\ < 0 & \left( {x > 1}\right) . \end{array}\right. $$
由此推出 $f\left( x\right) = \frac{g\left( x\right) }{{nx}\left( {1 - {x}^{2n}}\right) } > 0\left( {x \neq 1}\right)$ . 即当 $x \neq 1$ 时,要证的不等式成立. 而当 $x = 1$ 时,不等式左边显然等于右边. 故等号当且仅当 $x = 1$ 时成立.