📝 题目
例 11 (1) 设 $f\left( x\right)$ 是在 $\mathbf{R}$ 上的凹函数,求证: 对 $\forall {x}_{1},{x}_{2},\cdots$ , ${x}_{n} \in \mathbf{R}$ ,有
$$ \frac{f\left( {x}_{1}\right) + f\left( {x}_{2}\right) + \cdots + f\left( {x}_{n}\right) }{n} \geq f\left( \frac{{x}_{1} + {x}_{2} + \cdots + {x}_{n}}{n}\right) , $$
(6.16)
等号当且仅当 ${x}_{1} = {x}_{2} = \cdots = {x}_{n}$ 时成立.
(2)求证:当 ${x}_{k} > 0\left( {k = 1,2,\cdots }\right)$ 时,有
$$ \frac{{x}_{1} + {x}_{2} + \cdots + {x}_{n}}{n} \geq \sqrt[n]{{x}_{1}{x}_{2}\cdots {x}_{n}}. $$
💡 答案与解析
证(1)记 ${x}_{0} = \frac{{x}_{1} + {x}_{2} + \cdots + {x}_{n}}{n}$ . 因为 $f\left( x\right)$ 是凹函数,所以
$$ f\left( {x}_{i}\right) \geq f\left( {x}_{0}\right) + {f}^{\prime }\left( {x}_{0}\right) \left( {{x}_{i} - {x}_{0}}\right) \;\left( {i = 1,2,\cdots ,n}\right) , $$
此处 ${}^{a} =$ ”当且仅当 ${x}_{i} = {x}_{0}$ 时成立. 对上面 $n$ 个等式求和,即得
$$ \mathop{\sum }\limits_{{i = 1}}^{n}f\left( {x}_{i}\right) \geq {nf}\left( {x}_{0}\right) + {f}^{\prime }\left( {x}_{0}\right) \left( {\mathop{\sum }\limits_{{i = 1}}^{n}{x}_{i} - n{x}_{0}}\right) = {nf}\left( {x}_{0}\right) , $$
即 (6.16)式得证.
(2)对 $f\left( x\right) = - \ln x$ 利用第 (1) 小题结果即得结论.