📝 题目
例 2 设 $a,b > 0$ ,求证:
(1)当 $p > 1$ 时, ${a}^{p} + {b}^{p} \leq {\left( a + b\right) }^{p}$ ；
(2)当 $0 < p < 1$ 时, ${a}^{p} + {b}^{p} \geq {\left( a + b\right) }^{p}$ .
💡 答案与解析
证 (1) 当 $p$ 是正整数时,利用二项式公式
$$ {\left( a + b\right) }^{p} = {a}^{p} + {C}_{p}^{1}{a}^{p - 1}b + {C}_{p}^{2}{a}^{p - 2}{b}^{2} + \cdots + {b}^{p}. $$
当 $p$ 为一般实数时,不能用二项式公式,但借鉴 $p = 2$ 时的推导:
${\left( a + b\right) }^{2} = \left( {a + b}\right) \left( {a + b}\right) = a\left( {a + b}\right) + b\left( {a + b}\right) \geq {a}^{2} + {b}^{2},$ 我们可以令 $p = 1 + h\left( {h > 0}\right)$ ,则有
$$ {\left( a + b\right) }^{p} = \left( {a + b}\right) {\left( a + b\right) }^{h} = a{\left( a + b\right) }^{h} + b{\left( a + b\right) }^{h} $$
$$ \geq a \cdot {a}^{h} + b \cdot {b}^{h} = {a}^{p} + {b}^{p}. $$
(2)令 $p = 1 - h\left( {0 < h < 1}\right)$ ,则有
$$ {\left( a + b\right) }^{p} = \left( {a + b}\right) {\left( a + b\right) }^{-h} = a{\left( a + b\right) }^{-h} + b{\left( a + b\right) }^{-h} $$
$$ \leq a \cdot {a}^{-h} + b \cdot {b}^{-h} = {a}^{p} + {b}^{p}. $$