证明二元AM-GM不等式

hbghlyj

H. A. Priestley - Introduction to Integration - Clarendon Press; Oxford University Press, 1997
Exercise 2.10
(a) Let $ 0\lt \alpha\lt 1 $ and define $f(x)\coloneqq\alpha x - x^\alpha$.
Show that $f(x) \ge f(1) \  \forall x \ge 0$.
Hence show that $$ a^\alpha b ^{1-\alpha} \le \alpha a +(1-\alpha)b \ \ \forall a,b \ge 0$$
(b) Let $p$ and $q$ be such that $1<p<\infty$ and $p^{-1}+q^{-1}=1$, and let $A$ and $B$ be non-negative real numbers. Deduce from (a), with $\alpha=1/p$, that\[AB\le\frac{A^p}p+\frac{B^q}q,\]with equality if and only if $A^p=B^q$. [This inequality is needed in the theory of L^{$^p$}-spaces; see Chapter 28.]

hbghlyj

(a) $f'(x)=\alpha x^{\alpha-1} \left(x^{1-\alpha}-1\right)$
当$0<x<1$时$f'(x)<0$; 当$x>1$时$f'(x)>0$.
由2.26(3)Corollaries of the MVT知, $f(x)\ge f(1)\ \forall x\ge0$.
$\forall a,b\ge0$, 令$x=\frac ab$有$f\left(a\over b\right)\ge f(1)=\alpha-1$. 即\begin{equation}\label1 a^\alpha b ^{1-\alpha} \le \alpha a +(1-\alpha)b\end{equation}取等条件为$x=1$, 即$a=b$.
(b) 在\eqref{1}代入$\alpha=\frac1p,1-\alpha=\frac1q$得\[a^{\frac1p} b ^{\frac1q} \le \frac{a}p+\frac{b}q\]令$a=A^p,b=B^q$得\[AB\le\frac{A^p}p+\frac{B^q}q\]取等条件为$A^p=B^q$.

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又见Young's inequality for products–conjugate Hölder exponents

色k

常识，任意一本不等式书籍都会介绍，抄来干嘛。

威望 -2

hbghlyj

补充: 常识: 楊氏不等式
相关帖子: 一个显而易见的定积分恒等式

kuing

点评
hbghlyj    每天都要学一学 :)  发表于 2022-11-27 22:47

那你可以记录在你自己的笔记里，没必要抄上来。