古明地恋的数学科普——向前—向后数学归纳法

2022-11-19 23:50 作者:量子少女-古明地恋 0人读过 | 我要投稿

前言：本文适合学习过基本不等式并且掌握较为熟练的读者.

一、什么是向前—向后（Forward and Backward）数学归纳法

考虑到部分读者并不熟悉数学归纳法（因为目前的中学教材已经淡化这部分内容），所以在此给出数学归纳法的最基本形式：

当一个关于正整数n的命题满足以下条件时：

1. $n%3D1$ 时命题成立；

2. $n%3Dk$ 时命题成立可推得 $n%3Dk%2B1$ 时命题成立.

可证得命题对任意正整数n成立.

上述形式也称作第一数学归纳法.事实上，数学归纳法有多种变形形式.如当 $n%3Dk$ 替换为 $n%5Cleqslant%20k$ 时的第二数学归纳法.

此次所述的向前—向后数学归纳法也是数学归纳法的变形形式.其形式如下：

当一个命题满足如下条件时：

1.命题对无穷多个自然数 $n$ 成立；

2. $n%3Dk%2B1$ 时命题成立可推得 $n%3Dk$ 时命题成立.

可证得命题对任意正整数 $n$ 成立.

二、向前—向后数学归纳法的应用实例（不等式的证明）

1. $n$ 元算术—几何平均值（AM—GM）不等式

$%5Cdfrac%7B%5Csum_%7Bi%3D1%7D%5E%7Bn%7Da_i%7D%7Bn%7D%5Cgeqslant%20%5Csqrt%5Bn%5D%7B%5Cprod_%7Bi%3D1%7D%5E%7Bn%7Da_i%7D(a_i%3E0)$ ，当且仅当 $a_1%3Da_2%3D%5Ccdots%3Da_n$ 时等号成立.

证明如下：

$n%3D2$ 时，即是读者在中学阶段学习过2元形式的均值不等式：

$%5Cfrac%7Ba%2Bb%7D%7B2%7D%5Cgeqslant%20%5Csqrt%7Bab%7D%20(a%2Cb%3E0)$ ，当且仅当a=b时等号成立.

此处证明从略.

设命题对 $n%3D2%5Ek$ 成立，则 $n%3D2%5E%7Bk%2B1%7D$ 时，

因为 $%5Cdfrac%7B%5Csum_%7Bi%3D1%7D%5E%7B2%5E%7Bk%2B1%7D%7Da_i%7D%7B2%5E%7Bk%2B1%7D%7D%3D%5Cdfrac%7B%5Cdfrac%7B%5Csum_%7Bi%3D1%7D%5E%7B2%5E%7Bk%7D%7Da_i%7D%7B2%5E%7Bk%7D%7D%2B%5Cdfrac%7B%5Csum_%7Bi%3D2%5E%7Bk%7D%2B1%7D%5E%7B2%5E%7Bk%2B1%7D%7Da_i%7D%7B2%5E%7Bk%7D%7D%7D%7B2%7D%5Cgeqslant%20%5Cdfrac%7B%5Csqrt%5B2%5Ek%5D%7B%5Cprod_%7Bi%3D1%7D%5E%7B2%5En%7Da_i%7D%2B%5Csqrt%5B2%5Ek%5D%7B%5Cprod_%7Bi%3D2%5Ek%2B1%7D%5E%7B2%5E%7Bk%2B1%7D%7Da_i%7D%7D%7B2%7D%5Cgeqslant%20%5Csqrt%5B2%5E%7Bk%2B1%7D%5D%7B%5Cprod_%7Bi%3D1%7D%5E%7B2%5E%7Bk%2B1%7D%7Da_i%7D$

所以命题对 $n%3D2%5E%7Bk%2B1%7D$ 也成立.

假设命题对 $n%3Dk%2B1$ 成立，则 $n%3Dk$ 时，

$%5Cdfrac%7B%5Csum_%7Bi%3D1%7D%5E%7Bk%7Da_i%7D%7Bk%7D%3D%5Cdfrac%7B1%7D%7Bk%2B1%7D(%5Csum_%7Bi%3D1%7D%5E%7Bk%7Da_i%2B%5Cdfrac%7B1%7D%7Bk%7D%5Csum_%7Bi%3D1%7D%5E%7Bk%7Da_i)%5Cgeqslant%20%5Csqrt%5Bk%2B1%5D%7B(%5Cprod_%7Bi%3D1%7D%5E%7Bk%7Da_i)(%5Cdfrac%7B1%7D%7Bk%7D%5Csum_%7Bi%3D1%7D%5E%7Bk%7Da_i)%7D$

从而有

$%5Cdfrac%7B%5Csum_%7Bi%3D1%7D%5E%7Bk%7Da_i%7D%7Bk%7D%5Cgeqslant%20%5Csqrt%5Bk%5D%7B%5Cprod_%7Bi%3D1%7D%5E%7Bk%7Da_i%7D$

综上所述，原不等式得证.

2.樊畿（Fan Ky）不等式

$%5Cdfrac%7B%5Cprod_%7Bi%3D1%7D%5E%7Bn%7Dx_i%7D%7B(%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dx_i)%5En%7D%5Cleqslant%20%5Cdfrac%7B%5Cprod_%7Bi%3D1%7D%5E%7Bn%7D(1-x_i)%7D%7B%5B%5Csum_%7Bi%3D1%7D%5E%7Bn%7D(1-x_i)%5D%5En%7D%20%EF%BC%880%3Cx_i%5Cleqslant%20%5Cdfrac%7B1%7D%7B2%7D)$

n=1时，显然成立;n=2时，即证 $%5Cdfrac%7Bx_1x_2%7D%7B(x_1%2Bx_2)%5E2%7D%5Cleqslant%20%5Cdfrac%7B(1-x_1)(1-x_2)%7D%7B(1-x_1%2B1-x_2)%5E2%7D$

也就是 $(x_1-x_2)%5E2%20(1-x_1-x_2)%5Cgeqslant%200$ ，而这是显然的.

设结论对 $n%3D2%5Ek$ 成立，即 $%5Cdfrac%7B%5Cprod_%7Bi%3D1%7D%5E%7B2%5Ek%7Dx_i%7D%7B(%5Csum_%7Bi%3D1%7D%5E%7B2%5Ek%7Dx_i)%5En%7D%5Cleqslant%20%5Cdfrac%7B%5Cprod_%7Bi%3D1%7D%5E%7B2%5Ek%7D(1-x_i)%7D%7B%5B%5Csum_%7Bi%3D1%7D%5E%7B2%5Ek%7D(1-x_i)%5D%5E%7B2%5Ek%7D%7D$ 也就是 $%5Cdfrac%7B%5Cprod_%7Bi%3D1%7D%5E%7B2%5Ek%7Dx_i%7D%7B%5Cprod_%7Bi%3D1%7D%5E%7B2%5Ek%7D(1-x_i)%7D%5Cleqslant%20%5Cdfrac%7B(%5Csum_%7Bi%3D1%7D%5E%7B2%5Ek%7Dx_i)%5E%7B2%5Ek%7D%7D%7B%5B%5Csum_%7Bi%3D1%7D%5E%7B2%5Ek%7D(1-x_i)%5D%5E%7B2%5Ek%7D%7D$

$n%3D2%5E%7Bk%2B1%7D$ 时， $%5Cdfrac%7B%5Cprod_%7Bi%3D1%7D%5E%7B2%5E%7Bk%2B1%7D%7Dx_i%7D%7B%5Cprod_%7Bi%3D1%7D%5E%7B2%5E%7Bk%2B1%7D%7D(1-x_i)%7D%3D%5Cdfrac%7B%5Cprod_%7Bi%3D1%7D%5E%7B2%5Ek%7Dx_i%7D%7B%5Cprod_%7Bi%3D1%7D%5E%7B2%5Ek%7D(1-x_i)%7D%20%5Cdfrac%7B%5Cprod_%7Bi%3D2%5Ek%2B1%7D%5E%7B2%5E%7Bk%2B1%7D%7Dx_i%7D%7B%5Cprod_%7Bi%3D2%5Ek%2B1%7D%5E%7B2%5E%7Bk%2B1%7D%7D(1-x_i)%7D%5Cleqslant%20%5Cdfrac%7B(%5Csum_%7Bi%3D1%7D%5E%7B2%5Ek%7Dx_i)%5E%7B2%5Ek%7D%7D%7B%5B%5Csum_%7Bi%3D1%7D%5E%7B2%5Ek%7D(1-x_i)%5D%5E%7B2%5Ek%7D%7D%20%5Cdfrac%7B(%5Csum_%7Bi%3D2%5Ek%2B1%7D%5E%7B2%5E%7Bk%2B1%7D%7Dx_i)%5E%7B2%5Ek%7D%7D%7B%5B%5Csum_%7Bi%3D2%5Ek%2B1%7D%5E%7B2%5E%7Bk%2B1%7D%7D(1-x_i)%5D%5E%7B2%5Ek%7D%7D%20$

要证明上述式子成立，即证 $(%5Cdfrac%7Ba%7D%7B2%5Ek-a%7D%20%5Cdfrac%7Bb%7D%7B2%5Ek-b%7D)%5E%7B2%5Ek%7D%5Cleqslant%20(%5Cdfrac%7Ba%2Bb%7D%7B2%5E%7Bk%2B1%7D-a-b%7D)%5E%7B2%5E%7Bk%2B1%7D%7D$

也就是 $%20%5Cdfrac%7Bab%7D%7B(2%5Ek-a)(2%5Ek-b)%7D%5Cleqslant%20(%5Cdfrac%7Ba%2Bb%7D%7B2%5E%7Bk%2B1%7D-a-b%7D)%5E2$

即 $(2%5Ek-a-b)(a-b)%5E2%5Cgeqslant%200$ ，由于 $0%3Ca%2Cb%5Cleqslant%202%5E%7Bk-1%7D$ ，此式显然成立.

设结论对n=k+1成立，即 $%5Cdfrac%7B%5Cprod_%7Bi%3D1%7D%5E%7Bk%2B1%7Dx_i%7D%7B(%5Csum_%7Bi%3D1%7D%5E%7Bk%2B1%7Dx_i)%5E%7Bk%2B1%7D%7D%5Cleqslant%20%5Cdfrac%7B%5Cprod_%7Bi%3D1%7D%5E%7Bk%2B1%7D(1-x_i)%7D%7B%5B%5Csum_%7Bi%3D1%7D%5E%7Bk%2B1%7D(1-x_i)%5D%5E%7Bk%2B1%7D%7D$

考虑 $A%3D%5Cdfrac%7B1%7D%7Bk%7D%20%5Csum_%7Bi%3D1%7D%5E%7Bk%2B1%7Dx_i$ ，则 $(%5Cdfrac%7Bx_1%2Bx_2%2B%5Ccdots%2Bx_k%2BA%7D%7B(1-x_1)%2B(1-x_2)%2B%5Ccdots%2B(1-x_k)%2BA%7D)%5E%7Bk%2B1%7D%5Cgeqslant%20%5Cdfrac%7Bx_1x_2%5Ccdots%20x_kA%7D%7B(1-x_1)(1-x_2)%5Ccdots(1-x_k)(1-A)%7D$

由于 $(%5Cdfrac%7BA%7D%7B1-A%7D)%5E%7Bk%2B1%7D%3D(%5Cdfrac%7Bx_1%2Bx_2%2B%5Ccdots%2Bx_k%2BA%7D%7B(1-x_1)%2B(1-x_2)%2B%5Ccdots%2B(1-x_k)%2B(1-A)%7D)%5E%7Bk%2B1%7D%5Cgeqslant%20%5Cdfrac%7Bx_1x_2%5Ccdots%20x_kA%7D%7B(1-x_1)(1-x_2)%5Ccdots(1-x_k)(1-A)%7D$