古明地戀的數(shù)學(xué)科普——向前—向后數(shù)學(xué)歸納法

2022-11-19 23:50 作者:量子少女-古明地戀 3人讀過 | 我要投稿

前言：本文適合學(xué)習(xí)過基本不等式并且掌握較為熟練的讀者.

一、什么是向前—向后（Forward and Backward）數(shù)學(xué)歸納法

考慮到部分讀者并不熟悉數(shù)學(xué)歸納法（因為目前的中學(xué)教材已經(jīng)淡化這部分內(nèi)容），所以在此給出數(shù)學(xué)歸納法的最基本形式：

當(dāng)一個關(guān)于正整數(shù)n的命題滿足以下條件時：

1. $n%3D1$ 時命題成立；

2. $n%3Dk$ 時命題成立可推得 $n%3Dk%2B1$ 時命題成立.

可證得命題對任意正整數(shù)n成立.

上述形式也稱作第一數(shù)學(xué)歸納法.事實上，數(shù)學(xué)歸納法有多種變形形式.如當(dāng) $n%3Dk$ 替換為 $n%5Cleqslant%20k$ 時的第二數(shù)學(xué)歸納法.

此次所述的向前—向后數(shù)學(xué)歸納法也是數(shù)學(xué)歸納法的變形形式.其形式如下：

當(dāng)一個命題滿足如下條件時：

1.命題對無窮多個自然數(shù) $n$ 成立；

2. $n%3Dk%2B1$ 時命題成立可推得 $n%3Dk$ 時命題成立.

可證得命題對任意正整數(shù) $n$ 成立.

二、向前—向后數(shù)學(xué)歸納法的應(yīng)用實例（不等式的證明）

1. $n$ 元算術(shù)—幾何平均值（AM—GM）不等式

$%5Cdfrac%7B%5Csum_%7Bi%3D1%7D%5E%7Bn%7Da_i%7D%7Bn%7D%5Cgeqslant%20%5Csqrt%5Bn%5D%7B%5Cprod_%7Bi%3D1%7D%5E%7Bn%7Da_i%7D(a_i%3E0)$ ，當(dāng)且僅當(dāng) $a_1%3Da_2%3D%5Ccdots%3Da_n$ 時等號成立.

證明如下：

$n%3D2$ 時，即是讀者在中學(xué)階段學(xué)習(xí)過2元形式的均值不等式：

$%5Cfrac%7Ba%2Bb%7D%7B2%7D%5Cgeqslant%20%5Csqrt%7Bab%7D%20(a%2Cb%3E0)$ ，當(dāng)且僅當(dāng)a=b時等號成立.

此處證明從略.

設(shè)命題對 $n%3D2%5Ek$ 成立，則 $n%3D2%5E%7Bk%2B1%7D$ 時，

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

所以命題對 $n%3D2%5E%7Bk%2B1%7D$ 也成立.

假設(shè)命題對 $n%3Dk%2B1$ 成立，則 $n%3Dk$ 時，

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

從而有

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

綜上所述，原不等式得證.

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

也就是 $(x_1-x_2)%5E2%20(1-x_1-x_2)%5Cgeqslant%200$ ，而這是顯然的.

設(shè)結(jié)論對 $n%3D2%5Ek$ 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 $%5Cdfrac%7B%5Cprod_%7Bi%3D1%7D%5E%7B2%5Ek%7Dx_i%7D%7B%5Cprod_%7Bi%3D1%7D%5E%7B2%5Ek%7D(1-x_i)%7D%5Cleqslant%20%5Cdfrac%7B(%5Csum_%7Bi%3D1%7D%5E%7B2%5Ek%7Dx_i)%5E%7B2%5Ek%7D%7D%7B%5B%5Csum_%7Bi%3D1%7D%5E%7B2%5Ek%7D(1-x_i)%5D%5E%7B2%5Ek%7D%7D$

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

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

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 $0%3Ca%2Cb%5Cleqslant%202%5E%7Bk-1%7D$ ，此式顯然成立.

設(shè)結(jié)論對n=k+1成立，即 $%5Cdfrac%7B%5Cprod_%7Bi%3D1%7D%5E%7Bk%2B1%7Dx_i%7D%7B(%5Csum_%7Bi%3D1%7D%5E%7Bk%2B1%7Dx_i)%5E%7Bk%2B1%7D%7D%5Cleqslant%20%5Cdfrac%7B%5Cprod_%7Bi%3D1%7D%5E%7Bk%2B1%7D(1-x_i)%7D%7B%5B%5Csum_%7Bi%3D1%7D%5E%7Bk%2B1%7D(1-x_i)%5D%5E%7Bk%2B1%7D%7D$

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

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