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對(duì)數(shù)(指數(shù))平均不等式·證明過程

2023-04-01 02:37 作者:明月星塵ST 0人讀過 | 我要投稿

??對(duì)數(shù)平均不等式：對(duì)于兩個(gè)正數(shù)a和b且a $%5Cneq%20$ b，我們有 $%5Csqrt%7Bab%7D$ < $e%5Em$ < $%5Cfrac%7Ba%2Bb%7D%7B2%7D%20$ ，下面我們來(lái)證明

我們先證 $%5Cfrac%7Ba%2Bb%7D%7B%5Cln%20a%20-%5Cln%20b%20%20%7D%20%3C%5Cfrac%7Ba%2Bb%7D%7B2%7D%20$

因?yàn)閍,b均大于0

我們不妨設(shè)b>a

此時(shí)要證 $%5Cfrac%7Ba%2Bb%7D%7B%5Cln%20a%20-%5Cln%20b%20%20%7D%20%3C%5Cfrac%7Ba%2Bb%7D%7B2%7D%20$ 成立，即證 $%5Cfrac%7B%5Cfrac%7Ba%7D%7Bb%7D-1%20%7D%7B%5Cln%20%5Cfrac%7Ba%7D%7Bb%7D%20%20%7D%20%3C%5Cfrac%7B%5Cfrac%7Ba%7D%7Bb%7D%2B1%20%7D%7B2%7D%20$ 成立

此時(shí)有 $0%3C%5Cfrac%7Ba%7D%7Bb%7D%20%3C1$

那么上式可化為 $2(%5Cfrac%7Ba%7D%7Bb%7D%20-1)%3E%5Cln%20%5Cfrac%7Ba%7D%7Bb%7D%20%20(%5Cfrac%7Ba%7D%7Bb%7D%20%2B1)$

$%5Cimplies%202(%5Cfrac%7Ba%7D%7Bb%7D%20-1)-%5Cln%20%5Cfrac%7Ba%7D%7Bb%7D%20%20(%5Cfrac%7Ba%7D%7Bb%7D%20%2B1)%3E0$

令x= $%5Cfrac%7Ba%7D%7Bb%7D%20$ ?(0<x<1)

則上式可化為 $2(x-1)-%5Cln%20x%20(x%2B1)%3E0$

此時(shí)要證 $%5Cfrac%7Ba%2Bb%7D%7B%5Cln%20a%20-%5Cln%20b%20%20%7D%20%3C%5Cfrac%7Ba%2Bb%7D%7B2%7D%20$ 成立，即證 $2(x-1)-%5Cln%20x%20(x%2B1)%3E0$ 成立

令 $g(x)%3D2(x-1)-%5Cln%20x%20(x%2B1)$

則 $g'(x)$ = $1-%5Cln%20x%20-%5Cfrac1x$

即g'(x)= $%5Cfrac%7Bx-x%5Cln%20x%20-1%7D%7Bx%7D%20$

令h(x)=- $x-x%5Cln%20x%20-1$

顯然g'(x)與h(x)的增減區(qū)間相同

那么h'(x)= $-%5Cln%20x%20$

因?yàn)楫?dāng) $x%5Cin%20(0%2C1)$ 時(shí)， $-%5Cln%20x%20%3E0$

即有當(dāng) $x%5Cin%20(0%2C1)$ 時(shí), $h'(x)%3E0$

此時(shí)h(x)在(0,1)上單調(diào)遞增

所以有h(x)<h(1)? ?(0<x<1)

所以有g(shù)'(x)<g'(1)? (o<x<1)

由g'(x)= $1-%5Cln%20x%20-%5Cfrac1x$ ,得g'(1)=0

所以有g(shù)'(x)<0? ?(0<x<1)

所以g(x)在(0,1)上單調(diào)遞減

所以有g(shù)(x)>g(1)? (0<x<1)

因?yàn)橛?img type="latex" class="latex" src="http://api.bilibili.com/x/web-frontend/mathjax/tex?formula=g(x)%3D2(x-1)-%5Cln%20x%20(x%2B1)" alt="g(x)%3D2(x-1)-%5Cln%20x%20(x%2B1)"/>,知g(1)=0

所以g(x)>0 (0<x<1)

所以 $2(x-1)-%5Cln%20x%20(x%2B1)%3E0$ ?(0<x<1)得證

所以 $%5Cfrac%7Ba%2Bb%7D%7B%5Cln%20a%20-%5Cln%20b%20%20%7D%20%3C%5Cfrac%7Ba%2Bb%7D%7B2%7D%20$ (a,b>0且a $%5Cneq%20$ b)得證

接下來(lái)我們證 $%5Csqrt%7Bab%7D%20%3C%5Cfrac%7Ba-b%7D%7B%5Cln%20a%20-%5Cln%20b%20%7D%20$

因?yàn)閍,b均大于0

我們不妨設(shè)b>a

則原命題可化為 $%5Csqrt%7B%5Cfrac%7Ba%7D%7Bb%7D%20%7D%20%3C%5Cfrac%7B%5Cfrac%7Ba%7D%7Bb%7D-1%20%7D%7B%5Cln%20%5Cfrac%7Ba%7D%7Bb%7D%20%20%7D%20$

$%5Cimplies%20%5Cfrac%7Ba%7D%7Bb%7D%20-%5Csqrt%7B%5Cfrac%7Ba%7D%7Bb%7D%20%7D%20%5Cln%20%5Cfrac%7Ba%7D%7Bb%7D%20%20-1%3C0$

令x= $%5Csqrt%7B%5Cfrac%7Ba%7D%7Bb%7D%20%7D%20$ ?(0<x<1)

則有 $x%5E2%20-x%5Cln%20x%5E2%20%20-1%3C0$

此時(shí)要證原命題成立，即證上式成立即可

設(shè) $m(x)%3Dx%5E2%20-2x%5Cln%20x%20-1$

則 $m'(x)%3D2x-2%5Cln%20x%20-2$

則 $m''(x)%3D%5Cfrac%7B2(x-1)%7D%7Bx%7D%20$

顯然當(dāng) $x%5Cin%20(0%2C1)$ 時(shí)，m''(x)<0

所以m'(x)在（0,1)上單調(diào)遞減

所以此時(shí)m'(x)>m'(1),因?yàn)閙'(1)=0

即有m'(x)>0

此時(shí)有m(x)在(0,1)上單調(diào)遞增

此時(shí)m(x)<m(1)

因?yàn)閙(1)=0

所以此時(shí)g(x)<0

即 $x%5E2%20-x%5Cln%20x%5E2%20%20-1%3C0$

所以 $%5Csqrt%7Bab%7D%20%3C%5Cfrac%7Ba-b%7D%7B%5Cln%20a%20-%5Cln%20b%20%7D%20$ （a,b大于0且a不等于b)得證

至此對(duì)數(shù)平均不等式的證明完畢，接下來(lái)我們來(lái)證明指數(shù)平均不等式

對(duì)于任意實(shí)數(shù)m,n(m $%5Cneq%20$ n),有 $e%5E%5Cfrac%7Bm%2Bn%7D%7B2%7D%20%20%3C%5Cfrac%7Be%5Em%20-e%5En%20%7D%7Bm-n%7D%20%3C%5Cfrac%7B%20e%5Em%20%2Be%5En%20%7D%7B2%7D%20$

其實(shí)此時(shí)由對(duì)數(shù)平均不等式 $%5Csqrt%7Bab%7D$ < $%5Cfrac%7Ba-b%7D%7B%5Cln%20a%20-%20%5Cln%20b%20%7D%20$ < $%5Cfrac%7Ba%2Bb%7D%7B2%7D%20$ (a,b>0且a $%5Cneq%20$ b)中a取 $e%5Em%20$ ,b取 $e%5En%20$ ,即可得到指數(shù)平均不等式