对数(指数)平均不等式·证明过程

2023-04-01 02:37 作者:明月星尘ST 0人读过 | 我要投稿

对数平均不等式：对于两个正数a和b且a $%5Cneq%20$ b，我们有 $%5Csqrt%7Bab%7D$ < $e%5Em$ < $%5Cfrac%7Ba%2Bb%7D%7B2%7D%20$ ，下面我们来证明

我们先证 $%5Cfrac%7Ba%2Bb%7D%7B%5Cln%20a%20-%5Cln%20b%20%20%7D%20%3C%5Cfrac%7Ba%2Bb%7D%7B2%7D%20$

因为a,b均大于0

我们不妨设b>a

此时要证 $%5Cfrac%7Ba%2Bb%7D%7B%5Cln%20a%20-%5Cln%20b%20%20%7D%20%3C%5Cfrac%7Ba%2Bb%7D%7B2%7D%20$ 成立，即证 $%5Cfrac%7B%5Cfrac%7Ba%7D%7Bb%7D-1%20%7D%7B%5Cln%20%5Cfrac%7Ba%7D%7Bb%7D%20%20%7D%20%3C%5Cfrac%7B%5Cfrac%7Ba%7D%7Bb%7D%2B1%20%7D%7B2%7D%20$ 成立

此时有 $0%3C%5Cfrac%7Ba%7D%7Bb%7D%20%3C1$

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

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

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

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

此时要证 $%5Cfrac%7Ba%2Bb%7D%7B%5Cln%20a%20-%5Cln%20b%20%20%7D%20%3C%5Cfrac%7Ba%2Bb%7D%7B2%7D%20$ 成立，即证 $2(x-1)-%5Cln%20x%20(x%2B1)%3E0$ 成立

令 $g(x)%3D2(x-1)-%5Cln%20x%20(x%2B1)$

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

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

令h(x)=- $x-x%5Cln%20x%20-1$

显然g'(x)与h(x)的增减区间相同

那么h'(x)= $-%5Cln%20x%20$

因为当 $x%5Cin%20(0%2C1)$ 时， $-%5Cln%20x%20%3E0$

即有当 $x%5Cin%20(0%2C1)$ 时, $h'(x)%3E0$

此时h(x)在(0,1)上单调递增

所以有h(x)<h(1) (0<x<1)

所以有g'(x)<g'(1) (o<x<1)

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

所以有g'(x)<0 (0<x<1)

所以g(x)在(0,1)上单调递减

所以有g(x)>g(1) (0<x<1)

因为由 $g(x)%3D2(x-1)-%5Cln%20x%20(x%2B1)$ ,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)得证

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)得证

接下来我们证 $%5Csqrt%7Bab%7D%20%3C%5Cfrac%7Ba-b%7D%7B%5Cln%20a%20-%5Cln%20b%20%7D%20$

因为a,b均大于0

我们不妨设b>a

则原命题可化为 $%5Csqrt%7B%5Cfrac%7Ba%7D%7Bb%7D%20%7D%20%3C%5Cfrac%7B%5Cfrac%7Ba%7D%7Bb%7D-1%20%7D%7B%5Cln%20%5Cfrac%7Ba%7D%7Bb%7D%20%20%7D%20$

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

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

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

此时要证原命题成立，即证上式成立即可

设 $m(x)%3Dx%5E2%20-2x%5Cln%20x%20-1$

则 $m'(x)%3D2x-2%5Cln%20x%20-2$

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

显然当 $x%5Cin%20(0%2C1)$ 时，m''(x)<0

所以m'(x)在（0,1)上单调递减

所以此时m'(x)>m'(1),因为m'(1)=0

即有m'(x)>0

此时有m(x)在(0,1)上单调递增

此时m(x)<m(1)

因为m(1)=0

所以此时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)得证

至此对数平均不等式的证明完毕，接下来我们来证明指数平均不等式

对于任意实数m,n(m $%5Cneq%20$ n),有 $e%5E%5Cfrac%7Bm%2Bn%7D%7B2%7D%20%20%3C%5Cfrac%7Be%5Em%20-e%5En%20%7D%7Bm-n%7D%20%3C%5Cfrac%7B%20e%5Em%20%2Be%5En%20%7D%7B2%7D%20$

其实此时由对数平均不等式 $%5Csqrt%7Bab%7D$ < $%5Cfrac%7Ba-b%7D%7B%5Cln%20a%20-%20%5Cln%20b%20%7D%20$ < $%5Cfrac%7Ba%2Bb%7D%7B2%7D%20$ (a,b>0且a $%5Cneq%20$ b)中a取 $e%5Em%20$ ,b取 $e%5En%20$ ,即可得到指数平均不等式