一些基本不等式的证明

2023-06-03 09:18 作者:~Sakuno酱 0人读过 | 我要投稿

首先证明不等式

$(%5Cfrac%7Bx%2By%7D%7B2%7D)%5E2%20%5Cge%20xy$

$(%5Cfrac%7Bx%2By%7D%7B2%7D)%5E2%20-%20xy%20%3D%20%5Cfrac%7Bx%5E2%2By%5E2%7D%7B2%7D%20%5Cge%200%20$ 当 $x%3Dy%3D0$ 时等号成立

这个在证明柯西不等式时候会用到

柯西不等式

命题

$(%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dx_iy_i)%5E2%5Cle(%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dx_i%5E2)(%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dy_i%5E2)$

数学归纳法证明:

$n%3D1$ 时显然成立

假设 $n%3Dn$ 时成立，考虑 $n%2B1$

为了直观我们记

$%5Calpha%20%3D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dx_i%5E2$

$%5Cbeta%20%3D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dx_i%5E2$

$%5Cgamma%20%3D%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dx_iy_i$

直接右减去左

$(%5Csum_%7Bi%3D1%7D%5E%7Bn%2B1%7Dx_i%5E2)(%5Csum_%7Bi%3D1%7D%5E%7Bn%2B1%7Dy_i%5E2)%20-%20(%5Csum_%7Bi%3D1%7D%5E%7Bn%2B1%7Dx_iy_i)%5E2%20$

代换

$%3D%20(%5Calpha%2Bx_%7Bn%2B1%7D%5E2)(%5Cbeta%20%2B%20y_%7Bn%2B1%7D%5E2)-(%5Cgamma%20%2Bx_%7Bn%2B1%7Dy_%7Bn%2B1%7D)%5E2%20$

$%3D%20(%5Calpha%5Cbeta%20-%20%5Cgamma%5E2)%20%20%20%2B%20%5Calpha%20y%5E2_%7Bn%2B1%7D%2B%5Cbeta%20x%5E2_%7Bn%2B1%7D-2%5Cgamma%20x_%7Bn%2B1%7Dy_%7Bn%2B1%7D%20%20$

其中 $(%5Calpha%5Cbeta%20-%20%5Cgamma%5E2)%20%20%5Cge%200$

$%5Calpha%20y%5E2_%7Bn%2B1%7D%2B%5Cbeta%20x%5E2_%7Bn%2B1%7D-2%5Cgamma%20x_%7Bn%2B1%7Dy_%7Bn%2B1%7D%20%20$ 使用均值不等式

$%5Cge%202%5Csqrt%7B%5Calpha%5Cbeta%20x%5E2_%7Bn%2B1%7D%20y%5E2_%7Bn%2B1%7D%7D-2%5Cgamma%20x_%7Bn%2B1%7Dy_%7Bn%2B1%7D%20%20$ 使用 $%5Calpha%20%5Cbeta%20%5Cge%20%5Cgamma%5E2$ 得到 $%5Csqrt%7B%5Calpha%20%5Cbeta%7D%20%5Cge%20%7C%5Cgamma%7C$

$%5Cge2%20(%7C%5Cgamma%20x_%7Bn%2B1%7Dy_%7Bn%2B1%7D%7C-%5Cgamma%20x_%7Bn%2B1%7Dy_%7Bn%2B1%7D)$

$%5Cge%200$

平方平均大于等于算数平均

代入柯西不等式

$(%5Csum_%7Bi%3D1%7D%5Enx_i%5Ccdot%201)%5E2%20%5Cle%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dx_i%5E2%5Csum_%7Bi%3D1%7D%5En1%5E2%20%5Cle%20n%5Csum_%7Bi%3D1%7D%5Enx_i%5E2$

$%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dx_i%20%5Cle%5Csqrt%7Bn%20%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dx_i%5E2%7D$

同时除以 $n$

$%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5Enx_i%7D%7Bn%7D%20%5Cle%20%5Csqrt%7B%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dx_i%5E2%7D%7Bn%7D%7D$

算数平均大于几何平均

$%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5E%7Bn%7Dx_i%7D%7Bn%7D%5Cge%20(%5Cprod_%7Bi%3D1%7D%5E%7Bn%7Dx_i)%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D$

首先归纳法证明 $n%3D2%5Ek$ 时成立

$k%3D1$ 时易证

考虑 $k%2B1$

令

$%5Calpha%20%3D%20%5Csum_%7Bi%3D1%7D%5E%7B2%5Ek%7Dx_i$

$%5Cbeta%20%3D%20%5Csum_%7Bi%3D2%5Ek%2B1%7D%5E%7B2%5E%7Bk%2B1%7D%7Dx_i$

$%5Csum_%7Bi%3D1%7D%5E%7B2%5E%7Bk%2B1%7D%7Dx_i%3D%5Calpha%2B%5Cbeta%20%5Cge%202%5Csqrt%7B%5Calpha%20%5Cbeta%7D$

其中

$%5Calpha%20%3D%20%5Csum_%7Bi%3D1%7D%5E%7B2%5Ek%7Dx_i%20%5Cge%202%5Ek%20(%5Cprod_%7Bi%3D1%7D%5E%7B2%5Ek%7Dx_i)%5E%7B%5Cfrac%7B1%7D%7B2%5Ek%7D%7D$

$%5Cbeta%20%3D%20%5Csum_%7Bi%3D2%5Ek%2B1%7D%5E%7B2%5E%7Bk%2B1%7D%7Dx_i%20%5Cge%202%5Ek%20%0A%20(%5Cprod_%7Bi%3D2%5Ek%2B1%7D%5E%7B2%5E%7Bk%2B1%7D%7Dx_i)%5E%7B%5Cfrac%7B1%7D%7B2%5Ek%7D%7D$

所以

$%5Calpha%20%5Cbeta%20%5Cge%202%5E%7B2k%7D%20(%5Cprod_%7Bi%3D1%7D%5E%7B2%5E%7Bk%2B1%7D%7Dx_i)%5E%7B%5Cfrac%7B1%7D%7B2%5Ek%7D%7D$

$2%5Csqrt%7B%5Calpha%20%5Cbeta%7D%20%5Cge%202%5E%7Bk%2B1%7D%20%20(%5Cprod_%7Bi%3D1%7D%5E%7B2%5E%7Bk%2B1%7D%7Dx_i)%5E%7B%5Cfrac%7B1%7D%7B2%5E%7Bk%2B1%7D%7D%7D$

尝试证明更一般的 $n%3Dk$ , $n%20%5Cne%202%5Em$ 时成立

令 $p%3D2%5Ek$ $%5Coverline%7Bx%7D%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5E%7Bk%7D%7Bx_i%7D%7D%7Bk%7D$

令 $y_i%3D%20%5Cbegin%7Bcases%7D%0Ax_i%2C%5Cquad%20%26x%5Cleq%20k%20%5C%5C%0A%5Coverline%7Bx%7D%2C%5Cquad%20%26%20k%20%3C%20x%20%5Cle%20p%0A%5Cend%7Bcases%7D%20$

$%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5E%7Bp%7Dy_i%7D%7Bp%7D%20%3D%20%5Cfrac%7B%5Csum_%7Bi%3D1%7D%5E%7Bk%7Dx_i%20%2B%20%5Csum_%7Bi%3Dk%2B1%7D%5E%7Bp%7D%5Coverline%7Bx%7D%20%7D%7Bp%7D%20%5Cge%20(%5Cprod_%7Bi%3D1%7D%5E%7Bp%7Dy_i)%5E%5Cfrac%7B1%7D%7Bp%7D$

$%5Cfrac%7Bk%5Coverline%7Bx%7D%20%2B%20(p-k)%5Coverline%7Bx%7D%20%7D%7Bp%7D%20%5Cge%20(%5Cprod_%7Bi%3D1%7D%5E%7Bp%7Dy_i)%5E%5Cfrac%7B1%7D%7Bp%7D%20%5Cge%20(%5Cprod_%7Bi%3D1%7D%5E%7Bk%7Dx_i)%5E%7B%5Cfrac%7B1%7D%7Bp%7D%7D((%5Coverline%7Bx%7D)%5E%7Bp-k%7D)%5E%7B%5Cfrac%7B1%7D%7Bp%7D%7D$

$%5Coverline%7Bx%7D%20%5Cge%20(%5Cprod_%7Bi%3D1%7D%5E%7Bk%7Dx_i)%5E%7B%5Cfrac%7B1%7D%7Bp%7D%7D((%5Coverline%7Bx%7D)%5E%7Bp-k%7D)%5E%7B%5Cfrac%7B1%7D%7Bp%7D%7D$ 两边同时作 $p$ 次幂

$(%5Coverline%7Bx%7D)%5Ep%20%5Cge%20(%5Cprod_%7Bi%3D1%7D%5E%7Bk%7Dx_i)%5E%7B%7D((%5Coverline%7Bx%7D)%5E%7Bp-k%7D)$ 两边同时乘以 $(%5Coverline%7Bx%7D)%5E%7Bk-p%7D$

$(%5Coverline%7Bx%7D)%5Ek%20%5Cge%20%5Cprod_%7Bi%3D1%7D%5E%7Bk%7Dx_i$ 两边同时作 $%5Cfrac%7B1%7D%7Bk%7D$ 次幂

$%5Coverline%7Bx%7D%20%5Cge%20(%5Cprod_%7Bi%3D1%7D%5E%7Bk%7Dx_i)%5E%7B%5Cfrac%7B1%7D%7Bk%7D%7D$ 得证

几何平均大于等于调和平均

把上面不等式里面的 $x_i$ 换成 $%5Cfrac%7B1%7D%7Bx_i%7D$ 得到

$%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Bx_i%7D%20%5Cge%20n(%5Cprod_%7Bi%3D1%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Bx_i%7D)%5E%5Cfrac%7B1%7D%7Bn%7D%20%5Cge%20n%20%5Cfrac%7B1%7D%7B(%5Cprod_%7Bi%3D1%7D%5E%7Bn%7Dx_i)%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%7D$

两边同时取倒数得到

$(%5Cprod_%7Bi%3D1%7D%5E%7Bn%7Dx_i)%5E%7B%5Cfrac%7B1%7D%7Bn%7D%7D%20%5Cge%20%5Cfrac%7Bn%7D%7B%5Csum_%7Bi%3D1%7D%5E%7Bn%7D%5Cfrac%7B1%7D%7Bx_i%7D%7D$

三角不等式

$%7C%5C%7C%5Cmathrm%7Bx%7D%5C%7C-%5C%7C%5Cmathrm%7By%7D%5C%7C%7C%20%5Cle%20%5C%7C%5Cmathrm%7B%5Cmathrm%7Bx%7D-%5Cmathrm%7By%7D%7D%5C%7C%20%5Cle%20%5C%7C%5Cmathrm%7B%5Cmathrm%7Bx%7D%7D%5C%7C%20%20%2B%20%20%5C%7C%5Cmathrm%7B%5Cmathrm%7By%7D%7D%5C%7C%20%0A$

先证明左边的部分

记 $L%3D%7C%5C%7C%5Cmathrm%7Bx%7D%5C%7C-%5C%7C%5Cmathrm%7By%7D%5C%7C%7C%20$

$R%3D%5C%7C%5Cmathrm%7B%5Cmathrm%7Bx%7D-%5Cmathrm%7By%7D%7D%5C%7C$

$L%5E2%3D%5Csum_%7Bi%7Dx_i%5E2%20%2B%20%5Csum_%7Bi%7Dy_i%5E2-2%5Csqrt%7B(%5Csum_%7Bi%7Dx_i%5E2)%20(%5Csum_%7Bi%7Dy_i%5E2)%7D$

$R%5E2%3D%5Csum_%7Bi%7Dx_i%5E2%20%2B%20%5Csum_%7Bi%7Dy_i%5E2-2%5Csum_%7Bi%7Dx_iy_i$

$%5Cfrac%7BR%5E2-L%5E2%7D%7B2%7D%3D%5Csqrt%7B(%5Csum_%7Bi%7Dx_i%5E2)%20(%5Csum_%7Bi%7Dy_i%5E2)%7D%20-%5Csum_%7Bi%7Dx_iy_i$

应用柯西不等式得到

$%5Csqrt%7B(%5Csum_%7Bi%7Dx_i%5E2)%20(%5Csum_%7Bi%7Dy_i%5E2)%7D%20-%5Csum_%7Bi%7Dx_iy_i%20%5Cge%20%7C%5Csum_%7Bi%7Dx_iy_i%7C%20-%5Csum_%7Bi%7Dx_iy_i$

所以 $R%5E2%3EL%5E2$

$R%3EL$

再证明右边部分

记 $R%3D%5C%7C%5Cmathrm%7Bx%7D%5C%7C%2B%5C%7C%5Cmathrm%7By%7D%5C%7C$

$L%3D%5C%7C%5Cmathrm%7B%5Cmathrm%7Bx%7D-%5Cmathrm%7By%7D%7D%5C%7C$

同理有

$R%5E2-L%5E2%3D2%5Csqrt%7B(%5Csum_%7Bi%7Dx_i%5E2)%20(%5Csum_%7Bi%7Dy_i%5E2)%7D-2%5Csum_%7Bi%7Dx_iy_i$

$%5Cge2(%20%7C%5Csum_%7Bi%7Dx_iy_i%7C%20-%5Csum_%7Bi%7Dx_iy_i)%20%5Cge%200$

三角不等式的一个重要变形

在多元函数微分学里面经常用到

$%20%5C%7C%5Cmathrm%7B%5Cmathrm%7Bx%7D%7D%5C%7C%3D%5C%7C%20%5Csum_%7Bi%7Dx_ie_i%20%5C%7C%20%5Cle%20%20%5Csum_i%20%5C%7C%20x_ie_i%5C%7C%20%5Cle%20%5Csum_%7Bi%7D%20%7Cx_i%7C$

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一些基本不等式的证明

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