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一个不等式的证明

2023-07-28 09:39 作者:小王工作室SCIENCE  | 我要投稿

证明:当x-y%5Cin%20R时,(sinx-siny)%5E2%20%2B(cosx-cosy)%5E2%20%5Cleq%20(x-y)%5E2%20

首先我们先来解决函数f(x)%3Dx%5E2%2B2cosx的取值范围

先求导:f'(x)%3D2x-2sinx

所以我们需要解方程:2x-2sinx%3D0,即g(x)%3Dx-sinx%3D0

我们再次求导:g'(x)%3D1-cosx%5Cgeq%200(x%5Cin%20R)

所以g(x)%3Dx-sinx在实数范围内是单调递增的

所以当x%3C0时,g(x)%3Cg(0)%3D0;当x%3E0时,g(x)%3Eg(0)%3D0

所以当g(x)%3Dx-sinx%3D0时,x%3D0

所以f(x)%3Dx%5E2%2B2cosx%20%5Cgeq%200%5E2%20%2B2cos0%3D2

换元x%3Dz-y

所以(z-y)%5E2%2B2cos(z-y)%5Cgeq%20%202

(z-y)%5E2%5Cgeq%20%202-2cos(z-y)

(z-y)%5E2%5Cgeq%20sin%5E2z%2Bcos%5E2z%2Bsin%5E2y%2Bcos%5E2y%20%20-2coszcosy-2sinzsiny

(sinz-siny)%5E2%20%2B(cosz-cosy)%5E2%20%5Cleq%20(z-y)%5E2%20(z-y%5Cin%20R)

其他证明方法:

我们都知道柯西积分不等式:(%5Cint_%7Ba%7D%5E%7Bb%7D%20f(x)g(x)dx)%5E2%5Cleq%20(%5Cint_%7Ba%7D%5E%7Bb%7D%20%5Bf(x)%5D%5E2dx)(%5Cint_%7Ba%7D%5E%7Bb%7D%20%5Bg(x)%5D%5E2dx)

f(x)%3Dsinx%20%20%20%2Cg(x)%3Dcosx,得:

(%5Cint_%7Ba%7D%5E%7Bb%7Dsinxcosxdx)%5E2%5Cleq%20(%5Cint_%7Ba%7D%5E%7Bb%7D%20sin%5E2xdx)(%5Cint_%7Ba%7D%5E%7Bb%7D%20cos%5E2xdx)

计算积分得:

%5Cfrac%7B(cos2a-cos2b)%5E2%7D%7B16%7D%20%5Cleq%20%5Cfrac%7B(b-a)%5E2%7D%7B4%7D%20%2B%5Cfrac%7Bsinbcosb(sinacosa-sinbcosb)%2Bsinacosa(sinbcosb-sinacosa)%7D%7B4%7D%20

化简:

(cos2a-cos2b)%5E2%5Cleq%20(2a-2b)%5E2-(2sinacosa-2sinbcosb)%5E2

最终得:

(sin2a-sin2b)%5E2%20%2B(cos2a-cos2b)%5E2%20%5Cleq%20(2a-2b)%5E2%20

再让x%3D2a%2Cy%3D2b,得:

(sinx-siny)%5E2%20%2B(cosx-cosy)%5E2%20%5Cleq%20(x-y)%5E2%20




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