x1,x2都大于1时且x1不等于x2，证明|InX1-InX2|小于|X1-X2|
f(x)=x^2-x+c\u5b9a\u4e49\u5728\u533a\u95f4[0,1]\u4e0a,x1\u3001x2\u5747\u5c5e\u4e8e[0.1],\u4e14x1\u4e0d\u7b49\u4e8ex2\u3002\u8bc1\u660e|f(x2)-f(x1)<|x1-x2||f(x2)-f(x1)|=|x2^2-x2+c-x1^2+x1-c|=|(x2+x1)(x2-x1)+(x1-x2)|=|(1-x1-x2)(x1-x2)|
=|x1-x2|*|1-x1-x2|
\u56e0\u4e3a0<|1-x1-x2|<1
\u6240\u4ee5\u4e0a\u5f0f\u6210\u7acb
\u8bb0g(x)=f(x1)+f(x2)/2 -f(x),
\u5219g(x1)=[f(x1)+f(x2)]/2-f(x1)=[f(x2)-f(x1)]/2;
g(x2)=[f(x1)+f(x2)]/2-f(x2)=[f(x1)-f(x2)]/2=-g(x1).
\u5373g(x1)\u4e0eg(x2)\u5f02\u53f7.
\u56e0\u4e3ag(x)\u4e3a\u8fde\u7eed\u589e\u51fd\u6570,\u6545\u5fc5\u5b58\u5728\u4e00\u70b9x\u2208(x1,x2),\u4f7f\u5f97g(x)=0
\u4ea6\u5373
\u5fc5\u5b58\u5728\u4e00\u5b9e\u6570x0\u5c5e\u4e8e(x1,x2)\u4f7f\u5f97f(x0)=1/2[f(x1)+f(x2)]
那么|InX1-InX2|=lnx1-lnx2
|X1-X2| =x1-x2
等价于证明
lnx1-lnx2<x1-x2
等价于证明
lnx1-x1<lnx2-x2
等价于证明
y=lnx-x为减函数
y'=(1/x)-1 <0 当(x>1)
不用导数，初等方法不会。
不妨令1<x1<x2
显然lnx1<lnx2 x1<x2
故|lnx1-lnx2|=lnx2-lnx1
|x1-x2|=x2-x1
故|lnx1-lnx2|-|x1-x2|
=lnx2-lnx1-(x2-x1)
=ln(x2/x1)-lne^(x2-x1)
=ln[(x2e^x1)/(x1e^x2)]
这道题完全不用导数恐怕有点困难
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