Question

用函数的单调性证明下列不等式

Answer 1

欲证1/x+1/ln(1-x)＜1
即证[ln(1-x)+x]/[xln(1-x)]＜1
即证ln(1-x)+x＞1*[xln(1-x)]
(当0＜x＜1时，ln(1-x)＜0,xln(1-x)＜0.当x＜0时，ln(1-x)＞0,xln(1-x)＜0.∴分母[xln(1-x)]＜0乘到不等式右边应变号)
即证 ln(1-x)+x＞xln(1-x)
即证 ln(1-x)+x－xln(1-x)＞0
设f(x)= ln(1-x)+x－xln(1-x)
f′(x)= -1/(1-x)+1-[ln(1-x)-x/(1-x)]（ 注意复合函数求导。[ln(1-x)]′=[1/（1-x）]•(-x)′=-1/（1-x））
=-1/(1-x)+1-ln(1-x)+x/(1-x)
=[-1/(1-x)+x/(1-x) ]+1-ln(1-x)
=[(x-1)/(1-x) ]+1-ln(1-x)
=-1+1-ln(1-x)
= -ln(1-x)
令f′(x)=0得驻点x=0
当x＞0时，f′(x)＞0.f(x)单调递增
当x＜0时，f′(x)＜0.f(x)单调递减
∴x=0时，f(x)取极小值f(0)=ln(1-0)+0－0×ln(1-0)=0
∴f(x)＞f(0)=0 (x≠0)
∴1/x+1/ln(1-x)＜1 得证

用反证法写出即可