【用“等价证明”】证明：∵由题设知,x＜1.∴1-x＞0.又此时恒有e^x＞0.∴0＜e^x≤1/(1-x).0＜(1-x)×e^x≤1.构造函数f(x)=(1-x)e^x,(x＜1).求导得f'(x)=-e^x+(1-x)e^x=-xe^x.易知,当x＜0时,f'(x)=-xe^x＞0.当0＜x＜1时,f'(x)=-xe^x＜0.===>在(-∞,0)上,函数f(x)递增,在(0,1)上,函数f(x)递减,∴f(x)max=f(0)=1.即当x＜1时,恒有f(x)≤f(0)=1.===>(1-x)e^x≤1.∴当x＜1时,有e^x≤1/(1-x).等号仅当x=0时取得.证毕.
