【用“等價(jià)證明”】證明：∵由題設(shè)知,x＜1.∴1-x＞0.又此時(shí)恒有e^x＞0.∴0＜e^x≤1/(1-x).0＜(1-x)×e^x≤1.構(gòu)造函數(shù)f(x)=(1-x)e^x,(x＜1).求導(dǎo)得f'(x)=-e^x+(1-x)e^x=-xe^x.易知,當(dāng)x＜0時(shí),f'(x)=-xe^x＞0.當(dāng)0＜x＜1時(shí),f'(x)=-xe^x＜0.===>在(-∞,0)上,函數(shù)f(x)遞增,在(0,1)上,函數(shù)f(x)遞減,∴f(x)max=f(0)=1.即當(dāng)x＜1時(shí),恒有f(x)≤f(0)=1.===>(1-x)e^x≤1.∴當(dāng)x＜1時(shí),有e^x≤1/(1-x).等號(hào)僅當(dāng)x=0時(shí)取得.證畢.