令f(x)=ln(x+1) -x/(x+1)=ln(x+1) +1/(x+1) -1
則f'(x)=[1/(x+1)]•(x+1)'+[-1/(x+1)²]•(x+1)'=1/(x+1) -1/(x+1)²=x/(1+x)²
當(dāng)-10,f(x)是增函數(shù),
從而f(0)是最小值.所以
f(x)≥f(0)=0,即 當(dāng)x>-1時(shí),有l(wèi)n(x+1)≥x/(x+1)