当x>0时,(1+x)ln(1+x)>x

问题描述:

当x>0时,(1+x)ln(1+x)>x

构造函数f(x)=(x+1)㏑(x+1)-x.(x≥0).求导得f'(x)=㏑(x+1).∵x≥0.===>x+1≥1.===>㏑(x+1)≥0.即f'(x)≥0.∴在[0,+∞)上,f(x)递增.∴f(x)>f(0)=0.(x>0).∴(1+x)㏑(1+x)>x.(x>0)