Question

已知a≤1,x≥0,求证(x+1)ln(x+1)≥ax

Answer 1

令f(x)=(x+1)ln(x+1)-ax

对f(x)求导,即有f'(x)=ln(x+1)+(x+1)*[1/(x+1)]-a=ln(x+1)+1-a

因为a≤1,x≥0

故1-a≥0,ln(x+1)≥0,即f'(x)≥0,可得f(x)在定义域内单调递增

于是f(x)≥f(0),而f(0)=(0+1)ln(0+1）-a*0=0,故f(x)≥0

即有(x+1)ln(x+1)-ax≥0,即证(x+1)ln(x+1)≥ax