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,那么f'(x)=ln(x+1)+1-a,∵x≥0,a≤1,∴x+1≥1,ln(x+1)≥0,1-a≥0

∴f'(x)≥0,∴f(x)单调递增,∴f(x)≥f(0)=0,∴(x+1)ln(x+1)≥ax