若函数f(x)=x^2-2ax+ln(x+1)在定义域为增函数,则实数a的取值范围

定义域是x+1>0,即x>-1求导函数f'(x)=2x-2a+1/(x+1)在定义域为增函数即2x-2a+1/(x+1)>=02x+1/(x+1)>=2a2(x+1)+1/(x+1)-2>=2a2(x+1)+1/(x+1)利用均值不等式∵x+1>0∴2(x+1)+1/(x+1)>=2√22√2-2>=2a√2-1>=a实数a的取...